if I define a function with several args. as
----------
def func(foo, bar)
----------
and is used with the changed args.:
----------
func(bar='wrong', foo='order')*
----------
it will change the order of args. and this is not desirable, how to
solve it?
I imagine this with a lot of args. and could be danger
Kless wrote:
if I define a function with several args. as
----------
def func(foo, bar)
----------
and is used with the changed args.:
----------
func(bar='wrong', foo='order')*
----------
it will change the order of args. and this is not desirable, how to
solve it?I imagine this with a lot of args. and could be danger
Use a hash argument
def func(args={})
bar=args[:bar]
foo=args[:foo]
end
func(:bar=>1, :foo=>2)
--
vjoel : Joel VanderWerf : path berkeley edu : 510 665 3407
Kless wrote:
func(bar='wrong', foo='order')
Supplementing Joel's answer - that code is a Ruby fallacy and should be avoided. It does not name the arguments as they go in. It assigns two new variables into the calling scope - very confusingly.
--
Phlip
Kless wrote:
it will change the order of args. and this is not desirable, how to
solve it?
It will not. In Ruby, the order of passed arguments is the order of
received arguments, always. Your struct, as it was said, defines two
useless local variables, and in fact passes them to the function.
--
Posted via http://www.ruby-forum.com/\.