Initialize array of strings

Question

What is the right way to initialize char** ? I get coverity error - Uninitialized pointer read (UNINIT) when trying:

char **values = NULL; 

or

char **values = { NULL }; 

Answer 1

This example program illustrates initialization of an array of C strings.

#include <stdio.h>  const char * array[] = {     "First entry",     "Second entry",     "Third entry", };  #define n_array (sizeof (array) / sizeof (const char *))  int main () {     int i;      for (i = 0; i < n_array; i++) {         printf ("%d: %s\n", i, array[i]);     }     return 0; } 

It prints out the following:

0: First entry 1: Second entry 2: Third entry 

Answer 2

Its fine to just do char **strings;, char **strings = NULL, or char **strings = {NULL}

but to initialize it you'd have to use malloc:

#include <stdlib.h> #include <stdio.h> #include <string.h>  int main(){     // allocate space for 5 pointers to strings     char **strings = (char**)malloc(5*sizeof(char*));     int i = 0;     //allocate space for each string     // here allocate 50 bytes, which is more than enough for the strings     for(i = 0; i < 5; i++){         printf("%d\n", i);         strings[i] = (char*)malloc(50*sizeof(char));     }     //assign them all something     sprintf(strings[0], "bird goes tweet");     sprintf(strings[1], "mouse goes squeak");     sprintf(strings[2], "cow goes moo");     sprintf(strings[3], "frog goes croak");     sprintf(strings[4], "what does the fox say?");     // Print it out     for(i = 0; i < 5; i++){         printf("Line #%d(length: %lu): %s\n", i, strlen(strings[i]),strings[i]);     }      //Free each string     for(i = 0; i < 5; i++){         free(strings[i]);     }     //finally release the first string     free(strings);     return 0; }