iMX508 battery charge issue

   Our customer's iMX508 board battery charge has problem when battery's voltage lower than 3.4v. Their PMIC is MC13892. With USB adapter connected, the board will boot up, u-boot will check battery's voltage, if battery's voltage lower than 3.5V, it will set charge current to 400mA. And then check real charge current whether higher than 300mA, if not, that mean MC13892 in trickle charge mode, which means charge current is about 80mA. It is MC13892 default setting for handling dad battery. At this situation, u-boot will shut down system and MC13892 will charge the battery in trickle charge mode. When battery voltage higher than 3.4V, MC13892 will turn on system, u-boot will check battery voltage and charge current, at this moment, the battery voltage is lower than 3.4V and charge current is still 80mA. And u-boot shut down the system again and MC13892 will power on system when battery voltage higher than 3.4V. It falls into infinite loop.
    I got information from MC13892 expert, we should wait for 2 seconds to measure the charge current after we set 400mA charge current in u-boot code. I add 2 seconds delay between setting charge current and get charge current. I got two result. If  I shorted D+/D-, after 2 seconds, I can see charge current increase to 400mA which let u-boot run and load kernel, every thing works find with battery charge function. But if I keep D+/D- floating, I cannot see charge current increase to 400mA after 2 seconds, it keep 80mA.
   I dump all 64 PMIC registers which two conditions. Only one meaningful difference in reg[6], bit 10 CHRGSE1BS is "0" when D+/D- shorted, CHRGSE1BS is "1" when D+/D- floating. So that is the key point. I checked MC13892 spec, there is description:
   "No distinction can be made between a USB Host or a wall charger. Therefore, when attaching a wall charger, the CHRGSE1B pin must be forced low as a charger attach indicator. The CHRGSE1B pin has a built-in weak pull-up to VCORE. In the application, this pin is preferably pulled low"
   So if the CHRGSE1B is high, the charge current will limited to no more than 100mA which consider USB current limitation. Let's check customer's SCH.

  We can see CHRGSE1B connect to iMX508 CHRG_DETB_TO_PMIC pin, this mean iMX508 checked status of USB cable and inform PMIC by the result. We don't need it, re-work the SCH, removed R27 and C14, change R64 to 0 ohm. Tested it again, we can see no matter D+/D- shorted or floating, the charge current can increase to 400 mA. So the charge function is working now. We will consult this result with MC13892 expert later to know what happens.

"Killer" C Interview Questions

                          Steve Summit
                       Copyright 1997,1998
[DISCLAIMER:  This is, on balance, a very difficult test.
DO NOT use it in an actual interview situation; you would in all
likelihood only embarrass yourself and insult your interviewee.
In particular, you would obviously not want to ask a question
which you yourself did not know the answer to, but the answers
to many of these questions are not obvious or well-known.
There are several trick questions here, as well as a number which
are explicitly marked "poor".  The poor questions are, alas, not
uncommon in actual interviews, and are presented here only so
that their faults and wrong answers can be presented.
This test is intended for study purposes only.  Most of the
answers can be found in the comp.lang.c FAQ list.
The answers here were prepared before I read a draft of the
impending C9X standard.  I would word several of them
differently today in anticipation of that standard, and
a few of them will become wrong when that standard is adopted.
Acknowledgments: This test is an expanded version of
one I prepared for a training class held at the request
of Tony McNamara, at a now-defunct company called
SCS/Compute (no relation).] 
1.1:    How can you print a literal % with printf?
A:      %%

1.2:    Why doesn't \% print a literal % with printf?
A:      Backslash sequences are interpreted by the compiler
        (\n, \", \0, etc.), and \% is not one of the recognized
        backslash sequences.  It's not clear what the compiler
        would do with a \% sequence -- it might delete it, or
        replace it with a single %, or perhaps pass it through as
        \ %.  But it's printf's behavior we're trying to change,
        and printf's special character is %.  So it's a
        %-sequence we should be looking for to print a literal %,
        and printf defines the one we want as %%.
1.3:    Are the parentheses in a return statement mandatory?

A:      No.  The formal syntax of a return statement is
            return expression ;
        But it's legal to put parentheses around any expression,
        of course, whether they're needed or not.

1.4:    How can %f work for type double in printf if %lf is
        required in scanf?
A:      In variable-length argument lists such as printf's, the
        old "default argument promotions" apply, and type float
        is implicitly converted to double.  So printf always
        receives doubles, and defines %f to be the sequence that
        works whether you had passed a float or a double.
        (Strictly speaking, %lf is *not* a valid printf format
        specifier, although most versions of printf quietly
        accept it.)
        scanf, on the other hand, always accepts pointers, and
        the types pointer-to-float and pointer-to-double are very
        different (especially when you're using them for storing
        values).  No implicit promotions apply.

1.5:    If a machine uses some nonzero internal bit pattern for
        null pointers, how should the NULL macro be defined?
A:      As 0 (or (char *)0), as usual.  The *compiler* is
        responsible for translating null pointer constants into
        internal null pointer representations, not the
        preprocessor.

1.6:    If p is a pointer, is the test
            if(p)
        valid?  What if a machine uses some nonzero internal bit
        pattern for null pointers?
A:      The test is always valid.  Since the definition of "true"
        in C is "not equal to 0," the test is equivalent to
            if(p != 0)
        and the compiler then translates the 0 into the
        appropriate internal representation of a null pointer.

1.7:    What is the ANSI Standard definition of a null pointer
        constant?
A:      "An integral constant expression with the value 0, or
        such an expression cast to type (void *)".

1.8:    What does the auto keyword mean?  When is it needed?
A:      auto is a storage-class specifier, just like extern and
        static.  But since automatic duration is the default for
        local variables (and meaningless, in fact illegal, for
        global variables), the keyword is never needed.  (It's a
        relic from the dawn of C.)

1.9:    What does *p++ increment?
A:      The pointer p.  To increment what p points to, use (*p)++
        or ++*p.

1.10:   What's the value of the expression 5["abcdef"] ?
A:      'f'.
        (The string literal "abcdef" is an array, and the
        expression is equivalent to "abcdef"[5].  Why is the
        inside-out expression equivalent?  Because a[b] is
        equivalent to *(a + b) which is equivalent to *(b + a)
        which is equivalent to b[a].

