Haskell函数类型变量

Question

If you have

foo :: a -> a -> Bool

Does this enforce both types "a" are the same?

Answer 1

Yes. You can observe this with a function that otherwise ignores its arguments.

foo :: a -> a -> Bool
foo _ _ = True

Calling it with two arguments of the same type works.

Prelude> foo 1 1
True
Prelude> foo 'x' 'x'
True

Calling it with two arguments of different types produces a type error, the exact error depending on which types you choose.

Prelude> foo 1 'x'

<interactive>:5:5:
    No instance for (Num Char) arising from the literal ‘1’
    In the first argument of ‘foo’, namely ‘1’
    In the expression: foo 1 'x'
    In an equation for ‘it’: it = foo 1 'x'
Prelude> foo 'x' (1::Int)

<interactive>:8:10:
    Couldn't match expected type ‘Char’ with actual type ‘Int’
    In the second argument of ‘foo’, namely ‘(1 :: Int)’
    In the expression: foo 'x' (1 :: Int)
    In an equation for ‘it’: it = foo 'x' (1 :: Int)
Prelude> foo (1::Int) 'x'

<interactive>:9:14:
    Couldn't match expected type ‘Int’ with actual type ‘Char’
    In the second argument of ‘foo’, namely ‘'x'’
    In the expression: foo (1 :: Int) 'x'
    In an equation for ‘it’: it = foo (1 :: Int) 'x'