Date: Thu, 29 Feb 1996 22:31:20 -0700 From: Tony Jones <tony@rtd.com> To: questions@freebsd.org Subject: Re: C question: what is wrong with this code Message-ID: <199603010531.WAA17108@seagull.rtd.com>
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In article <Pine.BSF.3.91.960229164732.5332A-100000@lisa.rur.com> you wrote:
: I am having a problem with elementary C. The following stub
: reproduces the problem exactly. What am I doing wrong?
I'm not even going to ask why you are using char arrays rather than ints :-)
1) comparing integers lexically won't work
3468 is lexically < 400
#include <string.h>
main()
{
printf("%d\n", strcmp("3468", "400"));
}
zebedee:12% ./a.out
-1
2) From the man page:
>The contents of the array base are sorted in ascending order according to
>a comparison function pointed to by compar, which requires two arguments
>pointing to the objects being compared.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
If your array was made of int's, the comparison function would receive
int*. In your case, the arguments are char**, since your array contains
elements of type char *.
tony
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#define TABLESIZE 10
#define ITERATIONS 10
int f(const void *p1, const void *p2)
{
int a1 = atoi(*(char **)p1),
a2 = atoi(*(char **)p2);
if (a1 == a2)
return 0;
else return a1 < a2 ? -1 : 1;
}
main()
{
int i, j , n;
char *table[TABLESIZE], buff[1024];
for (i = 0; i < ITERATIONS; i++)
{
/*
* populate the table with some random strings
*/
n = random() % TABLESIZE + 1;
for (j = 0; j < n; j++)
{
sprintf(buff, "%d", random() % 10000);
table[j] = strdup(buff);
}
for (j = 0; j < n; j++)
printf("%s\t", table[j]);
printf("\n");
/* so far, so good. Now sort the table */
/* qsort(table, n, sizeof(char *), strcmp); */
qsort((void *)table, n, sizeof(char *), f);
for (j = 0; j < n; j++)
printf("%s\t", table[j]);
printf("\n");
/* huh!? Doesnt look sorted, does it? */
for (j = 0; j < n; j++)
free(table[j]);
}
exit(0);
}
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