1.11:   [POOR QUESTION] How can you swap two integer variables
        without using a temporary?
A:      The reason that this question is poor is that the answer
        ceased to be interesting when we came down out of the
        trees and stopped using assembly language.
        The "classic" solution, expressed in C, is
            a ^= b;
            b ^= a;
            a ^= b;
        Due to the marvels of the exclusive-OR operator, after
        these three operations, a's and b's values will be
        swapped.
        However, it is exactly as many lines, and (if we can
        spare one measly word on the stack) is likely to be more
        efficient, to write the obvious
            int t = a;
            a = b;
            b = t;
        No, this doesn't meet the stipulation of not using a
        temporary.  But the whole reason we're using C and not
        assembly language (well, one reason, anyway) is that
        we're not interested in keeping track of how many
        registers we have.
        If the processor happens to have an EXCH instruction, the
        compiler is more likely to recognize the possibility of
        using it if we use the three-assignment idiom, rather
        than the three-XOR.
        By the way, the even more seductively concise rendition
        of the "classic" trick in C, namely
            a ^= b ^= a ^= b
        is, strictly speaking, undefined, because it modifies a
        twice between sequence points.  Also, if an attempt is
        made to use the idiom (in any form) in a function which
        is supposed to swap the locations pointed to by two
        pointers, as in
            swap(int *p1, *p2)
            {
            *p1 ^= *p2;
            *p2 ^= *p1;
            *p1 ^= *p2;
            }
        then the function will fail if it is ever asked to swap a
        value with itself, as in
            swap(&a, &a);
        or
            swap(&a[i], &a[j]);
        when i == j.  (The latter case is not uncommon in sorting
        algorithms.  The effect when p1 == p2 is that the pointed-
        to value is set to 0.)

1.12:   What is sizeof('A') ?
A:      The same as sizeof(int).  Character constants have type
        int in C.  (This is one area in which C++ differs.)

1.13:   According to the ANSI Standard, how many bits are there
        in an int?  A char?  A short int?  A long int?  In other
        words, what is sizeof(int) ?  sizeof(char) ?
        sizeof(short int) ?  sizeof(long int) ?
A:      ANSI guarantees that the range of a signed char is at
        least +-127, of a short int is at least +-32767, of an int
        at least +-32767, and a long int at least +-2147483648.  So
        we can deduce that a char must be at least 8 bits, an int
        or a short int must be at least 16 bits, and a long int
        must be at least 32 bits.  The only guarantees about
        sizeof are that
            1 = sizeof(char) <= sizeof(short) <= sizeof(int) <= sizeof(long)

1.14:   If arr is an array, in an ordinary expression, what's
        the difference between arr and &arr ?
A:      If the array is of type T, the expression arr yields a
        pointer of type pointer-to-T pointing to the array's
        first element.  The expression &arr, on the other hand,
        yields a pointer of type pointer-to-array-of-T pointing
        to the entire array.  (The two pointers will likely have
        the same "value," but the types are distinct.  The
        difference would be visible if you assigned or
        incremented the resulting pointer.)

1.15:   What's the difference between
            char *p = malloc(n);
        and
            char *p = malloc(n * sizeof(char));
        ?
A:      There is little or no difference, since sizeof(char) is
        by definition exactly 1.

1.16:   What's the difference between these three declarations?
            char *a = "abc";
            char b[] = "abc";
            char c[3] = "abc";
A:      The first declares a pointer-to-char, initialized to
        point to a four-character array somewhere in (possibly
        read-only) memory containing the four characters
        a b c \0.  The second declares an array (a writable
        array) of 4 characters, initially containing the
        characters a b c \0.  The third declares an array of 3
        characters, initially containing a b c.  (The third array
        is therefore not an immediately valid string.)

1.17:   The first line of a source file contains the line
            extern int f(struct x *);
        The compiler warns about "struct x declared inside
        parameter list".  What is the compiler worried about?
A:      For two structures to be compatible, they must not only
        have the same tag name but be defined in the same scope.
        A function prototype, however, introduces a new, nested
        scope for its parameters.  Therefore, the structure tag x
        is defined in this narrow scope, which almost immediately
        disappears.  No other struct x pointer in this
        translation unit can therefore be compatible with f's
        first parameter, so it will be impossible to call f
        correctly (at least, without drawing more warnings).  The
        warning alluded to in the question is trying to tell you
        that you shouldn't mention struct tags for the first time
        in function prototypes.
        (The warning message in the question is actually produced
        by gcc, and the message runs on for two more lines,
        explaining that the scope of the structure declared "is
        only this definition or declaration, which is probably
        not what you want.")

1.18:   List several ways for a function to safely return a
        string.
A:      It can return a pointer to a static array, or it can
        return a pointer obtained from malloc, or it can fill in
        a buffer supplied by the caller.

1.19:   [hard] How would you implement the va_arg() macro in
        ? 
A:      A straightforward implementation, assuming a conventional
        stack-based architecture, is
            #define va_arg(argp, type) (((type *)(argp += sizeof(type)))[-1])
        This assumes that type va_list is char *.

1.20:   Under what circumstances is the declaration
            typedef xxx int16;
        where xxx is replaced with an appropriate type for a
        particular machine, useful?
A:      It is potentially useful if the int16 typedef is used to
        declare variables or structures which will be read from
        or written to some external data file or stream in some
        fixed, "binary" format.  (However, the typedef can at
        most ensure that the internal type is the same size as
        the external representation; it cannot correct for any
        byte order discrepancies.)
        Such a typedef may also be useful for allowing
        precompiled object files or libraries to be used with
        different compilers (compilers which define basic types
        such as int differently), without recompilation.

1.21:   Suppose that you declare
            struct x *xp;
        without any definition of struct x.  Is this legal?
        Under what circumstances would it be useful?
A:      It is perfectly legal to refer to a structure which has
        not been "fleshed out," as long as the compiler is never
        asked to compute the size of the structure or generate
        offsets to any members.  Passing around pointers to
        otherwise undefined structures is quite acceptable, and
        is a good way of implementing "opaque" data types in C.

1.22:   What's the difference between
            struct x1 { ... };
            typedef struct { ... } x2;
        ?
A:      The first declaration declares a structure tag x1; the
        second declares a typedef name x2.  The difference
        becomes clear when you declare actual variables of the
        two structure types:
            struct x1 a, b;
        but
            x2 a, b;
        (This distinction is insignificant in C++, where all
        structure and class tags automatically become full-
        fledged types, as if via typedef.)

