What's the difference between var++ and ++var . revisited

**Dick Guertin** · 08-29-2017

In 2007, the original title, without "revisited", was posted on this forum.
It was only a two line program, initializing x, and then printing it using either x++ or ++x. I believed there should have been a third line to show what x becomes after line 2. Here's my sample program:

Code:

    #include <stdio.h>
    int main (void)
    {  int x;
       x = 0;
       printf("1: First x:   X = %d\n" , x);
       printf("2: Using x++, X = %d\n" , x++);
       printf("3: Final x:   X = %d\n" , x);
       x = 0;
       printf("1: First x:   X = %d\n" , x);
       printf("2: Using ++x, X = %d\n" , ++x);
       printf("3: Final x:   X = %d\n" , x);
       return(0);
    }

And here is the result:

1: First x: X = 0
2: Using x++, X = 0
3: Final x: X = 1
1: First x: X = 0
2: Using ++x, X = 1
3: Final x: X = 1

Now it's clear that x transitions from 0 to 1 in both cases.

**whiteflags** · 08-29-2017

It's important to remember that expressions as function arguments are evaluated before the function is entered, so you aren't really using the return value for x++ or ++x. Whereas if you use the x++ or ++x in other places, it will make a difference.

Like for example this implementation of strncpy:

Code:

char * strncpy(char *a, char *b, size_t n) {
    char *rv = a;
    while (n-- > 0 && *b != '\0') {
        *a++ = *b++;
    }
    while (n > 0) { /* technically strncpy pads longer buffers with \0 */
        *a++ = '\0';
        --n;
    }
    return rv;
}

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