Method interface - hash argument vs. named parameters

I’ve never run into this issue before, but I’m working on a method
interface that is quite ‘dense’, and I need to be able to
differentiate between a hash passed to a method and named parameters.
i.e. All of these are valid:

foo(hash, :opt=>val)
foo(hash)
foo(:opt=>val)

However, there seems to be no way to distinguish that last the two
forms.

Is there any way? Or am I stuck?

On Thu, Jun 17, 2010 at 9:24 AM, Intransition [email protected]
wrote:

forms.

Is there any way? Or am I stuck?

I would expect stuck, because the difference seems nothing more than
visual.
(ie if the options hash was assigned to a var, rather than created
within
the arg list, then the last two would be exactly the same) though it
does
seem unusual that there is not a method like options_given? the way
there is
a block_given?

I suppose if this method interface is absolutely necessary, then a hack
you
might be able to get away with would be to check the keys to see if they
match up with your options. But I would be embarrassed to show anyone
such a
solution :wink:

Something to consider, though, whatever param the hash is an arg for, is
optional (ie omitted in your third invocation). So perhaps it would fit
better into the options hash, which you could then extract it from,
inside
the method:

def foo( options = Hash.new )
the_hash = options.delete(:the_hash) { Hash.new }
[ the_hash , options ]
end

hash = Hash[*1…4]
val = :val

foo( :the_hash => hash , :opt => val ) # => [{1=>2, 3=>4},
{:opt=>:val}]
foo( :the_hash => hash ) # => [{1=>2, 3=>4}, {}]
foo( :opt => val ) # => [{}, {:opt=>:val}]

foo(hash)
foo(:opt=>val)

However, there seems to be no way to distinguish that last the two
forms.

One possibility would be to check for known keys in the hash. If
they’re there, assume it’s an options hash.
Another might be to allow them to input it like
:hash => hash
if they ever want it that way.
But you’re probably just stuck.
-r
ref: http://github.com/rdp/arguments

On Jun 17, 11:25 am, Josh C. [email protected] wrote:

match up with your options. But I would be embarrassed to show anyone such a
end

hash = Hash[*1…4]
val = :val

foo( :the_hash => hash , :opt => val ) # => [{1=>2, 3=>4}, {:opt=>:val}]
foo( :the_hash => hash ) # => [{1=>2, 3=>4}, {}]
foo( :opt => val ) # => [{}, {:opt=>:val}]

Yep. That’s what I thought. Appreciate the suggestion. In my case I’ve
decided I’ll just have to do without the options hash.

I was hoping Ruby 1.9 at least would have had this licked. Early on
there was talk of named parameters as:

def foo(hash=nil, **opts)

But it never came to pass.

Thanks.

On 2010-06-17 07:24:47 -0700, Intransition said:

forms.

Is there any way? Or am I stuck?

If both the second and third option are valid then there is no way to
distinguish between the two hashes. If you could narrow it down to one
of them, you would be ok. If the hash is optional but the options
aren’t, you can use:

def foo(hash_or_options, options=nil)
hash = options ? hash_or_options : nil
options = hash_or_options unless options

end

Alternatively, you can pass the original argument as part of the options
hash

foo(:hash => hash, …)

I assume you’re using a more meaningful variable name than ‘hash’ in
the real code, so this may suffice. I would not recommend mixing the
two styles in the same method though (allowing a hash argument OR a
:hash option).