in reply to (SOLVED) parsing problems with prototype blocks (precedence)

«+ foo {$t}» is found where an argument would be. An expression is expected, so «+» is treated as a unary operator, so 

foo {$Q} + foo {$t}

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is the same as 

foo {$Q} foo {$t}

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To get the desired result, you can use the following: 

( foo {$Q} ) + foo {$t}

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Update: Replaced "Fixed:" with clearer wording.

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Re^2: (SOLVED) parsing problems with prototype blocks (precedence)
by LanX (Saint) on Nov 03, 2021 at 16:25 UTC
    UPDATE: I may have misunderstood your post, and FIXED is meant as workaround.

    > Fixed: ( foo {$Q} ) + foo {$t}

    no, it's not, because this was what I expected.

    > + is treated as the unary operator, so

    And it would be binary

    use strict;
use warnings;

sub foo (&){ shift->()}
sub baz (&){ shift->()}
sub bar {3}

my ($Q,$t,$res) = (10,3);

sub tst { ( foo {$Q} ) + baz {$t}; }


use B::Deparse;

print B::Deparse->new('-p')->coderef2text(\&tst);

print tst();

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    C:/Strawberry/perl/bin\perl.exe -w d:/tmp/pm/sym.pl 
{
    use warnings;
    use strict;
    (foo(sub {
        $Q;
    }
    ) + baz(sub {
        $t;
    }
    ));
}13

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    update
    FWIW I chose + for an example which at least compiles b/c of the ambiguity.

    I originally used / which fails without having any unary meaning (it's seen as the start of a m//)

    Cheers Rolf
    (addicted to the Perl Programming Language :)
    Wikisyntax for the Monastery

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      Yes, by "Fixed: ( foo {$Q} ) + foo {$t}", I meant "You can fix the problem by using ( foo {$Q} ) + foo {$t}"

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