Change pointer of pointer in C

Question

I'm trying to write a void function that gets a pointer to a pointer of a string (char**) as a parameter, and changes the original char* so it pointed to another string, lets say "hello".

Below is the code I've written:

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

void change_ptr(char **str_ptr)
{
    char hello[] = "hello";
    char *hello_ptr = hello;
    *str_ptr = hello_ptr;
}



int main()
{
    char str[] = "yay";
    char *str_ptr = str;
    char **ptr_to_str_ptr = &str_ptr;
    change_ptr(ptr_to_str_ptr);
    printf("%s\n", str);

    return 0;
}

As you can see, Im getting the pointer to the pointer of the char* "yay", delivering it to the function, and in the function I'm getting the pointer to "hello", and changes *str_ptr (which is the pointer to the original string) to the pointer to hello. But when I print str at the end, it prints "yay".

What am I doing wrong?

(when I debug with printing the addresses, everything seems fine.)

You need to print str_ptr. Note, however, that hello only exists within change_ptr(). — rveerd
– rveerd, Commented Oct 23, 2022 at 18:38
str is an array. str_ptr no longer points to it (after the call to change_ptr, it contains the value of an address which is no longer valid), but str still contains the contents "yay". — William Pursell
– William Pursell, Commented Oct 23, 2022 at 18:44
oh I get it. So I defined "yay" as char* str = "yay" instead, and now it worked. Thanks! — nat34
– nat34, Commented Oct 23, 2022 at 18:46
The char hello[] = "hello"; needs to be static char hello[] = "hello"; — Weather Vane
– Weather Vane, Commented Oct 23, 2022 at 18:47
You're accidentally printing out str, which you never changed, instead of str_ptr which you attempted to change. — David Grayson
– David Grayson, Commented Oct 23, 2022 at 18:56

David Grayson · Accepted Answer · 2022-10-23 18:57:36Z

1

This works:

#include <stdio.h>

void change_ptr(const char ** str_ptr)
{
  *str_ptr = "hello";
}

int main()
{
  char str[] = "yay";
  const char * str_ptr = str;
  const char ** ptr_to_str_ptr = &str_ptr;
  change_ptr(ptr_to_str_ptr);
  printf("%s\n", str_ptr);
}

Note that the string "hello" in this example is read-only because it is a string literal. I added const in a few places so you are reminded of this and the compiler will warn you if you try to write to the string. The pointer str_ptr can be modified, but not the string itself.

answered Oct 23, 2022 at 18:57

David Grayson

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