Re: Re: Did I match or didn't I?

re: "The two OR operators actually behave quite differently"

That is a whole new issue, since perlop says, "It's equivalent to || except for the very low precedence."

What are the differences? And lets let the doc pumpking know.

Updated: The precidence is indeed the only difference. The "as previously noted"

my ($result) = m[pattern(group)] or "default";
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is not correct.

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Re: Re: Re: Did I match or didn't I? by chipmunk (Parson) on Oct 18, 2001 at 20:17 UTC
The only difference between or and \|\| is precedence. \|\| has high precedence; or has low precedence. `my @list = func_returning_list() \|\| 'default';` This is parsed as `my @list = (func_returning_list() \|\| 'default');`, which fails because \|\| forces func_returning_list() into scalar context. `my @list = func_returning_list() or 'default';` This is parsed as `(my @list = func_returning_list()) or 'default';`, which fails because 'default' is not assigned to @list when func_returning_list() returns an empty list. Using ?: is the right solution. In perl6, the logic operators will allow their context to propagate to both their operands, so the first snippet will work as intended.	[reply] [d/l] [select]

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Re: Re: Re: Did I match or didn't I?
by chipmunk (Parson) on Oct 18, 2001 at 20:17 UTC

my @list = func_returning_list() || 'default'; This is parsed as my @list = (func_returning_list() || 'default');, which fails because || forces func_returning_list() into scalar context. my @list = func_returning_list() or 'default'; This is parsed as (my @list = func_returning_list()) or 'default';, which fails because 'default' is not assigned to @list when func_returning_list() returns an empty list.

Using ?: is the right solution.

In perl6, the logic operators will allow their context to propagate to both their operands, so the first snippet will work as intended.

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[d/l]
[select]