Lexical scope in Ruby

Ruby
1

From the “Progamming Languages” Coursera course
(https://class.coursera.org/proglang-003), this is a characteristic of a
lexically scoped language:
val x = 1
fun f y = x + y
val x = 2
val y = 3
val z = f (x+y)
z should result in 6

However, when I run it in Ruby 2.1.1 irb, it produces 7. What am I
missing?

2.1.1 :001 > x = 1
=> 1
2.1.1 :002 > b = Proc.new {|y| x + y}
=> #Proc:0x000001030a3aa8@:2(irb)
2.1.1 :003 > b.call(3)
=> 4
2.1.1 :004 > x = 2
=> 2
2.1.1 :005 > y = 3
=> 3
2.1.1 :006 > z = b.call(x+y)
=> 7

2

2.1.1 :001 > x = 1
=> 1
2.1.1 :002 > b = Proc.new {|y| x + y}
=> #Proc:0x000001030a3aa8@:2(irb)
2.1.1 :003 > b.call(3)
=> 4
2.1.1 :004 > x = 2 # you override x, now is 2
=> 2
2.1.1 :005 > y = 3 # this doesn’t matter, has no point
=> 3
2.1.1 :006 > z = b.call(x+y) # you pass 5 here, 2 + 5 = 7
=> 7

What happens is that if you use the same name for a local variable, as
an argument name inside #Proc b, inside the #Proc the argument take
relevance over the local variable outside its scope, “inner” scopes put
relevance to things created inside it first, for example: #Proc first
looks if there’s an argument name called x, if not looks outside the
scope, and so on. As y was defined inside the scope of the proc that’s,
no more searching.

3

Block arguments “shadow” variables defined in an outer scope and are
local to that block (at least since Ruby 1.9). Other refer to enclosing
scope, so x inside b = Proc.new {|y| x + y} can be changed outside
the block.

You are probably looking for a (lexical) closure, which “saves”
variables at the point of definition of the closure.
You may “catch” an environment generally by a method/proc/lambda
returning your proc with x applied or use “currying” ie. partial proc
application:

x = 1

now we define a closure using a method

def get_closure(x_cl); proc {|y| x_cl + y}; end

and assign returned proc to b1

b1 = get_closure(x)

another way with currying

b2 = proc {|x_cl, y| x_cl + y}.curry[x]
p b1[3] # 4
p b2[3] # 4
x = 2; y = 3
p b1[x+y] # 6
p b2[x+y] # 6