That holds true for the given ruby implementation as well, since n is incremented only in the scope of foo and not in the calling scope.

def foo (n)

    lambda {|i| n += i } 

end

n=5

a = foo(n)

puts a.call(3) #prints 8

puts n # still prints 5. n is not incremented

I agree that the python one is strictly not according to the problem definition, but for all practical purposes both python equivalents are the same, aren't they? or am I missing something?