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Haskell uses type inference, but that's not the same as implicit coercion, which Haskell explicitly avoids. That is: if you type 4 + 4.2 Then the compiler wi
by rawatson 15y ago
Haskell uses type inference, but that's not the same as implicit coercion, which Haskell explicitly avoids. That is: if you type
4 + 4.2
Then the compiler will infer that you mean
(4::Fractional a=>a) + (4.2::Fractional a => a)
However, you cannot add an integer and a float:
(4::Int) + (4.2::Float)
Couldn't match expected type `Int' with actual type `Float'
In the second argument of `(+)', namely `(4.2 :: Float)'
In the expression: (4 :: Int) + (4.2 :: Float)
In an equation for `it': it = (4 :: Int) + (4.2 :: Float)
This follows from the type of (+) :: Num a => a -> a -> a
- derleth 15y agoI guess I still don't get how there could be a difference between Float and Fractional, then.
- rawatson 15y agoHaskell uses type classes to support ad hoc polymorphism, or overloading. Consider the statement: (4.2::Fractional a=> a) + (4::Float) Float is an instance of the type class Fractional. That is: methods which are defined for all fractional types must be defined for floats. The compiler infers that (4.2::Fractional a=>a) must have type float, as it is being added to a float. This is compatible with the original type of the expression, as Float is an instance of fractional, so it is valid to read 4.2 as a Float. This diagram might help: http://www.haskell.org/onlinereport/basic.html#sect6.3 http://www.haskell.org/onlinereport/basic.html#sect6.3