On Thu, Aug 6, 2009 at 12:10, Bryan R Harris<bryan_r_har...@raytheon.com> wrote:
>
>
> A question about the comma operator:
>
> (John and Chas deserve a rest from my questions, if they want it =).
>
> The binary comma operator in scalar context supposedly "evaluates its left
> argument, throws that value away, then evaluates its right argument and
> returns that value."
>
> So this:
>
> $_ = "dogs and cats";
> $r = s/o/i/g, s/s/y/g;
> print "$r: $_\n";
>
> ... prints "1: digy and caty".
>
> Why doesn't it print a "2" instead of a "1"? It did 2 replaces of s to y...
> If I change the second line to read:
Because assignment binds more closely than comma. Perl saw this
($r = s/o/i/g), s/s/y/g;
Try this instead:
$r = (s/o/i/g, s/s/y/g);
snip
> ... not evaluated as a binary comma operator? Because it's in list context?
> Then why can't I do this with a list (without the parenthesis)?
>
> �...@a = 1, 2, 3;
snip
Because that is equivalent to
(@a = 1), 2, 3;
You might want to look at the [precedence chart][1]:
left terms and list operators (leftward)
left −>
nonassoc ++ −−
right **
right ! ~ \ and unary + and −
left =~ !~
left * / % x
left + − .
left << >>
nonassoc named unary operators
nonassoc < > <= >= lt gt le ge
nonassoc == != <=> eq ne cmp ~~
left &
left | ^
left &&
left || //
nonassoc .. ...
right ?:
= is here => right = += −= *= etc.
, is here => left , =>
nonassoc list operators (rightward)
right not
left and
left or xor
[1] : http://perldoc.perl.org/perlop.html#Operator-Precedence-and-Associativity
--
Chas. Owens
wonkden.net
The most important skill a programmer can have is the ability to read.
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