array of strings with pointers in C

Question

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

int main()
{
    char **pps;

    int i, n = 10;

    char name[256];

    pps = (char **)calloc(n,sizeof(char **));

    for (i=0; i<n; i++)
    {
        printf("\n Enter name[%d]: ",i+1);
        gets(name);
        printf("\n Name=%s len=%d",name,strlen(name)+1 );

        pps[i] = (char *)malloc(strlen(name)+1);
        printf("\n pps[i]=%u",pps[i]);
        if (pps[i] = NULL)
        {
            perror("\nerror in malloc");
            exit(1);
        }

        printf("\n pps[i]=%u",pps[i]);
        strncpy(pps[i],name,strlen(name)+1);
    }

}

/* input/output from program: Enter name[1]: abcdef

 Name=abcdef len=7
 pps[i]=13311184
 pps[i]=0

*/

The program gives Runtime Error ??? Please help to find what is wrong and why PPS[i] become currupted ??? I am using visual studio 2010

And don't put a printf between your failing malloc and the related perror. (printf may change errno and make the perror less than useless.) — William Pursell
– William Pursell, Commented Feb 13, 2016 at 18:22
Try to adjust warning level higher, because if (pps[i] = NULL) should give you a warning! Or if you already got a warning, then consider not ignoring the warnings, but actually fixing them... They're there for a reason! — hyde
– hyde, Commented Feb 13, 2016 at 19:18

machine_1 · Accepted Answer · 2016-02-13 18:35:02Z

4

Your problem is here:

if (pps[i] = NULL)

It should be:

if (pps[i] == NULL)

Remember that = is an assignment operator while == is comparation

Also remember to use free at the end of your program to avoid memory leak.

edited Feb 13, 2016 at 18:35

machine_1

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answered Feb 13, 2016 at 18:19

Martín Muñoz del Río

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