Thread: Total beginner questions

No, it doesn't cause a memory leak, but only because you're trashing memory your program most likely doesn't own.

To allocate memory dynamically use something like malloc().

Code:

int doSomeSillyThing(char **msg)
{
  static char *bleh = "Are you sure this is not causing a memory leak?";
  *msg = malloc(strlen(bleh) + 1);
  strcpy(*msg, bleh);
  return 10;
}

Note: You must free() the memory you allocate. However, you have a different problem to address first. You should be declaring msg in main() as a char *, and then passing its address. So in main()....:

Code:

int main()
{
  char *msg;
  int code = doSomeSillyThing(&msg);
  printf("msg = %s\n", msg);
  free(msg);
  return 0;
}

Note all the changes. They are important.

Edit: And don't forget to include the proper header files. stdio.h is needed for printf() and stdlib.h is needed for malloc() and free().

Incidentally, bleh should be a const char*. You could declare it as a const char[] instead, upon which you could use sizeof instead of strlen(), e.g.,

Code:

static const char bleh[] = "Are you sure this is not causing a memory leak?";
*msg = malloc(sizeof bleh);
strcpy(*msg, bleh);

And don't forget to include the proper header files. stdio.h is needed for printf() and stdlib.h is needed for malloc() and free().

Also, <string.h> should be included for strcpy() and strlen().

More correctly:

Code:

void setMsg(const char** msg)
{
    *msg = "Some string";
}

int main(void)
{
    const char* msg;
    setMsg(&msg);
}

When assigning string literals, always use const. They are unmodifiable (most of the time anyway).
However, when applying const here, we see the limit of this approach: we can't actually modify the msg string later due to it being const.
If you need to change it, you'll have to retort to something along the lines of the solutions above.