Rob Hanson wrote:
> I am guessing the problem is (as stated in the perldoc) that "the
> optimizer might have optimized call frames away before "caller" had a
> chance to get
> the information.". Furthermore it states that ""caller(N)" might not
> return
> information about the call frame you expect it do, for "N > 1"". ...But
> in my case "N" never exceeds 1.
>
actually, the optimizer didn't optimized the frames. the number you pass to
caller tells Perl how many levels atop of the current call for the info.
for example:
my @stuff = caller(0);
is the same as:
my @stuff = caller;
which means 0 level up of the call tree so you get what you expect, ie, it
gives you the information of the function's direct caller.
when you say:
my @stuff = caller(1);
you said give me the parent of the parent (the 1 means one level up of the
direct caller) and this could return nothing. consider:
#!/usr/bin/perl -w
use strict;
sub fun{
my @stuff = caller(1);
print join("\n",@stuff),"\n" if(@stuff);
}
fun
__END__
prints nothing because there is no 2 level caller tree. you can visualize
the number as the index to the caller tree. a 1 means the second level of
the caller. 0 means the first level of the caller tree. in fact, Perl's
default array index machanism works the same way, 0 means the first
element, 1 means the second element, etc.
david
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