Convert Python to Haskell / Lambda calculus
Wha开发者_如何学编程t is the Python code in Haskell and Lambda calculus?
def f1():
x = 77
def f2():
print x
f2
f1
My attempt in lambda calculus
\x. 77 (\x.x)
In Haskell:
f1 = f2
where x = 77
f2 = print x
Refactoring, since IO just confuses the issue:
f1 = f2
where x = 77
f2 = x
Refactoring:
f1 = x
where x = 77
Refactoring, since you want the variable?
f1 = (\x -> x) 77
Beta reduce:
f1 = 77
And you have your program in Haskell.
Lambda calculus doesn't have numeric literals (unlike Haskell), so we must use Church numerals. So compiling "77" to lambda calculus:
f1 = \f.\x. f (f (f (f ( ... ) x)))
And there's your Python program in Lambda Calculus.
Haskell:
f1 :: IO ()
f1 = let x = 77
f2 = print x
in f2
main :: IO ()
main = f1
Or to be more like your lambda calculus:
f1 :: Int
f1 = let f2 = x
x = 77
in f2
main :: IO ()
main = print f1
I don't know python, so I could be completely wrong, but this is my Haskell interpretation.
f1 = let x = 77 in show x
Or, since you've got a constant there
f1 = show 77
In lambda calculus:
λprint. print 77
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