Prelude> let myprint = putStrLn . show
Prelude> :t myprint
myprint :: () -> IO ()
OK, nothing too unusual here. Just GHCi type defaulting rules, I guess...
Prelude> let myprint = (putStrLn . show) :: Show x => x -> IO ()
Prelude> :t myprint
myprint :: () -> IO ()
What sorcery is this?? You're point-blank ignoring my type declaration?! O_O
Is there some way I can convince GHCi to do what I actually intended?
(:==) is not a valid symbol for a function or variable identifier in Haskell.
Open a command window and navigate to the directory where you want to keep your Haskell source files. Run Haskell by typing ghci or ghci MyFile. hs. (The "i" in "GHCi" stands for "interactive", as opposed to compiling and producing an executable file.)
The version number of your copy of GHC can be found by invoking ghc with the --version flag (see Verbosity options). The compiler version can be tested within compiled code with the MIN_VERSION_GLASGOW_HASKELL CPP macro (defined only when CPP is used).
Adding a type annotation to an expression as in
e :: type
makes the compiler check that e
has that type
, as well as use that type
to drive type variables instantiation and instance selection. However, if the type
is polymorphic it can still be instantiated later on. Consider e.g.
(id :: a -> a) "hello"
Above, a
will be instantiated to String
, despite my annotation. Further,
foo :: Int -> Int
foo = (id :: a -> a)
will make a
to be instantiated to Int
later on. The above id
annotation does not give any information to GHC: it already knows that id
has that type.
We could remove it without affecting the type checking at all. That is, the expressions id
and id :: a->a
are not only dynamically equivalent, but also statically such.
Similarly, the expressions
putStrLn . show
and
(putStrLn . show) :: Show x => x -> IO ()
are statically equivalent: we are just annotating the code with the type GHC can infer. In other words, we are not providing any information to GHC it does not already know.
After the annotation is type checked, GHC can then instantiate x
further. The monomorphism restriction does that in your example. To prevent that, use an annotation for the binding you are introducing, not for the expression:
myprint :: Show x => x -> IO ()
myprint = (putStrLn . show)
We can do the following, with monomorphism restriction on:
>let myprint :: Show x => x -> IO (); myprint = putStrLn . show
>:t myprint
myprint :: Show x => x -> IO ()
This is not the same as let myprint = putStrLn . show :: Show x => x -> IO ()
. In the former case we have a binding with a type signature, in the latter case we a have a let
binding with a type annotation inside the right hand side. Monomorphism checks top-level type signatures, but not local annotations.
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