Re: Strange Regex Behavior

[quester]

in reply to Strange Regex Behavior

It might be a little more obvious like this:

$b =  'test 100';

%hash = (
        a => ($b =~ /(\d+)/ ? $1 : 0),
        b => ($b =~ /(\w+)/ ? 1  : 0),
);

print "$hash{a}\n";

which prints

test

So, it did the second pattern match first, and interpreted $1 to be the result of that pattern match. It's the same sort of ambiguity that is found in, say, ($i++)+$i.

[BrowserUk]

So, it did the second pattern match first, and interpreted $1 to be the result of that pattern match

Then why does quoting $1 'fix' it?

[quester]

AFAIK, the order in which these subexpressions are evaluated is not defined. Adding one more operation (more or less any operation, "" or - or sqrt all work) does change the order of evaluation. But in the absence of some rule requiring the second and third operands of ?: to be evaluated after the first one rather than before, that's merely a detail of the implementation. I don't see any rule about it offhand in "conditional operator" in perlop.

[BrowserUk]

Hm. I believe the fact that the entire ternary expression is wrapped in parens means that the contents of those parens must be evaluated before anything outside them per perlop:

A TERM has the highest precedence in Perl. They include variables, quote and quote-like operators, any expression in parentheses, and any function whose arguments are parenthesized.

I also believe that as the two ternary expressions are part of a (4-term) comma separated list, those four terms should be evaluated in strictly left to right order per perlop:

"In list context, it's just the list argument separator, and inserts both its arguments into the list. These arguments are also evaluated from left to right."

.

[remiah]

I saw the same outputs in my perl 5.12.3. It seems perl confusing for $1 because this also prints "100" without warnings;

%hash =(
  a => ($b =~ /(\d+)/ ? $1 : 0),
  b => "test b",
);
print "$_=#$hash{$_}#\n" for keys %hash;

And named capture seems to work fine.

%hash = (
  a => (($b =~ /(?<tag>\d+)/) ? $+{tag} : 0),
  b => (($b =~ /(?<tag>test)/) ? $+{tag} : 0),
);
print "$_=#$hash{$_}#\n" for keys %hash;

But I have no idea for why named capture doesn't confuse...