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Re: using pop and shift to find the sum of an array

by johngg (Abbot)
on Sep 22, 2012 at 12:25 UTC ( #995098=note: print w/ replies, xml ) Need Help??


in reply to using pop and shift to find the sum of an array

You can randomly use pop and shift to remove elements one at a time from either end of your array and add them to the sum, doing this in a loop while there are still elements in the array. However, this obviously destroys your array.

$ perl -Mstrict -Mwarnings -E ' > my @arr = ( 1 .. 10 ); > my $sum; > $sum += int rand 2 ? shift @arr : pop @arr while @arr; > say $sum; > say qq{-> @arr <-};' 55 -> <- $

If, on the other hand, you wish to preserve your array, do the same thing in an on-the-fly subroutine with your array as the argument, which is passed into the subroutine as the @_ array (see perlvar and perlsub). The pop and shift functions inside a subroutine operate on @_ by default, leaving your @arr array untouched.

$ perl -Mstrict -Mwarnings -E ' > my @arr = ( 1 .. 10 ); > my $sum = sub { > my $sum; > $sum += int rand 2 ? shift : pop while @_; > return $sum; > }->( @arr ); > say $sum; > say qq{-> @arr <-};' 55 -> 1 2 3 4 5 6 7 8 9 10 <- $

I hope this is helpful.

Update: Clarified the wording a little.

Cheers,

JohnGG


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