Question

When I try to run the following code I get a seg fault. I’ve tried running it through gdb, and I understand that the error is occurring as part of calling printf, but I’m lost as to why exactly it isn’t working.

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

int main() {
  char c[5] = "Test";
  char *type = NULL;

  type = &c[0];
  printf("%s\n", *type);
}

If I replace printf("%s\n", *type);
with printf("%s\n", c); I get “Test” printed as I expected. Why doesn’t it work with a pointer to the char array?

Answer 1

You’re passing a plain char and printf is trying to dereference it. Try this instead:

printf("%s\n", type);
              ^ 

If you pass *type it’s like telling printf “I’ve got a string at location T”.

Also type = &c[0] is kind of misleading. Why don’t you just:

type = c;