您如何在C中分配子字符串？

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

Two things:

• &orig is the address of the pointer . Perhaps you want str = orig . Your compiler should have warned you about a pointer type mismatch here; if it didn't, then turn up the warning level until it does.
• Modifying a constant literal string won't always work. Use char orig[] = "Hey you guys." , which copies the literal string into an array called orig that you can safely modify.

Answer 2

(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;
}

Answer 3

1. you miss the first element
2. you cannot change a string literal
3. Modified code:

 #include <stdio.h> int main () { char orig[] = "Hey you guys."; char *str; str = orig; int i; while(*str){ if (*str == 'y') *str = '@'; *str++; } puts(str); puts(orig); return 0; }