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K & R 5-5 asks for a strncat function ( concat n characters of t);

void mystrncat(char *s, char *t, int n)

{

while ( *s++); /* find end of s */ /* <<<<< 1 */

/* stops at '\0' */ /* <<<<<2 */

while ( *t && n-- > 0)

*s++ = *t++; /* /*<<<<< 3 */

while ( n -- > 0);
*s++ = '\0';

}

Now, with 1 I **Thought** that *s++ fails when *s == '\0', so that s
points to '\0'.
So, when 3 occurs, I thought the first char of t ( *t) is assigned to
the "Old" position of s, which should be '\0', but it is not.
What am I missing.
Thanks in advance.

On May 1, 8:22*pm, Eric Sosman <esos...@ieee-dot-org.invalid> wrote:
, but the increment happens regardless of the
> outcome of the test.

Ok...I understand. Thank you.

> * * *The loop stops when `s' points at '\0' *before* being
> incremented, but the increment happens regardless of the
> outcome of the test.

May I just ask this. I recently had an extensive explanation about
this from one of the members of the group. Is this then a fair
statement about what happens.
When the loop fails, it is still proceeds to the next sequence point.
In other words, it is not like a "break" statement in a loop.
thanks

mdh wrote:
>> The loop stops when `s' points at '\0' *before* being
>> incremented, but the increment happens regardless of the
>> outcome of the test.
>
> May I just ask this. I recently had an extensive explanation about
> this from one of the members of the group. Is this then a fair
> statement about what happens.
> When the loop fails, it is still proceeds to the next sequence point.
> In other words, it is not like a "break" statement in a loop.

The loop does not fail, the test yields false. The expression "s++" is
always evaluated before the result of the expression is tested.

If you want s to point to the end of the string, use

while( *s ) { s++; }

--
Ian Collins.

On May 1, 9:20*pm, Ian Collins <ian-n...@hotmail.com> wrote:

>
> The loop does not fail, the test yields false. *The expression "s++" is
> always evaluated before the result of the expression is tested.
>

thank you.

On May 2, 9:39 am, mdh <m...@comcast.net> wrote:
> On May 1, 9:20 pm, Ian Collins <ian-n...@hotmail.com> wrote:
>
>
>
> > The loop does not fail, the test yields false. The expression "s++" is
> > always evaluated before the result of the expression is tested.
>
> thank you.

and moreover the unary operators are right to left associative.since
here its post incrementation,s is incremented in the next statement.
probably you can overcome this by using

while(*s)s++;

jt wrote:
> On May 2, 9:39 am, mdh <m...@comcast.net> wrote:
>> On May 1, 9:20 pm, Ian Collins <ian-n...@hotmail.com> wrote:
>>
>>
>>
>>> The loop does not fail, the test yields false. The expression "s++" is
>>> always evaluated before the result of the expression is tested.
>> thank you.
>
> and moreover the unary operators are right to left associative.since
> here its post incrementation,s is incremented in the next statement.

s++ is the statement, there isn't a next one.

> probably you can overcome this by using
>
> while(*s)s++;

Which is exactly what I wrote....

--
Ian Collins.