From: Yukihiro M. [mailto:[email protected]]
end
ora = foo( (baz = 5), (bar = 2) )
??
Second one.
Dan
On Fri, 3 Mar 2006, Berger, Daniel wrote:
foo baz = 5, bar = 2
a, b = foo(baz = 5), (bar = 2)
or
a = foo( (baz = 5), (bar = 2) )
??
Second one.
but that’s a problem no? number two is the same as this
baz = 5
bar = 2
a = foo baz, bar
see, when on writes
foo baz = 5, bar = 2
it’s ambiguous if those assignments are part of a method call or not.
bar
escpecitally, it could easily be considered an assignment statement
(bar=2)
that’s part of a parrallel assignment statement.
or am i missing something?
-a
[email protected] wrote:
??
Second one.
but that’s a problem no? number two is the same as this
baz = 5
bar = 2
a = foo baz, bar
No, then they become positional. That’s the same as foo 2, 5.
see, when on writes
foo baz = 5, bar = 2
it’s ambiguous if those assignments are part of a method call or not. bar
escpecitally, it could easily be considered an assignment statement (bar=2)
that’s part of a parrallel assignment statement.or am i missing something?
The parser just has to be trained well enough. 
That being said, perhaps using ‘=’ as a keyword operator would just
cause too
much confusion amongst the Ruby community. I know some people hate
changing
behavior based on context.
Regards,
Dan