1.23:   What do these declarations mean?
            int **a();
            int (*b)();
            int (*c[3])();
            int (*d)[10];
A:      declare a as function returning pointer to pointer to int
        declare b as pointer to function returning int
        declare c as array of 3 pointers to functions returning int
        declare d as pointer to array of 10 ints
        The way to read these is "inside out," remembering that
        [] and () bind more tightly than *, unless overridden by
        explicit parentheses.

1.24:   State the declaration for a pointer to a function
        returning a pointer to char.
A:      char *(*f)();

1.25:   If sizeof(long int) is 4, why might sizeof report the
        size of the structure
            struct x {char c; long int i;};
        as 8 instead of 5?
A:      The compiler will typically allocate invisible padding
        between the two members of the structure, to keep i
        aligned on a longword boundary.

1.26:   If sizeof(long int) is 4, why might sizeof report the
        size of the structure
            struct y {long int i; char c;};
        as 8 instead of 5?
A:      The compiler will typically allocate invisible padding at
        the end of structure, so that if an array of these
        structures is allocated, the i's will all be aligned.

1.27:   [POOR QUESTION] If i starts out as 1, what does the
        expression
            i++ + i++
        evaluate to?  What is i's final value?
A:      This is a poor question because it has no answer.  The
        expression attempts to modify i twice between sequence
        points (not to mention modifying and inspecting i's
        value, where the inspection is for purposes other than
        determining the value to be stored), so the expression is
        undefined.  Different compilers can (and do) generate
        different results, and none of them is "wrong."

1.28:   Consider these definitions:
            #define Push(val)   (*stackp++ = (val))
            #define Pop()       (*--stackp)
            int stack[100];
            int *stackp = stack;
        Now consider the expression
            Push(Pop() + Pop())
        1. What is the expression trying to do?  In what sort of
        program might such an expression be found?
        2. What are some deficiencies of this implementation?
        Under what circumstances might it fail?
A:      The expression is apparently intended to pop two values
        from a stack, add them, and push the result.  This code
        might be found in a calculator program, or in the
        evaluation loop of the engine for a stack-based language.
        The implementation has at least four problems, however.
        The Push macro does not check for stack overflow; if more
        than 100 values are pushed, the results will be
        unpredictable.  Similarly, the the Pop macro does not
        check for stack underflow; an attempt to pop a value when
        the stack is empty will likewise result in undefined
        behavior.
        On a stylistic note, the stackp variable is global as far
        as the Push and Pop macros are concerned.  If it is
        certain that, in a particular program, only one stack
        will be used, this assumption may be a reasonable one,
        as it allows considerably more succinct invocations.
        If multiple stacks are a possibility, however, it might
        be preferable to pass the stack pointer as an argument
        to the Push and Pop macros.
        Finally, the most serious problem is that the "add"
        operation as shown above is *not* guaranteed to work!
        After macro expansion, it becomes
            (*stackp++ = ((*--stackp) + (*--stackp)))
        This expression modifies a single object more than once
        between sequence points; specifically, it modifies stackp
        three times.  It is not guaranteed to work; moreover,
        there are popular compilers (one is gcc) under which it
        *will* *not* work as expected.  (The extra parentheses
        do nothing to affect the evaluation order; in particular,
        they do not make it any more defined.)

1.29:   [POOR QUESTION] Write a small function to sort an array
        of integers.
A:      This is a poor question because no one writes small
        functions to sort arrays of integers any more, except as
        pedagogical exercises.  If you have an array of integers
        that needs sorting, the thing to do is call your library
        sort routine -- in C, qsort().  So here is my "small
        function":
            static int intcmp(const void *, const void *);
            sortints(int a[], int n)
            {
                    qsort(a, n, sizeof(int), intcmp);
            }
            static int intcmp(const void *p1, const void *p2)
            {
                    int i1 = *(const int *)p1;
                    int i2 = *(const int *)p2;
                    if(i1 < i2)
                            return -1;
                    else if(i1 > i2)
                            return 1;
                    else    return 0;
            }
        (The reason for using two comparisons and three explicit
        return statements rather than the "more obvious"
        return i1 - i2; is that i1 - i2 can underflow, with
        unpredictable results.)

1.30:   State the ANSI rules for determining whether an
        expression is defined or undefined.
A:      An expression is undefined if, between sequence points,
        it attempts to modify the same location twice, or if it
        attempts to both read from and write to the same
        location.  It's permissible to read and write the same
        location only if the laws of causality (a higher
        authority even than X3.159) prove that the read must
        unfailingly precede the write, that is, if the write is
        of a value which was computed from the value which was
        read.  This exception means that old standbys such as
        i = i + 1 are still legal.
        Sequence points occur at the ends of full expressions
        (expression statements, and the expressions in if, while,
        for, do/while, switch, and return statements, and
        initializers), at the &&, ||, and comma operators, at the
        end of the first expression in a ?: expression, and just
        before the call of a function (after the arguments have
        all been evaluated).
        (The actual language from the ANSI Standard is
                Between the previous and next sequence point an
                object shall have its stored value modified at
                most once by the evaluation of an expression.
                Furthermore, the prior value shall be accessed
                only to determine the value to be stored.
        )

1.30a:  What's the difference between these two declarations?
            extern char x[];
            extern char *x;
A:      The first is an external declaration for an array of char
        named x, defined elsewhere.  The second is an external
        declaration for a pointer to char named x, also defined
        elsewhere.  These declarations could not both appear in
        the same program, because they specify incompatible types
        for x.

1.31:   What's the difference between these two declarations?
            int f1();
            extern int f2();
A:      There is no difference; the extern keyword is essentially
        optional in external function declarations.

1.32:   What's the difference between these two declarations?
            extern int f1();
            extern int f2(void);
A:      The first is an old-style function declaration declaring
        f1 as a function taking an unspecified (but fixed) number
        of arguments; the second is a prototype declaration
        declaring f2 as a function taking precisely zero
        arguments.

1.33:   What's the difference between these two definitions?
            int f1()
            {
            }
            int f2(void)
            {
            }
A:      There is no difference, other than that the first uses
        the old definition style and the second uses the
        prototype style.  Both functions take zero arguments.

