How to copy char *str to char c[] in C?

Question

Trying to copy a char *str to char c[] but getting segmentation fault or invalid initializer error.

Why is this code is giving me a seg fault?

char *token = "some random string";
char c[80];  
strcpy( c, token);
strncpy(c, token, sizeof c - 1); 
c[79] = '\0';
char *broken = strtok(c, "#");

Answer 1

use strncpy() rather than strcpy()

/* code not tested */
#include <string.h>

int main(void) {
  char *src = "gkjsdh fkdshfkjsdhfksdjghf ewi7tr weigrfdhf gsdjfsd jfgsdjf gsdjfgwe";
  char dst[10]; /* not enough for all of src */

  strcpy(dst, src); /* BANG!!! */
  strncpy(dst, src, sizeof dst - 1); /* OK ... but `dst` needs to be NUL terminated */
      dst[9] = '\0';
  return 0;
}

Answer 2

use strncpy to be sure to not copy more charachters than the char[] can contains

char *s = "abcdef";
char c[6];

strncpy(c, s, sizeof(c)-1);
// strncpy is not adding a \0 at the end of the string after copying it so you need to add it by yourself
c[sizeof(c)-1] = '\0';

Edit: Code added to question

Viewing your code the segmentation fault could be by the line

strcpy(c, token)

The problem is if token length is bigger than c length then memory is filled out of the c var and that cause troubles.