From:	neil AT robots DOT ox DOT ac DOT uk (Neil Townsend)
Newsgroups:	comp.os.msdos.djgpp
Subject:	Re: Optimizer and logical operators.
Date:	Fri, 4 Sep 1998 12:10:57 GMT
Organization:	None evident here.
Message-ID:	<1998Sep4.121057.1208@catorobots.ox.ac.uk>
References:	<4D580E8D0A378A4D DOT 13E50F10CD0F8074 DOT 909D1CC8A7C84CD0 AT library-proxy DOT airnews DOT net>
NNTP-Posting-Host:	cato.robots.ox.ac.uk
NNTP-Posting-Date:	4 Sep 1998 12:11:03 GMT
Originator:	neil AT cato
Lines:	32
To:	djgpp AT delorie DOT com
DJ-Gateway:	from newsgroup comp.os.msdos.djgpp

In comp.os.msdos.djgpp (SED AT ticnet DOT com) writes:
>So I should be able to rewrite:
>
>while( (*p1 != '\0') && isspace( *p1 ) )
>    ++p1;
>
>as
>
>while( (*p1 != '\0') && isspace( *(p1++) ) );

You could, but there would be a subtle change in functionality. In your
original code, p1 was incremented if the string hadn't ended and it pointed
to a spcae character. In the second line, p1 might be incremented one more
time: If the string hasn't ended and p1 point to a non-space character it
will still be incremented (because the increment is no longer conditional on
the success of the second clause). The loop will end in one of two ways:

1. The string is only space; p1 will point the the terminating chracter.
2. The string contains some non-space. The loop will terminate with p1
pointing to the character after the first non-space character (which will be
a '\0' if the string if there is only one non-space character followed by a
terminataing character.


eg

"       \0"  "     abcx\0"  "y\0"
        ^           ^         ^

Neil
-- 
Neil Townsend   +44 (1865) 273121   neil AT robots DOT ox DOT ac DOT uk

- Raw text -