Why can't read deduce the correct type?

Question

In Haskell, I can make Haskell value from a string with read.

Prelude> read \"1\" + 3
4

I can use fst to get t

Answer 1

As read is a polymorphic function, the read "3" + 4 works because the compiler know you want a Num because you applied + to read "3", If the compiler can't figure out what you want, you have to specify type of read, like this:

Prelude> let rd = read :: String->(Int,Int)
Prelude> rd "(1,2)"

Answer 2

First, let's see the type signature of read:

Prelude> :t read
read :: Read a => String -> a

As you can see, its return value must have a type of the Read typeclass. However, if we don't use the value the complier won't know what type it is.

In such cases, we can add :: to specify its type.

Prelude> read "5"
*** Exception: Prelude.read: no parse
Prelude> read "5" :: Int
5

Similarly,

Prelude> read "(1, 2)"
*** Exception: Prelude.read: no parse
Prelude> read "(1, 2)" :: (Int, Int)
(1,2)
Prelude> fst (read "(1,2)" :: (Int, Int))
1

The compiler can deduce the types most of the time, but when it meets read "5", it may get confused about whether the type should be Int or Float or anything else. Only after computing, Haskell can know its type. On the other hand, Haskell has static types, so it has to know all type before compiling.