Question

I’m trying to figure out why this doesn’t work:

#include <stdio.h>
int main ()
{
  char *orig = "Hey you guys.";
  char *str;
  str = &orig;
  while(*str++) {
    if (*str == 'y')
        *str = '@';
  }
  puts(orig);
  return 0;
}
// OUTPUT => "Hey you guys."
// Not "he@ @ou gu@s." as expected.

By assigning str = &orig, I thought that str would share the same memory address as orig.
What am I missing?

Answer 1

(1) for sharing memory you want to do str = orig, since str is already a pointer type.

(2) orig is defined as a string literal, a constant – so you cannot modify the value "Hey you guys.", even not when accessing it via str, it will result in a run time error.

EDIT: Issue #3: In your while loop, you first increase the pointer, and only then checks if it is ‘y’ and modify. By doing so – you will miss the first element. "yasdf" will become "yasdf" and not "@asdf", as you expect. [well I think that what you expect anyway…]

To achieve what you are after, you can follow this: [using strcpy and a buffer, to avoid writing on constant memory]

#include <stdio.h>
#include <string.h>
int main ()
{
  char *orig = "Hey you guys.";
  char buff[14]; //the length of orig + 1 byte for '\0'
  char *str = buff; //since str and buff are already pointers
  strcpy(str,orig);
  while(*str) {
    if (*str == 'y')
        *str = '@';
    str++;
  }
  puts(buff);
  return 0;
}