1.34:   How does operator precedence influence order of
        evaluation?
A:      Only partially.  Precedence affects the binding of
        operators to operands, but it does *not* control (or even
        influence) the order in which the operands themselves are
        evaluated.  For example, in
            a() + b() * c()
        we have no idea what order the three functions will be
        called in.  (The compiler might choose to call a first,
        even though its result will be needed last.)

1.35:   Will the expression in
            if(n != 0 && sum / n != 0)
        ever divide by 0?
A:      No.  The "short circuiting" behavior of the && operator
        guarantees that sum / n will not be evaluated if n is 0
        (because n != 0 is false).

1.36:   Will the expression
            x = ((n == 0) ? 0 : sum / n)
        ever divide by 0?
A:      No.  Only one of the pair of controlled expressions in a
        ?: expression is evaluated.  In this example, if n is 0,
        the third expression will not be evaluated at all.

1.37:   Explain these three fragments:
            if((p = malloc(10)) != NULL) ...
            if((fp = fopen(filename, "r")) == NULL) ...
            while((c = getc(fp)) != EOF) ...
A:      The first calls malloc, assigns the result to p, and does
        something if the just-assigned result is not NULL.
        The second calls fopen, assigns the result to fp, and
        does something if the just-assigned result is NULL.
        The third repeatedly calls getc, assigns the results in
        turn to c, and does something as long as each just-
        assigned result is not EOF.

1.38:   What's the difference between these two statements?
            ++i;
            i++;
A:      There is no difference.  The only difference between the
        prefix and postfix forms of the autoincrement operator is
        the value passed on to the surrounding expression, but
        since the expressions in the question stand alone as
        expression statements, the value is discarded, and each
        expression merely serves to increment i.

1.39:   Why might a compiler warn about conversions or
        assignments from char * to int * ?
A:      In general, compilers complain about assignments between
        pointers of different types (and are required by the
        Standard to so complain) because such assignments do not
        make sense.  A pointer to type T1 is supposed to point to
        objects of type T1, and presumably the only reason for
        assigning the pointer to a pointer of a different type,
        say pointer-to-T2, would be to try to access the pointed-
        to object as a value of type T2, but if the pointed-to
        object is of type T2, why were we pointing at it with a
        pointer-to-T1 in the first place?
        In the particular example cited in the question, the
        warning also implies the possibility of unaligned access.
        For example, this code:
            int a[2] = {0, 1};
            char *p = &a;           /* suspicious */
            int *ip = p + 1;            /* even more suspicious */
            printf("%d\n", *ip);
        is likely to crash (perhaps with a "Bus Error") because
        the programmer has contrived to make ip point to an odd,
        unaligned address.
        When it is desired to use pointers of the "wrong" type,
        explicit casts must generally be used.  One class of
        exceptions is exemplified by malloc: the memory it
        allocates, and hence the pointers it returns, are
        supposed to be usable as any type the programmer wishes,
        so malloc's return value will almost always be the
        "wrong" type.  To avoid the need for so much explicit,
        dangerous casting, ANSI invented the void * type, which
        quietly interconverts (i.e. without warning) between
        other pointer types.  Pointers of type void * are
        therefore used as containers to hold "generic" pointers
        which are known to be safely usable as pointers to other,
        more specific types.

1.40:   When do ANSI function prototype declarations *not*
        provide argument type checking, or implicit conversions?
A:      In the variable-length part of variable-length argument
        lists, and (perhaps obviously, perhaps not) when no
        prototype is in scope at all.  The point is that it is
        not safe to assume that since prototypes have been
        invented, programmers don't have to be careful about
        matching function-call arguments any more.  Care must
        still be exercised in variable-length argument lists, and
        if prototypes are to take care of the rest, care must be
        exercised to use prototypes correctly.

1.41:   State the rule(s) underlying the "equivalence" of arrays
        and pointers in C.
A:      Rule 1: When an array appears in an expression where its
        value is needed, the value generated is a pointer to the
        array's first element.  Rule 2: Array-like subscripts
        (integer expressions in brackets) may be used to
        subscript pointers as well as arrays; the expression p[i]
        is by definition equivalent to *(p+i).  (Actually, by
        rule 1, subscripts *always* find themselves applied to
        pointers, never arrays.)

1.42:   What's the difference between these two declarations?
            extern int f2(char []);
            extern int f1(char *);
A:      There is no difference.  The compiler always quietly
        rewrites function declarations so that any array
        parameters are actually declared as pointers, because
        (by the equivalence of arrays and pointers) a pointer
        is what the function will actually receive.

1.43:   Rewrite the parameter declaration in
            f(int x[5][7])
            {
            }
        to explicitly show the pointer type which the compiler
        will assume for x.
A:          f(int (*x)[7])
            {
            }
        Note that the type int (*)[7] is *not* the same as
        int **.

1.44:   A program uses a fixed-size array, and in response to
        user complaints you have been asked to replace it with a
        dynamically-allocated "array," obtained from malloc.
        Which parts of the program will need attention?  What
        "gotchas" must you be careful of?
A:      Ideally, you will merely have to change the declaration
        of the array from an array to a pointer, and add one call
        to malloc (with a check for a null return, of course) to
        initialize the pointer to a dynamically-allocated
        "array."  All of the code which accesses the array can
        remain unchanged, because expressions of the form x[i]
        are valid whether x is an array or a pointer.
        The only thing to be careful of is that if the existing
        code ever used the sizeof operator to determine the size
        of the array, that determination becomes grossly invalid,
        because after the change, sizeof will return only the
        size of the pointer.

1.45:   A program which uses a dynamically allocated array is
        still running into problems because the initial
        allocation is not always big enough.  Your task is now
        to use realloc to make the "array" bigger, if need be.
        What must you be careful of?
A:      The actual call to realloc is straightforward enough, to
        request that the base pointer now point at a larger block
        of memory.  The problem is that the larger block of
        memory may be in a different place; the base pointer may
        move.  Therefore, you must reassign not only the base
        pointer, but also any copies of the base pointer you may
        have made, and also any pointers which may have been set
        to point anywhere into the middle of the array.  (For
        pointers into the array, you must in general convert them
        temporarily into offsets from the base pointer, then call
        realloc, then recompute new pointers based on the offsets
        and the new base pointer.  See also question 2.10.)

1.46:   How can you you use sizeof to determine the number of
        elements in an array?
A:      The standard idiom is
            sizeof(array) / sizeof(array[0])
        (or, equivalently, sizeof(array) / sizeof(*array) ).

