I'd expect, because of constants folding, two fragments of code to behave the same.

I would not. Just looking at the code, without getting out the Devel toolbox or the debugger and such, the first variant gives a string and the second an expression to the reference-operator (backslash). The expression must be resolved in order to take a reference to its result. Constant folding takes place in either case, but the result of an expression is stored in its own (new) SV, whereas a string unaltered by constant folding is passed as is. Consider: 

sub parse{${$_[0]}=~/\Gfoo/gc or die}
parse \(''.'foo'.$_) for 1..2

[download]

Here, the terms '' and 'foo' are folded into foo by the compiler (constant folding), then (at runtime) it is concatenated with the loop variable. A reference is generated for the result of that expression.
The same steps are taken for (''.'foo') - but without the concatenation (since no such op).

The result of constant folding needs to be stored somewhere. It is not shoehorned back into either PV of '' or 'foo' (which one?), and its storage is volatile.

That's my view of it... I'd be glad to be corrected by somebody more familiar with perl internals.

perl -le'print map{pack c,($-++?1:13)+ord}split//,ESEL'

In reply to Re: What's going on with either constants folding or B::Deparse output in this case? by shmem
in thread What's going on with either constants folding or B::Deparse output in this case? by vr

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