Currying a proc with keyword arguments in Ruby

Question

Say I have a generic Proc, Lambda or method which takes an optional second argument:

pow = -> (base, exp: 2) { base**exp }

Now I want to curry this function, giving it an exp of 3.

cube = pow.curry.call(exp: 3)

There's an ambiguity here, arising from the keyword arguments and the new hash syntax, where Ruby interprets exp: 3 as a hash being passed as the first argument, base. This results in the function immediately being invoked, rendering a NoMethodError when #** is sent to the hash.

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Setting a default value for the first argument will similarly result in the function being immediately invoked when currying, and if I mark the first argument as required, without providing a default:

pow = -> (base:, exp: 2) { base**exp }

the interpreter will complain that I'm missing argument base when I attempt to curry the Proc.

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How can I curry a function with the second argument?

Answer 1

You could build your own keyword-flavored curry method that collects keyword arguments until the required parameters are present. Something like:

def kw_curry(method)
  -> (**kw_args) {
    required = method.parameters.select { |type, _| type == :keyreq }
    if required.all? { |_, name| kw_args.has_key?(name) }
      method.call(**kw_args)
    else
      -> (**other_kw_args) { kw_curry(method)[**kw_args, **other_kw_args] }
    end
  }
end

def foo(a:, b:, c: nil)
  { a: a, b: b, c: c }
end

proc = kw_curry(method(:foo))
proc[a: 1]              #=> #<Proc:0x007f9a1c0891f8 (lambda)>
proc[b: 1]              #=> #<Proc:0x007f9a1c088f28 (lambda)>
proc[a: 1, b: 2]        #=> {:a=>1, :b=>2, :c=>nil}
proc[b: 2][a: 1]        #=> {:a=>1, :b=>2, :c=>nil}
proc[a: 1, c: 3][b: 2]  #=> {:a=>1, :b=>2, :c=>3}

The example above is limited to keyword arguments only, but you can certainly extend it to support both, keyword arguments and positional arguments.

Answer 2

I don't think you can do it with Proc.curry, but there is always the longhand way

cube = -> (base) {pow.(base, exp: 3)}

You could also create a factory function

pow_factory = -> (exp) {-> (base) {pow.(base, exp: exp)}}
cube = pow_factory.(3)