1.47:   When sizeof doesn't work (when the array is declared
        extern, or is a parameter to a function), what are some
        strategies for determining the size of an array?
A:      Use a sentinel value as the last element of the array;
        pass the size around in a separate variable or as a
        separate function parameter; use a preprocessor macro
        to define the size.

1.48:   Why might explicit casts on malloc's return value, as in
            int *ip = (int *)malloc(10 * sizeof(int));
        be a bad idea?
A:      Although such casts used to be required (before the
        void * type, which converts quietly and implicitly, was
        invented), they can now be considered poor style, because
        they will probably muzzle the compiler's attempts to warn
        you on those occasions when you forget to #include
        or otherwise declare malloc, such that malloc
        will be incorrectly assumed to be a function returning
        int. 

1.49:   How are Boolean true/false values defined in C?  What
        values can the == (and other logical and comparison
        operators) yield?
A:      The value 0 is considered "false," and any nonzero value
        is considered "true."  The relational and logical
        operators all yield 0 for false, 1 for true.

1.50:   x is an integer, having some value.  What is the value of
        the expression
            0 <= !x && !!x < 2
        ?
A:      1.

1.51:   In your opinion, is it acceptable for a header file to
        contain #include directives for other header files?
A:      The argument in favor of "nested #include files" is that
        they allow each header to arrange to have any subsidiary
        definitions, upon which its own definitions depend, made
        automatically.  (For example, a file containing a
        prototype for a function that accepts an argument of type
        FILE * could #include to define FILE.)  The
        alternative is to potentially require everyone who
        includes a particular header file to include one or
        several others first, or risk cryptic errors. 
        The argument against is that nested headers can be
        confusing, can make definitions difficult to find,
        and can in some circumstances even make it difficult
        to determine which file(s) is/are being included.

1.52:   How can a header file be protected against being
        included multiple times (perhaps due to nested
        #include directives)?
A:      The standard trick is to place lines like
            #ifndef headerfilename_H
            #define headerfilename_H
        at the beginning of the file, and an extra #endif at the
        end.

1.53:   A source file contains as its first two lines:
            #include "a.h"
            int i;
        The compiler complains about an invalid declaration on
        line 2.  What's probably happening?
A:      It's likely that the last declaration in a.h is missing
        its trailing semicolon, causing that declaration to merge
        into "int i", with meaningless results.  (That is, the
        merged declaration is probably something along the lines
        of
            extern int f() int i;
        or
            struct x { int y; } int i;
        .)

1.54:   What's the difference between a header file and a
        library?
A:      A header file typically contains declarations and
        definitions, but it never contains executable code.
        (A header file arguably shouldn't even contain any
        function bodies which would compile into executable
        code.)  A library, on the other hand, contains only
        compiled, executable code and data.
        A third-party library is often delivered as a library and
        a header file.  Both pieces are important.  The header
        file is included during compilation, and the library is
        included during linking.

1.55:   What are the acceptable declaration(s) for main()? A:      The most common declarations, all legal, are:
            main()
            int main()
            int main(void)
            int main(int argc, char **argv)
            int main(int argc, char *argv[])
            int main(argc, argv) int argc; char *argv[];
        (Basically: the return type must be an implicit or
        explicit int; the parameter list must either be empty, or
        void, or one int plus one array of strings; and the
        function may be declared using either old-style or
        prototyped syntax.  The actual names of the two
        parameters are arbitrary, although of course argc and
        argv are traditional.)

1.56:   You wish to use ANSI function prototypes to guard against
        errors due to accidentally calling functions with
        incorrect arguments.  Where should you place the
        prototype declarations?  How can you ensure that the
        prototypes will be maximally effective?
A:      The prototype for a global function should be placed in a
        header file, and the header file should be included in
        all files where the function is called, *and* in the file
        where the function is defined.  The prototype for a
        static function should be placed at the top of the file
        where the function is defined.
        Since following these rules is only slightly less hard
        than getting all function calls right by hand (i.e.
        without the aid of prototypes), the compiler should be
        configured to warn about functions called without
        prototypes in scope, *and* about functions defined
        without prototypes in scope.

1.57:   Why must the variable used to hold getchar's return value
        be declared as int?
A:      Because getchar can return, besides all values of type
        char, the additional "out of band" value EOF, and there
        obviously isn't room in a variable of type char to hold
        one more than the number of values which can be
        unambiguously stored in a variable of type char.

1.58:   You must write code to read and write "binary" data
        files.  How do you proceed?  How will you actually open,
        read, and write the files?
A:      When calling fopen, the files must be opened using the b
        modifier (e.g. "rb", "wb").  Binary data files are
        generally read and written a byte at a time using getc
        and putc, or a data structure at a time using fread and
        fwrite.

1.59:   Write the function
            void error(const char *message, ...);
        which accepts a message string, possibly containing %
        sequences, along with optional extra arguments
        corresponding to the % sequences, and prints the string
        "error: ", followed by the message as printf would print
        it, followed by a newline, all to stderr.
A:          #include
            #include  
            void error(char *fmt, ...)
            {
                    va_list argp;
                    fprintf(stderr, "error: ");
                    va_start(argp, fmt);
                    vfprintf(stderr, fmt, argp);
                    va_end(argp);
                    fprintf(stderr, "\n");
            }

1.60:   Write the function
            char *vstrcat(char *, ...);
        which accepts a variable number of strings and
        concatenates them all together into a block of malloc'ed
        memory just big enough for the result.  The end of the
        list of strings will be indicated with a null pointer.
        For example, the call
            char *p = vstrcat("Hello, ", "world!", (char *)NULL);
        should return the string "Hello, world!".
A:          #include           /* for malloc, NULL, size_t */
            #include           /* for va_ stuff */
            #include           /* for strcat et al. */ 
            char *vstrcat(char *first, ...)
            {
                    size_t len;
                    char *retbuf;
                    va_list argp;
                    char *p;
                    if(first == NULL)
                            return NULL;
                    len = strlen(first);
                    va_start(argp, first);
                    while((p = va_arg(argp, char *)) != NULL)
                            len += strlen(p);
                    va_end(argp);
                    retbuf = malloc(len + 1);   /* +1 for trailing \0 */
                    if(retbuf == NULL)
                            return NULL;        /* error */
                    (void)strcpy(retbuf, first);
                    va_start(argp, first);      /* restart; 2nd scan */
                    while((p = va_arg(argp, char *)) != NULL)
                            (void)strcat(retbuf, p);
                    va_end(argp);
                    return retbuf;
            }

