I would like to reallocate an array of string with a function. I write a very simple program to demonstrate here. I expect to get the letter "b" to be output but I get NULL.
void gain_memory(char ***ptr) {
*ptr = (char **) realloc(*ptr, sizeof(char*) * 2);
*ptr[1] = "b\0";
}
int main()
{
char **ptr = malloc(sizeof(char*));
gain_memory(&ptr);
printf("%s", ptr[1]); // get NULL instead of "b"
return 0;
}
Thank you very much!
The [] operator has high priority than *, so change the code like this will work right.
(*ptr)[1] = "b";
P.S. "\0" is unnecessary.
You should put parentheses around *ptr in gain_memory:
(*ptr)[1] = "b\0";
You're not allocating any memory for the actual strings in your array of strings, you need to do something like this:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void gain_memory(char ***ptr, int elem) {
*ptr = (char**)realloc(*ptr, 2*elem*sizeof(char*));
(*ptr)[1] = "b";
}
int main()
{
//How many strings in your array?
//Lets say we want 10 strings
int elem = 10;
char **ptr = malloc(sizeof(char*) * elem);
//Now we allocate memory for each string
for(int i = 0; i < elem; i++)
//Lets say we allocate 255 characters for each string
//plus one for the final '\0'
ptr[i] = malloc(sizeof(char) * 256);
//Now we grow the array
gain_memory(&ptr, elem);
printf("%s", ptr[1]);
return 0;
}
realloc, this is C after all. (Doing so might hide problems that the compiler would tell you otherwise.)reallocimmediately to the pointer you're reallocating. Ifreallocfails, you've lost the original pointer and have leaked memory. (Oh, and check for allocation failure too.)