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Postings from March 2001
On Mon, Mar 12, 2001 at 12:30:43PM +0000, David Mitchell wrote:
> Konovalov, Vadim" <vkonovalov@lucent.com> wrote:
>
> > print 'foo'=~/(.*)/ && $1, 'bar'=~/(.*)/ && $1, "\n";
> > print 'foo'=~/(.*)/ && "$1", 'bar'=~/(.*)/ && "$1", "\n";
> >
> >
> > results in:
> >
> > barbar
> > foobar
> >
> > which was unexpected to me.
> > While there may be found a way to explain such behaviour, it may result in
> > errors that are quite hard to find.
>
> This is not a bug.
>
> It occurs due to the way parameters to functions are handled.
>
> if you call foo($var), then within foo(), $_[0] does not contain a *copy*
> of the current value of $var; instead $_[0] is an *alias* of $var - ie
> modifying $_[0] actually modifies $var.
>
> Or to put is another way: behind the scenes, foo($var) causes the address
> of $var to be passed to foo, rather than its current value being passed.
Yes, for calling functions. And that's documented. But perlop says:
The `||' and `&&' operators differ from C's in that, rather than
returning 0 or 1, they return the last value evaluated.
It clearly states *value*. Not address. Not alias. Value.
And before you say it's because print is being called and that's a function,
you are wrong. That's irrelevant; it's about what && returns.
$a = (($foo = 2) && $foo) + (($foo = 3) && $foo);
print $a;
prints 6, not 5 as you would expect if && returns a value.
Camel III agrees with perlop that && and || return values.
Abigail
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