C: Trouble understanding pointers in the while loop of this code

Question

I'm using a debugger to read through this code, and I'm a little confused by while ((*d++ = *s2++)); - in the debugger variables, d seems to shorten after each loop (goes from 'Hello hello' to 'ello hello' while s1 changes to 'cello hello'). What is the while loop looping through (shouldn't it be while(condition); do(something))?

Why aren't the variable values of d and s1 the same (isn't d is a pointer to s1)? And when they return to the main function, is curdst = the pointer of dst?

/*    
  Input: char pointers for source (s2) and destination (s1)

  Output: returns the pointer to the destination (s1)
*/

char *my_strcpy(char * , const char * );

int main()
{

  char src[] = "cs23!";
  char dst[]="Hello hello";
  char *curdst;
  int len=0;

  while(src[len++]);

  // do the copy

  curdst= my_strcpy(dst, src);

  // check to see if the NULL char is copied too.

  printf("dst array %s and last element %d\n", dst, atoi(&dst[len]));

  return 0;

}

char *my_strcpy(char *s1, const char *s2) {

  register char *d = s1;

  // print the pointer variables address and their contents, and first char

  printf("s2 address %p, its contents is a pointer %p to first char %c \n", (void *)&s2, (void *)s2, *s2);
  printf("s1 address %p, its contents is a pointer %p to first char %c \n", (void *)&s1, (void *)s1, *s1);

  while ((*d++ = *s2++));
  return(s1);

}

Note that `*d++ = *s2++` is very different from `*d++ == *s2++`. Here, you're overwriting the values in `*d` with values from `*s2`, until you reach the null terminator in `s2`. — hnefatl, Dec 20 '17 at 19:11
The condition here is that the `*s2` is other than zero (string null-terminator). — Eugene Sh., Dec 20 '17 at 19:12
See [this answer](https://stackoverflow.com/a/20957532/2201041) for a breakdown of a similar thing. — , Dec 20 '17 at 19:13

score 2 · Accepted Answer · answered Dec 20 '17 at 19:27

This is a rather typical didactic implementation of strcpy(). It works like this:

my_strcpy() takes two arguments. The first argument is a pointer to the first element of a destination character array. The second argument is a pointer to the first element of a source character string, that is, a character array terminated by a NUL (aka \0) character. The function copies the characters from the source string to the destination buffer, including the NUL terminator, and returns a pointer to the first element of the destination buffer.
```
char *my_strcpy(char *s1, const char *s2) {
```
First things first, make a copy of the first argument, because we need to return it after the copying is done.
```
char *d = s1;
```
Then, copy the characters; this is done in a tight loop, working like this:
- Copy the current character * s2 to the place pointed to by d, as if executing * d = * s2; then
- Increment d to point to the next place in the destination buffer as if executing d++ and increment s2 to point to the next character to be copied as if executing s2++; and
- Exit the loop if the last character which was copied was NUL.
This is written in a very concise manner:
```
while (* d++ = * s2++);
```
* s2++ means "take the character pointed to by s2 then increment s2". Similarily, * d++ as a left-hand value means "use the variable pointed to by d then increment d". The order of precedence of operators helps dispense with parentheses because ++ has higher priority than * which has higher priority than =. The value of the assignment is the assigned value, so the loop ends when the assigned character has the value 0.
Finally, return s1, which was not changed by the function.
```
return s1;
}
```

Thank you!! Is changing `*d` changing `s1` so that when you return `s1`, you have copied s2 to s1? — hello world, Dec 20 '17 at 21:35
@helloworld: You have copied the character string to which `s2` points into the buffer to which `s1` points. — AlexP, Dec 20 '17 at 22:12

C: Trouble understanding pointers in the while loop of this code

1 Answers1