Given the above, here's a step by step breakdown of what Perl does.

c: When the code is compiled.
r: When the code is run.

     my @list = x() | (1,2,3);
c => my @list = scalar(x()) | scalar(1,2,3);
c => my @list = scalar(x()) | scalar(2,3);  <<void warn>>
c => my @list = scalar(x()) | scalar(3);  <<void warn (supressed)>>
c => my @list = scalar(x()) | 3;
r => my @list = undef | 3;
r => my @list = 0 | 3; <<uninit warning>>
r => my @list = 3;

     my @list0 = x() || (1,2,3);
c => my @list0 = scalar(x()) || (1,2,3);
r => my @list0 = undef || (1,2,3);
r => my @list0 = (1,2,3);

     my @list1 = x() or (1,2,3);
c => ( my @list1 = x() ) or (1,2,3);
c => ( my @list1 = x() ) or (2,3);  <<void warn>>
c => ( my @list1 = x() ) or 3;  <<void warn (supressed)>>
c => my @list1 = x();  <<void warn>>
r => my @list1 = ();

     my @list2 = $c->bbox('all') || (1,2,3);
c => my @list2 = scalar($c->bbox('all')) || (1,2,3);
r => my @list2 = [ ... ] || (1,2,3);
r => my @list2 = [ ... ];

     my @list3 = $c->bbox('all') or (1,2,3);
c => ( my @list3 = $c->bbox('all') ) or (1,2,3);
c => ( my @list3 = $c->bbox('all') ) or (2,3);  <<void warn>>
c => ( my @list3 = $c->bbox('all') ) or 3;  <<void warn (supressed)>>
c => my @list3 = $c->bbox('all');  <<void warn>>
r => my @list3 = ( ... );
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My areas of confusion span the far reaches of the known universe, however for this problem, I can live with just 3 :-)

I'm interested to understand why the '|' math operation decided to return a 3 for me, when given two non-numbers to operate on. This was the fun part.

Thanks for the precedence pointer - I always seem to forget about that (duh - basic stuff)...

The context confusion is the key - thanks for untangling my knotted mind. I see that this works:

my @list2 = @{$c->bbox('all') || (1,2,3)};
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Thanks

-Craig

I'm interested to understand why the '|' math operation decided to return a 3 for me, when given two non-numbers to operate on. This was the fun part.

    (3,4,5) | (7,8,9)
=== ((3,4),5) | ((7,8),9)
=== (4,5) | ((7,8),9)
=== 5 | ((7,8),9)
=== 5 | (8,9)
=== 5 | 9
=== 13
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It's a common misconception that (..., ..., ...) is a list. It's simply two instances of the binary operator ",". In list context, "," returns a list consisting of both operands (which are evaluated in list context). In scalar context, "," returns its RHS operand (after evaluating both sides in scalar context).

Is there a way to write this that will work for both empty and non-empty canvas?

Close. Since you're dereferencing ("@{}") the result of the "||", you need to return an array reference on both sides of the "||".

my @list2 = @{ $c->bbox('all') || [ 1,2,3 ] };
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Or if you want to avoid creating an array and a reference to it just so it can be derefenced and listed,

my @list2 = $c->bbox('all');
@list2 = (1,2,3) if !@list2;
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In scalar context, "," returns its RHS operand (after evaluating both sides in scalar context).

That's true except when it's not. See Re^2: ||= oddity

my @list2 = @{ $c->bbox('all') || [ 1,2,3 ] }; does not work because [] evaluates as true in boolean context.

As I posted below, '||' will not return the (true) left-hand-side in list context; you can use:


my @t; my @list = (@t = $c->bbox()) ? @t : (1, 2, 3);
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and it'll work.