simple strcpy from string to string crashes entire programme

Question

sorry if this is a commonly asked question but I'm not sure as to why this code will not output anything:

#include <stdio.h>
#include <string.h>

int main()
{
char source[] = "hello world!";

char **destination;

strcpy(destination[0], source);

puts(destination[0]);
puts("string copied!");

return 0;
}

it looks like it is crashing at strcpy() as "string copied!" does not appear in the terminal either

Where do you think that the copied string will be placed? What memory space does `destination[0]` point to? You need to allocate or define a buffer and give that address to your target. — Adrian Mole, Aug 03 '21 at 09:35
Far as that goes, what does `destination` even point to ? I suggest a [decent book](https://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list) and some quality time with the sections on pointers. Pointers are pretty fundamental to doing almost anything in C. You don't just want to learn it; you need to master it. — WhozCraig, Aug 03 '21 at 09:36
Welcome to SO. Does your compiler tell you something about using `destination` without being initialized? You should always take care about all warnings from your compiler. — Gerhardh, Aug 03 '21 at 09:40
@Nagev Comments aren't exactly the right place for such code. — Gerhardh, Aug 03 '21 at 11:32

score 0 · Accepted Answer · answered Aug 03 '21 at 09:43

Your problem is that char **destination is an empty pointer to a pointer. destination[0] points to absolutely nothing.

Here is how you would actually do it:

#include <stdio.h>
#include <string.h>

int main()
{
  char source[] = "hello world!";

  char destination[100];

  strcpy(destination, source);

  puts(destination);
  puts("string copied!");

  return 0;
}

Or if you want to determine the size of destination dynamically:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main()
{
  char source[] = "hello world!";

  char *destination = malloc(strlen(source) * sizeof(char)+1); // +1 because of null character

  strcpy(destination, source);

  puts(destination);
  puts("string copied!");


  free(destination); // IMPORTANT!!!!
  return 0;
}

Make sure you free(destination) because it is allocated on the heap and thus must be freed manually.

simple strcpy from string to string crashes entire programme

1 Answers1