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) lamb

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?