Re: calls and parens

[Luke Palmer]

Juerd writes:
> Which assumptions are wrong?
> 
> foo (3) + 4;# foo(7)
> foo(3) + 4; # foo(3)
> foo.(3) + 4;# foo(3)
> foo .(3) + 4;   # foo(3)
> 
> $foo (3) + 4;   # syntax error
> $foo(3) + 4;# $foo(3)
> $foo.(3) + 4;   # $foo(3)
> $foo .(3) + 4;  # $foo(3)
> 
> $o.m (3) + 4;   # syntax error
> $o.m(3) + 4;# m(3)

none(@above)

> What do these mean?
> 
> $o.m .(foo) # m(foo) or m().(foo) ???
> $o.m.(foo)  # m(foo) or m().(foo) ???
> 
> In the case of m(foo), m().(foo) is the obvious way to call the returned
> sub.
> 
> In the case of m().(foo), I would not have any idea how to put
> whitespace in between method and opening paren.
> 
> This leads me to believe that $o.m.(foo) and $o.m .(foo) are $o.m(foo).

Yep.

> Parens cannot be used to group an expression which is then
> used as a method name:
> 
> $o.("on_" ~ %methods{$event}).();  # $o(...)

Well, you can't do that anyway.  It has to be:

$o.::("on_" ~ %methods{$event}).()

Which I believe does the right thing anyway.

Luke

calls and parens

[Juerd]

Which assumptions are wrong?

foo (3) + 4;# foo(7)
foo(3) + 4; # foo(3)
foo.(3) + 4;# foo(3)
foo .(3) + 4;   # foo(3)

$foo (3) + 4;   # syntax error
$foo(3) + 4;# $foo(3)
$foo.(3) + 4;   # $foo(3)
$foo .(3) + 4;  # $foo(3)

$o.m (3) + 4;   # syntax error
$o.m(3) + 4;# m(3)

What do these mean?

$o.m .(foo) # m(foo) or m().(foo) ???
$o.m.(foo)  # m(foo) or m().(foo) ???

In the case of m(foo), m().(foo) is the obvious way to call the returned
sub.

In the case of m().(foo), I would not have any idea how to put
whitespace in between method and opening paren.

This leads me to believe that $o.m.(foo) and $o.m .(foo) are $o.m(foo).

-

Parens cannot be used to group an expression which is then
used as a method name:

$o.("on_" ~ %methods{$event}).();  # $o(...)

Is there a way to do this without temporary variable?


Juerd
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