Thread: An interesting problem of qsort()

  1. #1
    Registered User
    Join Date
    Apr 2007

    An interesting problem of qsort()

    As you know, the qsort has the signature:
    void qsort ( void * base, size_t num, size_t size, int ( * comparator ) ( const void *, const void * ) );

    The following code is from
    Is it wrong? Seems
    qsort (values, 6, sizeof(int), compare);
    should be
    qsort (values, 6, sizeof(int), &compare);
    /* qsort example */
    #include <stdio.h>
    #include <stdlib.h>
    int values[] = { 40, 10, 100, 90, 20, 25 };
    int compare (const void * a, const void * b)
      return ( *(int*)a - *(int*)b );
    int main ()
      int n;
      qsort (values, 6, sizeof(int), compare);
      for (n=0; n<6; n++)
         printf ("%d ",values[n]);
      return 0;

  2. #2
    C++ Witch laserlight's Avatar
    Join Date
    Oct 2003
    It is not wrong. When passed as an argument, a function is converted to a function pointer.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
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  3. #3
    Registered User hk_mp5kpdw's Avatar
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    Jan 2002
    Northern Virginia/Washington DC Metropolitan Area
    In the above context, "compare" is the address of the function, you do not therefore need to prefix that with '&'.
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
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  4. #4
    and the hat of sweating
    Join Date
    Aug 2007
    Toronto, ON
    Is this C++?
    If so, you should consider using std::sort() instead of the old qsort() function.

  5. #5
    Just Lurking Dave_Sinkula's Avatar
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    Oct 2002
    Quote Originally Posted by meili100 View Post
    int compare (const void * a, const void * b)
      return ( *(int*)a - *(int*)b );
    Sidebar: the comparison function has issues with integer over/underflow and is not a good example of well-defined behavior.
    7. It is easier to write an incorrect program than understand a correct one.
    40. There are two ways to write error-free programs; only the third one works.*

  6. #6
    Registered User MacNilly's Avatar
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    Oct 2005
    CA, USA
    the name of a function, like the name of an array, can be assigned to a pointer of the same type. No need for the address-of operator.

  7. #7
    Frequently Quite Prolix dwks's Avatar
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    Apr 2005
    . . . but unlike arrays, you can use either way with function pointers. And there are two ways to call function pointers as well.
    #include <iostream>
    void func() {
        std::cout << "func()\n";
    int main() {
        void (*f)();
        f = func;
        f = &func;
        return 0;
    The reason for this is that some people thought one way was better, and others the other way. Since both arguments made sense, ANSI incorporated both into C. And C++ got it from there.

    (All of these arguments are from memory, they may be wrong . . . .)

    The argument for func: when you call a function you use parentheses, so leaving off the parentheses is like leaving off the [] array index operators: you should get a pointer. A similar argument applies for func().

    The argument for &func: func is a single object. With arrays, it's true, you can use "array" to get a pointer; but all single objects, be they structures or numbers, you need to use an & to get the address. A similar argument applies for (*func)().

    For more detailed information, see:
    (I know it says &func is better than func, but I don't think that's so.)

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