1.61:   Write a stripped-down version of printf which accepts
        only the %c, %d, %o, %s, %x, and %% format specifiers.
        (Do not worry about width, precision, flags, or length
        modifiers.)
A:      [Although this question is obviously supposed to test one's
        familiarity with the va_ macros, a significant nuisance in
        composing a working answer is performing the sub-task of
        converting integers to digit strings.  For some reason,
        back when I composed this test, I felt it appropriate to
        defer that task to an "itoa" function; perhaps I had just
        presented an implementation of itoa to the same class for
        whom I first prepared this test.]
            #include
            #include  
            extern char *itoa(int, char *, int);
            myprintf(const char *fmt, ...)
            {
            const char *p;
            va_list argp;
            int i;
            char *s;
            char fmtbuf[256];
            va_start(argp, fmt);
            for(p = fmt; *p != '\0'; p++)
                    {
                    if(*p != '%')
                            {
                            putchar(*p);
                            continue;
                            }
                    switch(*++p)
                            {
                            case 'c':
                                    i = va_arg(argp, int);
                                    putchar(i);
                                    break;
                            case 'd':
                                    i = va_arg(argp, int);
                                    s = itoa(i, fmtbuf, 10);
                                    fputs(s, stdout);
                                    break;
                            case 's':
                                    s = va_arg(argp, char *);
                                    fputs(s, stdout);
                                    break;
                            case 'x':
                                    i = va_arg(argp, int);
                                    s = itoa(i, fmtbuf, 16);
                                    fputs(s, stdout);
                                    break;
                            case '%':
                                    putchar('%');
                                    break;
                            }
                    }
            va_end(argp);
            }

1.62:   You are to write a program which accepts single
        keystrokes from the user, without waiting for the RETURN
        key.  You are to restrict yourself only to features
        guaranteed by the ANSI/ISO C Standard.  How do you
        proceed?
A:      You proceed by pondering the sorrow of your fate, and
        perhaps by complaining to your boss/professor/psychologist
        that you've been given an impossible task.  There is no
        ANSI Standard function for reading one keystroke from the
        user without waiting for the RETURN key.  You'll have to
        use facilities specific to your operating system; you
        won't be able to write the code strictly portably.

1.63:   [POOR QUESTION] How do you convert an integer to binary
        or hexadecimal?
A:      The question is poor because an integer is a *number*; it
        doesn't make much sense to ask what base it's in.  If I'm
        holding eleven apples, what base is that in?
        (Of course, internal to the computer, an integer is
        almost certainly represented in binary, although it's not
        at all unreasonable to think of it as being hexadecimal,
        or decimal for that matter.)
        The only time the base of a number matters is when it's
        being read from or written to the outside world as a
        string of digits.  In those cases, and depending on just
        what you're doing, you can specify the base by picking
        the correct printf or scanf format specifier (%d, %o, or
        %x), or by picking the third argument to strtol.  (There
        isn't a Standard function to convert an integer to a
        string using an arbitrary base.  For that task, it's a
        straightforward exercise to write a function to do the
        conversion.  Some versions of the nonstandard itoa
        function also accept a base or radix argument.)

1.64:   You're trying to discover the sizes of the basic types
        under a certain compiler.  You write the code
            printf("sizeof(char) = %d\n", sizeof(char));
            printf("sizeof(short) = %d\n", sizeof(short));
            printf("sizeof(int) = %d\n", sizeof(int));
            printf("sizeof(long) = %d\n", sizeof(long));
        However, all four values are printed as 0.  What have you
        learned?
A:      You've learned that this compiler defines size_t, the
        type returned by sizeof, as an unsigned long int, and that
        the compiler also defines long integers as having a
        larger size than plain int.  (Furthermore, you've learned
        that the machine probably uses big-endian byte order.)
        Finally, you may have learned that the code you should
        have written is along the lines of either
            printf("sizeof(int) = %u\n", (unsigned)sizeof(int));
        or
            printf("sizeof(int) = %lu\n", (unsigned long)sizeof(int));

Section 2. What's wrong with...?
2.1:        main(int argc, char *argv[])
            {
            ...
            if(argv[i] == "-v")
                    ...
A:      Applied to pointers, the == operator compares only
        whether the pointers are equal.  To compare whether the
        strings are equal, you'll have to call strcmp.

2.2:        a ^= b ^= a ^= b
        (What is the expression trying to do?)
A:      The expression is undefined because it modifies the
        variable a twice between sequence points.  What it's
        trying to do is swap the variables a and b using a
        hoary old assembly programmer's trick.

2.3:        char* p1, p2;
A:      p2 will be declared as type char, *not* pointer-to-char.

2.4:        char c;
            while((c = getchar()) != EOF)
                    ...
A:      The variable used to contain getchar's return value must
        be declared as int if EOF is to be reliably detected.

2.5:        while(c = getchar() != EOF)
                    ...
A:      Parentheses are missing; the code will call getchar,
        compare the result to EOF, assign the result *of the
        comparison* to c, and take another trip around the loop
        if the condition was true (i.e. if the character read was
        not EOF).  (The net result will be that the input will be
        read as if it were a string of nothing but the character
        '\001'.  The loop would still halt properly on EOF,
        however.)

2.6:        int i, a[10];
            for(i = 0; i <= 10; i++)
                    a[i] = 0;
A:      The loop assigns to the nonexistent eleventh value of the
        array, a[10], because it uses a loop continuation
        condition of <= 10 instead of < 10 or <= 9.

2.7:        #include
            ...
            #define TRUE 1
            #define FALSE 0
            ...
            if(isalpha(c) == TRUE)
                    ... 
A:      Since *any* nonzero value is considered "true" in C; it's
        rarely if ever a good idea to compare explicitly against
        a single TRUE value (or FALSE, for that matter).  In
        particular, the macros, including isalpha(),
        tend to return nonzero values other than 1, so the test
        as written is likely to fail even for alphabetic
        characters.  The correct test is simply 
            if(isalpha(c))

2.8:        printf("%d\n", sizeof(int));
A:      The sizeof operator returns type size_t, which is an
        unsigned integral type *not* necessarily the same size as
        an int.  The correct code is either
            printf("sizeof(int) = %u\n", (unsigned int)sizeof(int));
        or
            printf("sizeof(int) = %lu\n", (unsigned long int)sizeof(int));

2.9:        p = realloc(p, newsize);
            if(p == NULL)
                    {
                    fprintf(stderr, "out of memory\n");
                    return;
                    }
A:      If realloc returns null, and assuming p used to point to
        some malloc'ed memory, the memory remains allocated,
        although having overwritten p, there may well be no way
        to use or free the memory.  The code is a potential
        memory leak.

