chop() and list assignments

[converter]

Someone on DALnet #perl challenged the channel to come up with the shortest subroutine to return the last character of its argument(s). I thought about it for a few seconds, and replied with:

sub foo {chop(@_=@_)}

This works as expected when foo() is passed a single scalar argument, but when foo() is passed a list, rather than returning the last character of the last list element as one would expect after reading the documentation on chop()*, it returns the last character of the first element in the list.

@a = ('ab','cd','ef');
print chop @a;     # f

print foo(@a);     # b

# the list assignment has the same effect here:
print chop(@a=@a); # b

Changing foo() to:

sub foo {chop(@_=reverse @_)}

makes foo() work as expected.

Is this a documented behavior? I am probably missing some simple concept involved in list assignments, but I'd like to know why the list assignment has this effect.

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*from perlfunc: If you chop a list, each element is the chopped. Only the value of the last `chop' is returned.

edit: chipmunk on 2001-03-05