develooper Front page | perl.beginners | Postings from February 2002

RE: "Learning Perl" Question

Thread Previous | Thread Next
Bob Showalter
February 13, 2002 09:09
RE: "Learning Perl" Question
Message ID:
> -----Original Message-----
> From: Hanson, Robert []
> Sent: Wednesday, February 13, 2002 11:06 AM
> To:
> Subject: "Learning Perl" Question
> [Sorry if this isn't the place to post this, but I thought it might be
> interesting.  Flaming will be accepted]
> I'm teaching a Perl class from the Learning Perl book, and noticed an
> inconsistency with the way certain constructs work.
> In chapter 2 it mentions a rule in Perl: "any time that you 
> need a variable
> in Perl, you can use an assignment instead.  First, Perl does the
> assignment.  Then it uses the variable in whatever way you requested".
> Ok, that makes sense.  It allows you to do things like this:
> chomp( $foo = <STDIN> );
> $x = $y = 10;
> Then in chapter 5 it discusses the each function, and why it 
> works in a
> while loop.  It says that (referring to the boolean value) 
> "Now, when Perl
> evaluates the each %hash there are no more key-value pairs 
> available.  So
> each has to return an empty list.  The empty list is assigned 
> to ($key,
> $value), so $key gets undef, and $value also gets undef.  But 
> that hardly
> matters, because the whole thing is being evaluated in the conditional
> expression of the while loop.  The value of a list assignment 
> is a scalar
> context is the number of elements in the **source list** -- 
> in this case,
> that's 0".
> So this adds to the first rule (sort of).  Basically in 
> scalar context the
> function takes the variable as it's argument, and in list 
> context it takes
> the count of items in the source list.
> ....Ok, that works for me so far, but then why would these work?
> $x = () = 10; # $x is 1
> $x = () = (10, 20); # $x is 2
> From this it sounds like the first rule does not exactly work 
> as stated.  
> It
> seems that the real rule is that the "=" operator returns a 
> value just like
> any other Perl operator.

True. But Randal's rule doesn't contradict that at all.

>  It seems to me that given "x = y" 
> that the "="
> operator returns the value from the right side of the 
> operator (in the given
> context of x).
> So "x = y = z" means that "y = z" returns the value of z, and 
> then "x = "
> stores that value. ...Or at least that is how I am conceptualizing it.

No. See below.

> ....And now the real question.  Why won't this work?
> ($x) = () = 10;

What do you mean "work?". $x becomes undef, which is consistent
with the documented behavior (see below).

> Is there a need for an extra rule to cover this?  Or is there 
> a single rule
> that covers the syntax from all of the above examples?

() = 10 is a list assignment, to an empty list.

The following is from perldoc perlop:

      Similarly, a list assignment in list context produces the list of
      lvalues assigned to, and a list assignment in scalar context returns
      the number of elements produced by the expression on the right hand
      side of the assignment.

This description covers the results you see from each

Thread Previous | Thread Next Perl Programming lists via nntp and http.
Comments to Ask Bjørn Hansen at | Group listing | About