2.10:       /* p points to a block of memory obtained from malloc; */
            /* p2 points somewhere within that block */
            newp = realloc(p, newsize);
            if(newp != NULL && newp != p)
                    {
                    int offset = newp - p;
                    p2 += offset;
                    }
A:      Pointer subtraction is well-defined only for pointers
        into the same block of memory.  If realloc moves the
        block of memory while changing its size, which is the
        very case the code is trying to test for, then the
        subtraction newp - p is invalid, because p and newp
        do not point into the same block of memory.  (The
        subtraction could overflow or otherwise produce
        nonsensical results, especially on segmented
        architectures.  Strictly speaking, *any* use of
        the old value of p after realloc moves the block
        is invalid, even comparing it to newp.)
        The correct way to relocate p2 within the possibly-moved
        block, correcting *all* the problems in the original
        (including a subtle one not mentioned), is:
            ptrdiff_t offset = p2 - p;
            newp = realloc(p, newsize);
            if(newp != NULL)
                    {
                    p = newp;
                    p2 = p + offset;
                    }

2.11:       int a[10], b[10];
            ...
            a = b;
A:      You can't assign arrays.

2.12:       int i = 0;
            char *p = i;
            if(p == 0)
                    ... A:      Assigning an integer 0 to a pointer does not reliably
        result in a null pointer.  (You must use a *constant* 0.) 

Priority inversion

When reading semaphore and Mutexes stuff, there is key word "Priority inversion", I am interesting in it. Find a document which looks good. Forward it below.
Also we can find definition from wiki http://en.wikipedia.org/wiki/Priority_inversion

优先级翻转与优先级继承

田海立

2006-3-7

摘要

本文描述操作系统中的优先级翻转（Priority Inversion，也有翻译为反转，逆转或倒置的）现象以及如何用优先级继承来解决此类问题的方法，并阐述了 Microsoft Platform Builder for Windows CE 环境下，如何查看此种现象。

目  录

摘要
1 优先级翻转（Priority Inversion）
2 优先级继承（Priority Inheritance）
       2.1 优先级继承
       2.2 WinCE中优先级继承
       2.3 优先级继承的传递性
3 总结
参考资料及进一步阅读
关于作者

1 优先级翻转（Priority Inversion）

优先级翻转（Priority Inversion），是指某同步资源被较低优先级的进程/线程所拥有，较高优先级的进程/线程竞争该同步资源未获得该资源，而使得较高优先级进程/线程反而推迟被调度执行的现象。

优先级翻转现象在普通的分时调度系统中对系统的影响不大，或者本来就对优先级高低的进程/线程被调度的顺序就不太在意的情况下，不用太多关注。但是对于基于优先级调度的实时系统，优先级高的进程/线程被优先调度是调度算法首要考虑的因素，调度程序对于较高优先级被调度单元总是最钟情的，所有其它相应资源分配的考量也是基于优先级做出的。SUN Solaris和Microsoft Windows CE等系统都对优先级翻转现象在操作系统级做了处理。

假定一个进程中有三个线程Thread1、Thread2和Thread3，它们的优先级顺序是Priority(Thread1) > Priority(Thread2) > Priority(Thread3)。考虑图一的执行情况。

图一、优先级翻转现象（无优先级继承）

       ◇ T0时刻，只有Thread3处于可运行状态，运行过程中，Thread3拥有了一个同步资源SYNCH1；
       ◇ T1时刻，Thread2就绪进入可运行状态，由于优先级高于正在运行的Thread3，Thread3被抢占（未释放同步资源SYNCH1），Thread2被调度执行；
       ◇ 同样地T2时刻，Thread1抢占Thread2；
       ◇ Thread1运行到T3时刻，需要同步资源SYNCH1，但SYNCH1被更低优先级的Thread3所拥有，Thread1被挂起等待该资源，而此时处于可运行状态的线程Thread2和Thread3中，Thread2的优先级大于Thread3的优先级，Thread2被调度执行。

上述现象中，优先级最高的Thread1既要等优先级低的Thread2运行完，还要等优先级更低的Thread3运行完之后才能被调度，如果Thread2和Thread3执行的很费时的操作，显然Thread1的被调度时机就不能保证，整个实时调度的性能就很差了。

2 优先级继承（Priority Inheritance）

2.1 优先级继承

为了解决上述由于优先级翻转引起的问题，Solaris和WinCE引入了优先级继承的解决方法。优先级继承也就是，高优先级进程T_H在等待低优先级的线程T_L继承占用的竞争资源时，为了使T_H能够尽快获得调度运行，由操作系统把T_L的优先级提高到T_H的优先级，从而让T_L以T_H的优先级参与调度，尽快让T_L执行并释放调T_H欲获得的竞争资源，然后T_L的优先级调整到继承前的水平，此时T_H可获得竞争资源而继续执行。

有了优先级继承之后的上述现象的执行情况如图二所示。

图二、优先级继承执行情况

与图一比较，图二中，到了T3时刻，Thread1需要Thread3占用的同步资源SYNCH1，操作系统检测到这种情况后，就把Thread3的优先级提高到Thread1的优先级。此时处于可运行状态的线程Thread2和Thread3中，Thread3的优先级大于Thread2的优先级，Thread3被调度执行。

Thread3执行到T4时刻，释放了同步资源SYNCH1，操作系统何时恢复了Thread3的优先级，Thread1获得了同步资源SYNCH1，重新进入可执行队列。处于可运行状态的线程Thread1和Thread2中，Thread1的优先级大于Thread2的优先级，所以Thread1被调度执行。

上述机制，使优先级最高的Thread1获得执行的时机提前。

2.2 WinCE中优先级继承

下面用在Microsoft Platform Builder for Windows CE 5.0环境的Emulator中，来做一个优先级继承的实验。

2.2.1 程序设计

主线程做了如下工作：
       ◇ 创建三个子线程，设置它们的优先级顺序：
              Priority(Thread1) > Priority(Thread2) > Priority(Thread3)；
       ◇ 只是启动Thread3。
       ◇ 创建一个互斥体g_Mutex1，初始状态是未被获得的。

三个子线程的执行体分别如下：

Thread3：
       ◇ 获得g_Mutex1；
       ◇ 循环执行一定次数的
              打印“Thread3: Executing BEFORE releasing the MUTEX...” 
       ◇ 释放g_Mutex1；
       ◇ 循环执行：
              打印“Thread3: Executing AFTER releasing the MUTEX...”

