Re: Can't I chop the lvalue of join?

Can I ask why you're using join(',', keys %$c) in your code, but you're not using join(',', map '?', keys %$c)?

Anyway, since chop() modifies its argument, it must be a variable. But do what I said above.

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Jeff[japhy]Pinyan: Perl, regex, and perl hacker, who'd like a job (NYC-area)
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

Comment on Re: Can't I chop the lvalue of join?

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Re: Re: Can't I chop the lvalue of join? by DapperDan (Pilgrim) on Sep 28, 2002 at 14:33 UTC
I hadn't used join with an in-lined map before. You're right, it's so damned obvious given the way i used `join` with `keys`, but I just couldn't see that until it was pointed out to me. Thanks a lot japhy. PS: not that it really matters any more, but is there a reason why chop only works on a variable rather than any scalar lvalue?	[reply]
Re: Re: Re: Can't I chop the lvalue of join? by japhy (Canon) on Sep 28, 2002 at 15:36 UTC
It does work on lvalues. But the return value of `join()` is not an lvalue. You can't assign to it. `join('x', qw( a b c )) = "foo";` is a syntax error, whereas `substr($x, 2, 4) = "ouch";` is not, so you can say `chop(substr($x, 2, 4))`. _____________________________________________________ Jeff`[japhy]`Pinyan: Perl, regex, and perl hacker, who'd like a job (NYC-area) `s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;`	[reply] [d/l] [select]
Re^4: Can't I chop the lvalue of join? by Aristotle (Chancellor) on Sep 28, 2002 at 18:07 UTC
(Which, of course, is in this case a silly way to say the same as:) `substr($x, 4, 1) = ''; # or substr($x, 4, 1, '');` [download] Makeshifts last the longest.	[reply] [d/l]