... because Ruby isn't terse enough yet

Re:... because Ruby isn't terse enough yet by mdemare on Sunday October 28, 2007 @12:19PM (#73454)

Actually it doesn't work - calling the returned lambda should increment n - your implementation doesn't.

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Re:... because Ruby isn't terse enough yet by neelesh on Sunday October 28, 2007 @11:02PM (#73622)

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?

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