Q63055: qsort() Compare Routine Must Have Equal Case Under C 6.00

Article: Q63055
Product(s): See article
Version(s): 6.00   | 6.00
Operating System(s): MS-DOS | OS/2
Keyword(s): ENDUSER | S_QUICKC | mspl13_c
Last Modified: 25-JUL-1990

When writing a compare routine for use with the qsort() function
provided in Microsoft C version 6.00, provision must be made to return
0 (zero) if the two elements passed to the routine are identical. This
is necessary even if it is that all elements of the array passed to
qsort are different.

The documentation on qsort() given in the online help specifies that
the compare routine should return 0 in the case of equality. However,
under earlier versions of Microsoft C, a routine will work without
this equality clause provided no two elements of the array are equal.

Under Microsoft C version 6.00, the qsort() function will actually
pass the same address as both parameters to the compare routine during
the execution of its sorting algorithm. If 0 (zero) is not returned in
this case, it will cause qsort() to try and rearrange the two array
elements that were passed. Since both elements were the same address,
qsort() breaks down at this point causing spurious and often
disastrous consequences. The following code, which compiles and runs
correctly under Microsoft C version 5.10, demonstrates the
differences.

Sample Code
-----------

#include<stdio.h>
#include<stdlib.h>
#include<search.h>

int arname[10]={ 10,9,8,7,6,5,4,3,2,1 };

void main(void)
{
    qsort(arname,10,sizeof(int),cmpr);
    printf("%d - %d",arname[0],arname[9]);
}

int cmpr(int *p,int *q)
{
    if (*p<*q) return(-1);
    else return(1);
}

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