Re: Regular expression

The 1 doesn't do anything. Really

Seems weird, but in perl, just like the trailing "1;" in the end of a "require"'d file, a line consisting of a thingie ( whether it may be a scalar, array, hash, whatever ) is totally valid

So the equivalent of above code is

    while( s/(\d)(\d\d\d)(?!d)/$1,$2/ ) {
          1; ## do nothing, let the regex go until it no longer matche
+s
    }
[download]

Comment on Re: Regular expression Download Code

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Re: Re: Regular expression by alfie (Pilgrim) on Jun 07, 2001 at 20:41 UTC
Please don't confuse the 1 in this example (which indeed is just there for there has to be something there) with the 1; at the end of a perl module!! The 1; statement at the end of a module is truly needed as a returnvalue to the require statement. Substitute it to a 0 and watch the difference. In this example here you could substitue the 1 with a 0, which would change nothing at all... -- use signature; signature(" So long\nAlfie");	[reply]
Re: Re: Re: Regular expression by lestrrat (Deacon) on Jun 08, 2001 at 00:45 UTC
Hmm, sorry if it was misleading, but I think it's a "statement" nonetheless, no? `package foo; sub foo { 1; } 1;` [download] Besides the fact that a required file needs a "true" value returned, I don't think the 1 at the end of the file is any different from the 1 in a sub or any block of code? Syntactically, yeah, they mean different things, but I believe the mechanism that goes behind there are the same... That was sort of where I was trying to get at, if it makes any sense	[reply] [d/l]