Initializing a char *tab[] with malloc/calloc

Question

I’m trying to initialize a char *tab[] and this causes segmentation fault, what am I doing wrong?

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <float.h>
int main()
{
    char **cle_par = NULL;
    cle_par = (char**) calloc (22, sizeof (char *)+1);
    int y=0;
    for (y=0; y<22; y++);
    {
        cle_par[y]=(char *) malloc(45*sizeof(char));
    }
    strcpy(cle_par[0], "this_is_a_test");
}

Typo: extra semicolon: `for (y=0; y<22; y++);` -> `for (y=0; y<22; y++)` — dbush, Oct 25 '18 at 15:15
Another example of why an opening brace should **not** be on its own line. — dbush, Oct 25 '18 at 15:16
Oh god you're right.. I'm surprised there was no warning though. Thanks a lot — Loukoum Mira, Oct 25 '18 at 15:20
Why an array that holds pointers should be a double pointer and why in the heap? — Tsakiroglou Fotis, Oct 25 '18 at 15:23

score 1 · Answer 1 · answered Oct 25 '18 at 17:53

1

Your code has a "typo" error which invalidates the for loop and makes it an empty loop.

for (y=0; y<22; y++);, the trailing semicolon makes the loop empty, and I believe it's not your intention. If you remove it. Your code should work fine.

answered Oct 25 '18 at 17:53

stensal

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Initializing a char *tab[] with malloc/calloc

1 Answers1