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[QUOTE="Ben Morrow, post: 4755312"]
<snip benchmark>

A tool I like for answering this sort of question is B::Graph, which
produces a graph of the optree. It seems that

print "a", "b";

becomes

nextstate -> pushmark -> const -> const -> print

(nextstate starts a new statement. pushmark makes a mark in the Perl
stack to show where the arguments for the print begin. The consts push
items onto the stack, the print takes them off down to the mark and
prints them) whereas

print "a" . "b";

becomes

nextstate -> pushmark -> const -> print

ie. perl will perform the concatenation at compile time, and thus it
must be more efficient. However,

print "a", $x;

becomes

nextstate -> pushmark -> const -> padsv -> print

as before (padsv gets the value of the variable and pushes it onto the
stack) whereas

print "a" . $x;

becomes

nextstate -> pushmark -> const -> padsv -> concat -> print

ie. the concatenation is done as a separate step before the print, so
it may well be slower (this depends on whether printing several items
as opposed to one is slower or faster than concatenating them).

Certainly it is not true that they are the same: I suspect Walter was
thinking of

print "a$x";

which *is* converted into

print "a" . $x;

internally.

Ben
[/QUOTE]

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