C syntax. Copying a character string

Question

Going through K&R I saw the following code snippet of a function, strcopy, which copies a character array to another.

If t is the pointer to the first array, and s is the pointer to the array which t is copied to, the code is:

void strcopy(char *s, char *t){

     while(*s++=*t++)
             ;
}

I'm confused by the while loop. I undersatnd that inside the condition t is copied to s, but I don't understand what condition is being tested here. When will *t++ be false (or zero)? Presumably, when the character string finishes. We can test whether the string is finished by checking if the character pointed at is '\0'. K&R says as much. But then the book rather blithely points out that this test isn't necessary. So I'm wondering what is being tested here?

such a classical code. but if you write something like this, maybe your colleague will complain... that expand the code into several lines will be helpful for reading easily. and the compiler will do the optimization for you. — Stan
– Stan, Commented Jul 24, 2011 at 13:44

vsz · Accepted Answer · 2011-07-24 14:09:28Z

6

*s++ = *t++;

is the same as:

*s = *t;
t++;
s++;

The condition always evaluates the left value, so in this case it is like testing

while(*s)

Of course, '\0' evaluates to false, so you don't need while(something!='\0'), because while(something) is enough.

edited Jul 24, 2011 at 14:09

answered Jul 24, 2011 at 12:53

vsz

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1 Comment

ShinTakezou Over a year ago

t++ and s++, the * is not needed in this case

Frédéric Hamidi · Accepted Answer · 2011-07-26 18:23:22Z

5

*s++ = *t++ will evaluate to the value that was assigned. At the end of the string, *t will be '\0', and upon assignment the expression will evaluate to '\0' (which C interprets as false).

I guess K&R are saying that an additional test is not necessary, since everything is handled in the while condition.

edited Jul 26, 2011 at 18:23

answered Jul 24, 2011 at 12:52

Frédéric Hamidi

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Comments

helios35 · Accepted Answer · 2011-07-25 10:50:38Z

3

You can test explicitly if the terminating character is reached by using

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

if you look up '\0' in an ascii table, you'll see that the character '\0' has an integer-value of 0. Writing something like printf("%d\n", '\0'); would also prove it.

So, the while-statement above can also be written as

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

However, a while-loop always checks whether the condition has a non-zero-value, so it's always redundant to compare a value in a while-loop against zero in this way. Therefore, you can simply skip the comparison.

edited Jul 25, 2011 at 10:50

answered Jul 24, 2011 at 13:10

helios35

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Andrew White · Accepted Answer · 2011-07-24 12:55:22Z

0

C assignments flow right-to-left and a while loop does an implicit "if" test. So what the compiles does from the perspective of the test is ...

while(*t != 0) // for the sake off testing if the loop should continue ONLY

Another example...

if (a = b) // note the = vs == so, this will only be true if b != 0

A kind of under-the-hood pseudo code version of that loop...

loop:
*s=*t;
if (*s == 0) should_break=1;
s++;
t++;
if (should_break==1) break;
goto loop;

answered Jul 24, 2011 at 12:55

Andrew White

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