On Wed, 2005-03-30 at 13:17, Thomas Sandla▀ wrote:
> HaloO Luke,
> 
> you wrote:
> > No, I think I agree with you here.  But what happens if you change
> > you're second-to-last line to:
> > 
> >     my $a = foo();
> >     $a.meth() = 8;
> > 
> > Perl 6 is both a statically typed language and a dynamically typed
> > language, and the problems that I am addressing are mostly about the
> > dynamic part.
> 
> My state of affairs is that these two lines of code lack declarations
> that clearly announce what the user of foo and X wanted:
> 1) &foo returns an X

No, that was most of the point. &foo did not declare a return type, and
while my code was simplistic, we obviously cannot be certain what &foo
might return in the general case.

Given that, Luke was making the point that $a had not explicit type.

-- 
Aaron Sherman <[EMAIL PROTECTED]>
Senior Systems Engineer and Toolsmith
"It's the sound of a satellite saying, 'get me down!'" -Shriekback


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