An interesting problem of qsort()

This is a discussion on An interesting problem of qsort() within the C++ Programming forums, part of the General Programming Boards category; As you know, the qsort has the signature: void qsort ( void * base, size_t num, size_t size, int ( ...

  1. #1
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    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 http://www.cplusplus.com/reference/c...lib/qsort.html
    Is it wrong? Seems
    Code:
    qsort (values, 6, sizeof(int), compare);
    should be
    Code:
    qsort (values, 6, sizeof(int), &compare);
    Code:
    /* 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
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    It is not wrong. When passed as an argument, a function is converted to a function pointer.
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  3. #3
    Registered User hk_mp5kpdw's Avatar
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    In the above context, "compare" is the address of the function, you do not therefore need to prefix that with '&'.
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  4. #4
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    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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    Quote Originally Posted by meili100 View Post
    Code:
    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.
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  6. #6
    Registered User MacNilly's Avatar
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    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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    . . . but unlike arrays, you can use either way with function pointers. And there are two ways to call function pointers as well.
    Code:
    #include <iostream>
    
    void func() {
        std::cout << "func()\n";
    }
    
    int main() {
        void (*f)();
    
        f = func;
        f = &func;
    
        f();
        (*f)();
    
        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: http://www.newty.de/fpt/index.html
    (I know it says &func is better than func, but I don't think that's so.)
    dwk

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