Re^4: What is happening ?

[vinoth.ree]

Is it possible to get each pair of key and value in foreach with each function ?

[ikegami]

each is an iterator. Foreach loops works on list. You're asking if it's possible to cut a piece of wood with a screwdriver.

A foreach loop (whether using the for or foreach keyword) calls the list building expression once, and iterates over the resulting list.

A C-style for loop (whether using the for or foreach keyword) could, since it's really a while loop with bells and whistles.

Since you can probably cut a piece of wood with a screwdriver if you tried hard enough,

for my $pair (
    sub {
        my @pairs;
        while (my ($k,$v) = each %h) { push @pairs, [$k,$v] }
        @pairs
    }->()
) {
    my ($k,$v) = @$pair;
    ...
}

But using the appropriate tool is better.

while (my ($k,$v) = each %h) {
    ...
}

or

for my $k (keys %h) {
    my $v = $h{$k};
    ...
}

[grizzley]

No.each works with while, and for works with keys. Corrections: Re^6: What is happening ? and Re^7: What is happening ?

[Anonymous Monk]

Actually :)

#!/usr/bin/perl --
use strict;
use warnings;

my %f = 1 .. 4;
for( ; my($k,$v) = each %f ; ){
   print qq!($k)=($v)\n!;
}
__END__
(1)=(2)
(3)=(4)

But B::Deparse reveals

use warnings;
use strict 'refs';
my(%f) = 1..4;
while (my($k, $v) = each %f) {
    do {
        print "($k)=($v)\n"
    };
}
__DATA__

So I'm not sure :D

[grizzley]

Ok, I give up, I will disprove the other half of my thesis by myself :)

%f=1..4;
while(($k)=keys%f)
{
  print qq!($k)=($f{$k})\n!;
  delete $f{$k}
}