Forcing a list context in Perl

Question

I'm trying to understand Perl context and I tripped over a rock;

Given the code;

#!/usr/bin/perl  my $b = (33,22,11); print "$b\n";  my $b = () = (33,22,11); print "$b\n";  my @b = (33,22,11); print "@b\n";  my @b = () = (33,22,11); print "@b\n"; 

The results are (the last line is blank);

 11     3    33 22 11   <> 

Since the 2nd print returned the length of the list I was assuming that somewhere an array was generated since an array in a scalar context evaluates to its length. But the 4th print seems to belie that assumption. I was expecting to see '33 22 11' printed but got nothing instead. What's happening here?

Answer 1

You're confused because you think = is a single operator while it can result in two different operators: a list assignment operator or a scalar assignment operator. Mini-Tutorial: Scalar vs List Assignment Operator explains the differences.

my $b = (33,22,11); ------------------            Scalar assign in void context.         ----------            List literal in scalar context. Returns last.  my @b = (33,22,11); ------------------            List assign in void context.         ----------            List literal in list context. Returns all.  my $b = ( () = (33,22,11) ); ---------------------------   Scalar assign in void context.         -------------------   List assign in scalar context. Returns count of RHS                ----------     List literal in list context. Returns all.  my @b = ( () = (33,22,11) ); ---------------------------   List assign in void context.         -------------------   List assign in list context. Returns LHS.                ----------     List literal in list context. Returns all. 

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As for your title, forcing list context is impossible per say. If a function returns a list when a scalar is expected, it results in extra values on the stack, which leads to operators getting the wrong arguments.

You can, however, do something like:

( EXPR )[0] 

or

( EXPR )[-1] 

EXPR will be called in list context, but the whole will return just one element of the returned list.

Answer 2

List assignment in scalar context returns the number of elements on the right hand side (case 2). The case 4 first assigns (33, 22, 11) to (), and then assigns () (whose value has not changed) to @b.