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Re: [MacPerl] Commify and "1 while" syntax
- To: Ronald J Kimball <rjk@linguist.dartmouth.edu>, Darryl Tang <darryl@yoshisonline.com>, macperl@macperl.org
- Subject: Re: [MacPerl] Commify and "1 while" syntax
- From: Kevin van Haaren <kevinv@hockey.net>
- Date: Mon, 31 May 1999 22:36:35 -0500
- In-Reply-To: <19990531132756.A1552474@linguist.dartmouth.edu>
- References: <v04020a00b3786b38cc37@[181.100.100.100]>; from Kevin vanHaaren on Mon, May 31, 1999 at 11:43:46AM -0500<l03130300b3784dc13f00@[24.221.1.199]><v04020a00b3786b38cc37@[181.100.100.100]>
At 1:27 PM -0400 1999/5/31, Ronald J Kimball wrote:
>
>Uh, no. The 1 in the above code is the body of the while loop, not the
>conditional. (Note: 'conditional', not 'argument'.)
>
>
>Yes. Notice where the 'this is true' occurs.
>
>
>
>No. The body of the loop doesn't have to return any specific value. The
>conditional is what is being tested for truth or falseness.
>
>undef while /foo/;
>
>is just as valid as
>
>1 while /foo/;
>
>
>Just like
>while(/foo/) { 1 }
>and
>while(/foo/) { undef }
>
>
>If this construct tested 1 as the conditional, then it would always be true
>and the loop would never terminate!
>
>
>Ronald
Well I don't remember where my head was when I wrote all that wrong stuff,
but it was pretty dark 8-)
Sorry for the misdirections (I know better than that 8-(
Kevin
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