calloc and malloc question

[jonantice]

In C:
Is:

array = (char **)calloc(nelem, sizeof(char *)

equivalent to:

array = (char **)malloc(nelem*sizeof(char *))

I'm getting seg faults. I want to find a way to realloc array. (change it's size on the fly)

[Franklin52]

You can resize a previously allocated memory block with the realloc() function.
The function resizes memory previously obtained with a malloc() or calloc().
Try this (assuming that array is a pointer returned by malloc or calloc):

array = realloc(array, nelem*sizeof(char));

Regards

[jonantice]

That is what I am doing.
I allocated memory like this:

array = (char **)calloc(nelem, sizeof(char *))

but when my nelem changes and I do

array = (char **)realloc(nelem*sizeof(char *))

I get a segmentation fault after a few reallocations.
Should this work or is there something else I'm doing wrong?

[Franklin52]

Here's an example how to allocate and reallocate memory for an array of strings:

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

int main ()
{
    char **words;
    int i;

    words = malloc(10*sizeof(char*)); // allocate 10 pointers to array of string
    
    for( i = 0; i < 10; i++ ) {		// allocate memory for 25 chars for each string
	words[i] = malloc(25);
	sprintf(words[i], "&#37;s","abc");
	printf("%s\n", words[i]);
    }
    
    words = realloc(words, 11*sizeof(char*));	// reallocate a pointer to an array of string
    words[10] = malloc(25);			// reallocate memory for 25 chars for the string
    sprintf(words[10], "%s","efg");		//+of the 11th element
    printf("%s\n", words[10]);

    for( i = 0; i < 11; i++ ) {
      free(words[i]);
    }
    free(words);
}

Regards