How can I automatically forward all paramters from one method to another in Ruby?

Question

If I have methods:

def method_a(p1, p2)
  # do some stuff
  method_b(p1, p2)
end

def method_b(p1, p2)
  # do other stuff
end

Is there a way to call method_b and automatically pass all parameters to it? (Sort like how you can call super and it automatically forwards all params)

Personally; yuck. I'm not comfortable with the `super` magic, either, actually. This is one of those cases where IMO being explicit should be the default choice. — Dave Newton, Dec 13 '13 at 17:07

Малъ Скрылевъ · Accepted Answer · 2013-12-13T17:04:01.270

1

I know one appriximate method:

def method_a *args
   # do some stuff
   method_b *args
end

def method_b *args
   # do other stuff
end

or expanding arguments in the second method:

def method_a *args
   # do some stuff
   method_b *args
end

def method_b p1, p2
   # do other stuff
end

Since super is key-work method, the ruby interperter can treat it as of the same argument list as in the specific method you've called. But default from to call a method without argument is the same as for super method, just method name:

method_a # calls to :method_a without arguments, not as the same argument list for the caller method.

So, it will be strong omonim for the call method syntax.

edited Dec 13 '13 at 17:04

answered Dec 13 '13 at 16:58

Малъ Скрылевъ

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This is definitely a plausible way to go. Both methods will always have the same parameters (in my current case). I was hoping to name them explicitly for readability as I do now, but if this is the only option, might have to go for it. – ardavis Dec 13 '13 at 16:58
I added a short description why call to `super` isn't the same as call to other named method. – Малъ Скрылевъ Dec 13 '13 at 17:04

score 1 · Answer 2 · answered Dec 13 '13 at 17:18

1

Considering arbitrary number of arguments and a possibility of a block, the most general format is:

def method_a(*args, &pr)
  # do some stuff
  method_b(*args, &pr)
end

Then, in the definition of method_b, you can set a specific number of arguments and whether or not it takes a block.

answered Dec 13 '13 at 17:18

sawa

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1

This will not catch keyword parameters, unfortunately. – Jörg W Mittag Dec 14 '13 at 11:38
1

@JörgWMittag: It will catch keyword parameters, these will appear as a hash at the end of `args`. – Marc-André Lafortune Dec 14 '13 at 20:31

score 0 · Answer 3 · edited Dec 13 '13 at 17:15

0

Use *args like this:

def method_a(*args)
  ...
  method_b(*args)
end

def method_b(p1, p2)
  ...
end

You can process the arguments like an array in the method_a.

edited Dec 13 '13 at 17:15

Simone Carletti

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answered Dec 13 '13 at 16:59

Matzi

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You forgot to use the splat operator when invoking method_b. – Simone Carletti Dec 13 '13 at 17:15

score -1 · Answer 4 · answered Dec 13 '13 at 17:12

-1

Wow, this is hacky. But it works

def fwd_call b, meth
  send(meth, *b.eval('method(__method__).parameters.map { |p| eval(p.last.to_s) }'))
end

def method1 x, y
  fwd_call(binding, :method2)
end

def method2 x, y
  x+y
end

puts method1(3, 4)
# 7

answered Dec 13 '13 at 17:12

Max

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How can I automatically forward all paramters from one method to another in Ruby?

4 Answers4