Thanks a lot.
BTW what is the third argument($i)
nFor( $n-1, @_, $i ) doing here? | [reply] |
BTW what is the third argument($i) nFor( $n-1, @_, $i ) doing here?
As the sub recurses, one of the array references is remove from the front of @_ at each level, and the current value being iterated by the for loop at that level is added to the end.
Once $n == 0, all the array references have been removed and $n has been shifted off, all that is left in @_, is the set of elements to be printed.
BTW. Here is a cleaner implementation that takes a callback to which the results sets are passed:
#! perl -slw
use strict;
sub nForX(&@) {
my $code = shift;
my $n = shift;
return $code->( @_ ) unless $n;
for my $i ( @{ shift() } ) {
&nForX( $code, $n-1, @_, $i );
}
}
my @a = 1..10;
my @b = 'a'..'z';
my @c = map chr, 33 .. 47;
nForX {
print join ' ', @_;
} 3, \( @a, @b, @c );
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thanks a lot. yes i get it now.
In the same light.....i want to call nfor() with different number of arguments...
nfor(1,\@a_1);
nfor(2,\@a_1,\@a_2);
nfor(3,\@a_1,\@a_2,\@a_3);...so on
nfor(N,\@a_1,\@a_2,...\@a_N).
I have a way...but what to check if there is a much crispier way to do it. | [reply]
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