About strcpy() in TCPL

Question

I am reading The C Programming Language, and when it gets to Character Pointers and Functions (5.5) I get a problem.

In 5.5, the authors showed four versions of strcpy(). My problem lies in the version 3:

/*strcpy: copy t to s; pointer version 3*/
void strcpy(char *s, char *t)
{
    while (*s++ = *t++)
        ;
}

There is no comparison against '\0'. And how does the termination of the loop work under such a condition?

The C language was pretty much designed to make this type of code. — Zan Lynx, Dec 28 '17 at 03:04

score 4 · Answer 1 · answered Dec 28 '17 at 02:39

4

*s++ = *t++

is equivalent to

(*s++ = *t++) != 0

which is equivalent to

(*s++ = *t++) != '\0'

answered Dec 28 '17 at 02:39

Edward Karak

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1

So it's just an 'idiom', as the authors said? – 1MinLeft Dec 28 '17 at 02:40
@1MinLeft Yes [15 chars] – Edward Karak Dec 28 '17 at 02:51

user2736738 · Accepted Answer · 2017-12-28T02:51:40.187

There is. The value of the assignment statement is the value assigned. So it will be checking whether the assigned value is 0 ('\0') or not which is what is expected to be done over here.

Equivalently this code boils down to (this is how it would work). Think like this - atleast once the copy will happen. So it shows us that it will be a do-while loop.

char somechar;
do {
   somechar = *t;
   *s = somechar ;
   s++;
   t++;
} while( somechar );

score 1 · Answer 3 · answered Dec 28 '17 at 02:48

1

In C, if (var = expression) means 1) assign expression to var, then 2) check if var evaluates to TRUE. The same is for while (var = expression).

answered Dec 28 '17 at 02:48

About strcpy() in TCPL

3 Answers3