Path: utzoo!attcan!uunet!lll-winken!lll-tis!ames!mailrus!cornell!uw-beaver!uw-june!uw-entropy!dataio!bright
From: bright@Data-IO.COM (Walter Bright)
Newsgroups: comp.lang.c
Subject: (c == '\n') is 1 or 0
Message-ID: <1785@dataio.Data-IO.COM>
Date: 14 Dec 88 18:59:58 GMT
References: <gables.352@umigw.miami.edu> <349@greens.UUCP>
Reply-To: bright@dataio.Data-IO.COM (Walter Bright)
Organization: Data I/O Corporation; Redmond, WA
Lines: 16

In article <349@greens.UUCP> matthew@sunpix.UUCP ( Sun NCAA) writes:
<I'm not sure about the validity of the construct 'chcnt++ += (c == '\n');', but
<I'd write your program scrap as:
<        while ( (c = getchar) != EOF)
<        {   if (c == '\n')
<                chcnt++;
<            chcnt++;
<        }
<(Before closing, there is one logic flaw in your scrap of code. '(c == '\n')'
<does not have to evaluate to an integer 1. It must only be a non-zero value if
<true.)

I beg to differ. (c == '\n') is *guaranteed* to evaluate to 1 or 0, and
nothing else. Thus, this is guaranteed to work:
	int c;		/* c needs to be an int, not a char	*/
 	while ( (c = getchar()) != EOF)
 		chcnt += 1 + (c == '\n');