Thread2：
       ◇ 打印：“Thread2: Start executing...” 
       ◇ 循环执行：
              打印“Thread2: executing...”

Thread1：
       ◇ 打印：“Thread1: Start executing...” 
       ◇ 打印：“Thread1: Waiting on Mutex1...” 
       ◇ 执行：WaitForSingleObject(g_Mutex1, INFINITE); 
       ◇ 打印：“Thread1: GOT g_Mutex1, and will resume.” 
       ◇ 循环执行：
              打印“Thread1: Executing AFTER getting the MUTEX...”

2.2.2 Log结果分析

结果输出如下：

       TID:a3ecaa1e Thread1: Handler = a3c867b2, ThreadId = a3c867b2
       TID:a3ecaa1e Thread2: Handler = 83c8c002, ThreadId = 83c8c002
       TID:a3ecaa1e Thread3: Handler = 3c866c2, ThreadId = 3c866c2
       TID:3c866c2 Thread3 GOT the Mutex g_Mutex1
       TID:83c8c002 Thread2: Start executing... 
       TID:83c8c002 Thread2: executing... 
       TID:a3c867b2 Thread1: Start executing... 
       TID:a3c867b2 Thread1: Waiting on Mutex1... 
       TID:3c866c2 Thread3: Executing BEFORE releasing the MUTEX... 
       TID:3c866c2 Thread3: Executing BEFORE releasing the MUTEX... 
       ...... 
       TID:3c866c2 Thread3: Executing BEFORE releasing the MUTEX... 
       TID:3c866c2 Thread3: Executing BEFORE releasing the MUTEX... 
       TID:3c866c2 Thread3: Released the mutex1. 
       TID:a3c867b2 Thread1: GOT g_Mutex1, and will resume. 
       TID:a3c867b2 Thread1: Executing AFTER getting the MUTEX... 
       TID:a3c867b2 Thread1: Executing AFTER getting the MUTEX... 
       TID:a3c867b2 Thread1: Executing AFTER getting the MUTEX... 
       ......

结果中的斜体加粗部分标识了图二的T3－T4时刻的操作。Thread3由于拥有Thread1欲申请的同步资源g_Mutex1而继承了Thread1的优先级，获得了执行。

结果中的粗体加下划线部分标识了图二的T4时刻的操作。Thread3释放同步资源g_Mutex1，Thread3的优先级被恢复到T3之前；Thread1获得了g_Mutex1之后继续执行。

2.2.3 用Platform Builder的工具查看

优先级继承时的线程的优先级是无法在程序中通过诸如GetThreadPriority() / CEGetThreadPriority() 之类的函数来查看CurrPrio的，这些函数查看的是显式设置的BasePrio。（WinCE的早期版本里，这些函数返回的是CurrPrio）

我们可以通过Platform Builder的调试工具来查看，不过因为PB提供的Emulator与BP Host之间同步需要时间，所以我们在上面的程序中需要延长Thread3在T3－T4时间段的时间以获得足够的时间查看此时Thread3继承的优先级。仅仅为了查看，这里可以设置一个极限情况，即Thread3在释放g_Mutex1之前无限循环。

下图是做完上述设置之后，通过 Platform Builder 的菜单“View | Debug Windows | Threads”，调出的Process/Thread View。

图三、通过Thread View查看优先级继承

图中，第一行对应Thread3；第二行对应PriorityInversion进程的主线程；第三行对应Thread1；第四行对应Thread2。从Thread3的CurrPrio列可以看出，它的当前的优先级确实提升到了Thread1的优先级。

2.3 优先级继承的传递性

优先级继承的传递性，是指图二的优先级继承发生时，在Thread3的优先级已经提升到Thread1执行的T3－T4时间段里，如果Thread3竞争比Thread3的初始优先级还要低的线程Thread4拥有的同步资源，操作系统同样会把Thread4的优先级提升到Thread1的优先级。SUN的Solaris系统实现了优先级继承和继承的传递，Microsoft的Windows CE虽然实现了优先级继承，但是它只实现了一级，即，没有实现优先级继承的传递性。

下图是WinCE中按照上面描述增加了Thread4之后，参看Thread View得到的一个快照。

图四、通过Thread View查看优先级继承的传递性

图中，第一行对应Thread1；第二行对应Thread2；第三行对应PriorityInversion进程的主线程；第四行对应Thread4；第五行对应Thread3。从Thread3和Thread4所在行的CurrPrio列可以看出，Thread3由于等待Thread4拥有的竞争资源而被Blocked，优先级也没有得到提升；Thread4由于拥有Thread3所要竞争的同步资源，其优先级被提升到Thread3的优先级。这种现象不具有优先级继承的传递性。

3 总结

优先级翻转是多线程环境中应该尽力避免的现象，尤其是依赖线程优先级来调度的系统中。在设计多线程程序时候，避免优先级翻转的最简单方法，就是不要让不同优先级的线程去竞争同一个同步资源。

优先级继承可以减缓优先级翻转带来的问题，但是也不能保证优先级最高的线程绝对地被最先调度执行，还是有一定的延缓时间。如果有优先级继承的传递性，传递的级别很深时，对系统性能的影响还是很大的。

参考资料及进一步阅读

       ◇ Microsoft, MSDN, 2005.        ◇ Uresh Vahalia, UNIX Internals: The New Frontiers, PEARSON EDU / 人民邮电出版社影印, 2003/2003.7

关于作者

       田海立，硕士，国家系统分析师，中国系统分析员协会顾问团专业顾问。
       您可以通过 HaiLi.Tian(at)csai.cn 或 TianHaiLi(at)nju.org.cn 与他联系，到 http://blog.csdn.net/thl789/ 看他最新的文章。

版权声明：

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