def foo(n):
return lambda i:n+i
def foo(n):
n = [n]
def bar(i):
n[0] += i
return n[0]
return bar
def foo (n)
lambda {|i| n += i }
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?