Re: Re: Re: writing looped programs

[mrbbking]

So once the variable goes out of scope, its memory is flagged to be reused, iirc.

Yeah - that's what I was thinking.

If the sleep is at the top of the loop, before any my variables have been declared, then the variables can't be taking up any memory.

If the sleep is at the end of the loop, then all my variables declared in the loop are still in scope, and therefore must continue to exist.

...right?

[count0]

But if we're talking about a loop, this memory conservation only exists in the first iteration of it.


while (1) {
   # during the first iteration, no variables have been
   # declared yet.
   # However, if we're in the 2nd, 3rd, etc loop..
   # that 3ish KB of memory allocated for $variable
   # still belong to Perl, and are most likely not yet
   # free()'ed

   sleep 10;

   # Create roughly 3 KB
   { my $variable = 'foo' x 1024; }
}


So you see... we conserve memory for an extra 10 seconds (the first 10 seconds of this code's execution)... but that's it. After that, Perl holds on to that memory created for $variable... even when $variable goes out of scope. So the position of the sleep really makes no real difference.

Anyway, these sorts of attempts at micro-optimizations usually prove to be futile, as Perl does a great job of it on its own. =)