Larry Wall writes:
> On Wed, Jan 21, 2004 at 08:51:33PM -0500, Joe Gottman wrote:
>
> : Great, so
> : $x = foo(), bar();
> : means the same thing as
> : $x = ( foo(), bar() );
>
> No, we haven't changed the relative precedence of assignment and
> comma. I've been tempted to, but I alway
But that doesn't mean we can necessarily
assume that the list in parens wants to evaluate its args in scalar
context. Maybe it does, and maybe it doesn't.
That doesn't mean that we have to put the C comma operator back
in though. It might just mean that the default is wrong on ()[].
S
- Original Message -
From: "Larry Wall" <[EMAIL PROTECTED]>
To: "Perl6" <[EMAIL PROTECTED]>
Sent: Wednesday, January 21, 2004 2:51 PM
Subject: [perl] Re: Comma Operator
> On Tue, Jan 20, 2004 at 08:12:28PM -0800, Jonathan Lang wrote:
> : Joe Gottma
On Tue, Jan 20, 2004 at 08:12:28PM -0800, Jonathan Lang wrote:
: Joe Gottman wrote:
: >About a month ago, a thread here suggested that we change the meaning
: > of the comma operator. Currently, in scalar context the expression
: > foo(), bar()
: > means "evaluate f
Joe Gottman wrote:
>About a month ago, a thread here suggested that we change the meaning
> of the comma operator. Currently, in scalar context the expression
> foo(), bar()
> means "evaluate foo(), discard the result, then return the value of
> bar()".
> I
About a month ago, a thread here suggested that we change the meaning of
the comma operator. Currently, in scalar context the expression
foo(), bar()
means "evaluate foo(), discard the result, then return the value of bar()".
It was suggested that this be changed to return the
