Calling/applying lambda vs. function call - the sy

Question

I am kinda new to Ruby and still trying to understand some of the language design principles. IF I've got it right, the lambda expression call in Ruby must be with square braces, while the "regular" function call is with "regular"/round braces.

Is there a special reason that the syntax is different? Or, in other words, (why) should the caller be aware whether they call a function or apply a lambda expression?

Answer 1

Regular Ruby method calls use () not curly braces which are for blocks. If you don't like [] for calling a lambda, you can always use the call method.

Example:

>> by_two = lambda { |x| x * 2 } #=> #<Proc:0x0000000101304588@(irb):1>
>> by_two[5] #=> 10
>> by_two.call(5) #=> 10

Edit

In newer version of Ruby also:

>> by_two.(5) #=> 10

As to why you can't just do by_two(5), when Ruby sees a bareword it first tries to resolve it as a local variable and if that fails as a method.

Answer 2

Because in Ruby, methods are not lambdas (like, for example, in JavaScript).

Methods always belong to objects, can be inherited (by sub-classing or mixins), can be overwritten in an object's eigenclass and can be given a block (which is a lambda). They have their own scope for variables. Example method definition:

a = :some_variable
def some_method
  # do something, but not possible to access local variable a
end

# call with:
some_method

However lambdas/procs are plain closures, maybe stored in a variable - nothing else:

a = :some_variable
some_lambda = lambda{
  # do something, access local variable a if you want to
}

# call with:
some_lambda[]

Ruby combines both approaches with a powerful syntax, for example, passing blocks:

def some_method_with_block(a)
  # do something, call given block (which is a lambda) with:
  yield(a) ? 42 : 21
end

# example call:
some_method_with_block(1) do |x|
  x.odd?
end #=> 42

Answer 3

If you want brackets, you can do

by_two = lambda { |x| x * 2 }
by_two.(5) # => 10

Note the . between by_two and (5).