our @x;
sub foo {
  print "wantarray? ", wantarray?1:0, "\n";
  push @x, (1,2,3);
  return @x;
}
pop foo();
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generates

Type of arg 1 to pop must be array (not subroutine entry) at Monks/Sni
+ppet.pm line 7, near ");"
Execution of Monks/Snippet.pm aborted due to compilation errors.
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The same compilation error results if one tries to use a subroutine to generate the first parameter of push as well. Replacing pop foo() with push foo(), 4 results in

Type of arg 1 to push must be array (not subroutine entry) at Monks/Sn
+ippet.pm line 7, near "4;"
Execution of Monks/Snippet.pm aborted due to compilation errors.
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Best, beth

I'm not sure it even gets to the point of evaluation :-).

Are you saying it's a no-op?

push @{[ print "It definitely gets evaluated" ]}, 1;
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To revisit what I already said, it's useless to talk about the context in which the @a from push @a, ... is evaluated because it doesn't return what it normally returns in either list or scalar context.

$ perl -le'print prototype "CORE::pop"'
;\@
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A reference is a scalar and pop's first argument is a scalar reference to an array.

exactly, but that's different from "evaluating in scalar context".

like in $foo=\@bar, $foo is not the length of @bar!

IIRC in Perl 6 there is less confusion, since the scalar of an array is the array-reference.

Cheers Rolf

I never said it didn't result in a scalar. I said it wasn't evaluated in scalar context.

$ perl -MO=Concise -e'pop @a'
6  <@> leave[1 ref] vKP/REFC ->(end)
1     <0> enter ->2
2     <;> nextstate(main 1 -e:1) v ->3
5     <1> pop vK/1 ->6
4        <1> rv2av[t2] lKRM/1 ->5       <-- "l"ist context.
3           <#> gv[*a] s ->4
-e syntax OK
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I also said it's silly to consider the context of something subjected to the \@ prototype.