Er... no. If the original poster was assigning a list to a scalar, then the behaviour of the comma operator would ensure that he ended up with the last element in the list in the scalar variable - which would have been what he wanted.

my $scalar = ('this', 'is', 'a', 'list');
print $scalar; # prints 'list'
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The problem was caused because he was assigning an array, not a list.

Please read What is the difference between a list and an array? from perlfaq4 which explains this difference in more detail.