$ perl -MO=Deparse two.pl

BEGIN { $^W = 1; }
sub two {
    use strict 'refs';
    2;
}
use strict 'refs';
printf "two + 2 = %d\n", two(2);
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So, two + 2 is interpreted as two(2). Instead of evaluating your function, then adding 2, perl treats everything after your function call as a LIST of arguments when you omit the parentheses. From perlsub:

2. NAME LIST; # Parentheses optional if predeclared/imported.

While it is legal syntax to omit the parentheses on function calls, it is a good practice to use them, even if your function does not need arguments:

printf "two + 2 = %d\n",two() + 2;
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As a side note, your code would render better if you replace your "pre" tags with "code" tags: Writeup Formatting Tips

A couple of ways, (that amount to the same thing), to make it right:

use constant two => 2;;

printf "two + 2 = %d\n", two + 2;;
two + 2 = 4

sub two () { 2 };;

printf "two + 2 = %d\n", two + 2;;
two + 2 = 4
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#! /usr/bin/perl -w
use strict;
sub two {2}
printf "two + 2 = %d\n",(two) + 2;
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So, two + 2 is interpreted as two(2).

And that's because the following are all the same:

two + 2
two +2
two(+2)
two(2)
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Omitting parens around arguments can lead to weird problems. You've encountered one. You can disambiguate as follows:

two() + 2
(two) + 2
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You could also add a prototype to two.

printf "two + 2 = %d\n",&two + 2;
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(note the ampersand) you get "two + 2 = 4". So it seems the parser treats your "two" as a bareword and converts it to 0 in numeric context (therefore the result).

I wonder why this does not procuce a warning though...

So it seems the parser treats your "two" as a bareword and converts it to 0 in numeric context

No. A bareword would generate a compile error with strict. B::Deparse shows us what's going on.

If you want to know how it works clearly.You print the array @_ inside your subroutine.@_ is an array which contains the arguments passed to subroutine. If you call the subroutine without using &, then it will pass the data given after two subroutine will be passed.In your case, two + 2 2 will be passed as an argument. In the subroutine, you didn't handle anything with that argument.Then, 2 will be returned from that subroutine.That's why 2 gets printed. If you use &, then 2 from the subroutine will be returned and then it added with + 2 given.Then, it will print 4.

use strict;
use warnings;

sub two { print @_; 2 }

print "Two=%d\n", two + 2;
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If you want to have the array structure , use Dumper function which is in Data module.

perl -e 'sub two { $_[0] + 2 };print two + 2'
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