Function pointer in Ruby that doesn't violate access rules

Question

In C++ I can do this:

(condition ? sin : cos)(0.5);

or

typedef std::deque<int> T;
(T().*(condition ? &T::push_back : &T::push_front))(1);

What would be an equivalent of this in Ruby?

I know I can use send or method, but they allow me to call private methods.

# String#puts and String#print are private
("".method condition ? :puts : :print).call

`if` forces me to repeat arguments twice: `if condition a.f(1) else a.g(1) end`. I want to avoid that. — detunized, Nov 10 '12 at 14:00

score 2 · Accepted Answer · edited Nov 10 '12 at 15:15

2

send is the way to go in Ruby. If you don't like it because it allows to call private methods, use public_send instead:

Math.public_send(condition ? :sin : :cos, 0.5)

edited Nov 10 '12 at 15:15

Andrew Marshall

answered Nov 10 '12 at 14:10

tokland

Sounds kinda inefficient, though. I timed it and calling method directly is faster than calling it through `send`. In C++ there's no overhead, except for missed optimization opportunity. – detunized Nov 10 '12 at 14:32

1 Answers1