Nkuipers <[EMAIL PROTECTED]> wrote:
> Hello everyone,
>
> The following is from page 75 in the Camel:
>
> "List assignment in scalar context returns the number of elements
> produced by the expression on the <i>right</i> side of the
> assignment:
>
> $x = ( ($a, $b) = (7,7,7) ); #set $x to 3, not 2
>
> ...."
>
> It goes on to explain how this is useful but from previous reading
> in the same section I don't understand how or why this works. When
> I first saw this, I worked out the assignment where $a and $b each
> get a 7 and the third 7 is discarded, then $x gets the last value
> in the list of $a and $b because of the comma operator for list
> literals in a scalar context...
The result of scalar assignment is the lvalue on the
left hand side:
$x = $y = 2; # $x = 2
($x = 1) = 2; # $x = 2
print $x = 2; # prints 2
In list context, the result of list assignment is
the list of values on left hand side:
@x = () = 0..10; # @x = ()
@x = ($y) = 0..10; # @x = (0)
@x = ($y, $z) = 0..10; # @x = (0,1)
These are consistent, and it's easy to infer the
simple, but incomplete, rule:
"assigment returns its left hand side"
Which can lead a person to believe that compound
assignments can always be broken down into two or
more steps:
$x = $y = 1; # or: $y = 1; $x = $y
The problem is that list assignment in scalar
context breaks the pattern.
$x = ($y) = 0..10; # $x = 11
There's no way to explain why "$x = ($y)" ought to
yield 11 here. (In fact, it shouldn't.)
You just have to accept that list assigment in scalar
context *doesn't* return its LHS. It returns the
number of elements on the RHS.
HTH
--
Steve
perldoc -qa.j | perl -lpe '($_)=m("(.*)")'
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