Cant get first character in array of strings - c

Question

I want to get first char character of each string. Here a example:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main() {
int size = 2;
char** text = (char**) malloc(sizeof(char*) * size);
for(int i = 0; i < size; ++i) {
char buf[80];
fgets(buf, 80, stdin);
text[i] = (char*)malloc(strlen(buf));
strcpy(text[i], buf);
}
for(int i = 0; i < strlen(text[i]); ++i) {
printf("%c ", text[i][0]);
}
}
In last for loop, program falls in Segmentation fault. I dont know why.

Answer 1

The strlen function returns the number of characters in the given string not including the terminal nul character; however, the strcpy function copies all characters including that terminating nul!
So, your allocation for text[i] is not quite big enough and, by writing beyond the buffer's bounds, you are getting undefined behaviour.
Add an extra character to the malloc call:
for(int i = 0; i < size; ++i) {
char buf[80];
fgets(buf, 80, stdin);
text[i] = malloc(strlen(buf) + 1); // Need space for the terminal nul!
strcpy(text[i], buf);
}
Or, more simply, use the strdup function, which achieves the same result as your malloc and strcpy in one fell swoop:
for(int i = 0; i < size; ++i) {
char buf[80];
fgets(buf, 80, stdin);
text[i] = strdup(buf);
}
Either way, don't forget to call free on all the buffers you allocate.
EDIT: You are also using the wrong 'limit' in your final output loop; this:
for(int i = 0; i < strlen(text[i]); ++i) { // strlen() is not the # strings
printf("%c ", text[i][0]);
}
Should be:
for(int i = 0; i < size; ++i) { // "size" is your number of strings!
printf("%c ", text[i][0]);
}