perl foreach loop in subroutine

Question

I know the subroutine in perl pass arg by reference. But in the below code, the foreach loop in the subroutine should not be changing the value for @list because the my $i should be creating a new lexically scope var $i. Any assignment to $i should be lexically scope but not changing the @list value.

Can anyone explain what is happening inside the foreach loop that causes the value change to @list?

sub absList {    
    foreach my $i (@_) {
        $i = abs($i);
    }
}

@list = (-2,2,4,-4);
absList(@list);
print "@list";

Outputs:

2 2 4 4

mpapec · Accepted Answer · 2014-06-07 12:48:23Z

5

$i is aliased to elements of @_ array due foreach feature (and @_ elements are aliased to elements of the @list due @_ magic).

From perldoc

If any element of LIST is an lvalue, you can modify it by modifying VAR inside the loop. Conversely, if any element of LIST is NOT an lvalue, any attempt to modify that element will fail. In other words, the foreach loop index variable is an implicit alias for each item in the list that you're looping over.

If you want to avoid such behavior, don't iterate directly over @_ array, ie.

sub absList {
    my @arr = @_;
    foreach my $i (@arr) {
        $i = abs($i);
    }
}

edited Jun 7, 2014 at 12:48

answered Jun 7, 2014 at 12:27

mpapec

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5 Comments

user2763829 Over a year ago

Thanks for the links to the info. So it means that $i in foreach will always be referenced/aliased to the @arr regardless I am using it with/without my. And if the @arr is the special variable @_, it means that $i will be the actual value of the elements in @_. Assigning any value to $i means changing the value of the elements in $@. Am I right?

mpapec Over a year ago

@user2763829 yes, my defines only the scope for iteration variable.

user2763829 Over a year ago

Thanks. Even without my in foreach, the $i will always scoped within the foreach loop as the $i will be scope locally. So is there actually a need to declare the my to $i (in this scenario or any other scenario)?

mpapec Over a year ago

@user2763829 if you use strict which is recommended, then you have to declare $i anyway somewhere, and perl best practice is to limit the scope of every variable to smallest block in which it is actually used. Which leads to foreach my $i (..)

ikegami Over a year ago

Or: sub absList { foreach my $i (@_) { my $j = abs($i); ... } }

Chankey Pathak · Accepted Answer · 2014-06-07 12:27:29Z

2

$i here refers the list elements which it gets from @_, that's why changing $i causes change in respective list element.

answered Jun 7, 2014 at 12:27

Chankey Pathak

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