Question

I am currently trying to learn C and I have come to a problem that I’ve been unable to solve.

Consider:

#include <stdio.h> #include <stdlib.h> #include <string.h>  #define ELEMENTS 5  void make(char **array, int *array_size) {     int i;     char *t = 'Hello, World!';      array = malloc(ELEMENTS * sizeof(char *));      for (i = 0; i < ELEMENTS; ++i) {         array[i] = malloc(strlen(t) + 1 * sizeof(char));         array[i] = strdup(t);     } }  int main(int argc, char **argv) {     char **array;     int size;     int i;      make(array, &size);      for (i = 0; i < size; ++i) {         printf('%s\n', array[i]);     }      return 0; } 

I have no idea why the above fails to read back the contents of the array after creating it. I have literally spent an hour trying to understand why it fails but have come up empty handed. No doubt it’s something trivial.

Cheers,

Answer 1

Here is the working code:

#include <stdio.h> #include <stdlib.h> #include <string.h>  #define ELEMENTS 5  void make(char ***array) {     char *t = 'Hello, World!';      *array = malloc(ELEMENTS * sizeof(char *));      int i;     for (i = 0; i < ELEMENTS; ++i) {         (*array)[i] = strdup(t);     } }  int main(int argc, char **argv) {     char **array;     make(&array);      int i;     for (i = 0; i < ELEMENTS; ++i) {         printf('%s\n', array[i]);         free(array[i]);     }     free(array);     return 0; } 

As the other have posted – there was unused size, and strdup allocates memory by itself, and it is nice to free the memory afterwards…