regression: acceptable foreign result types

[dmwit]

The following short file is rejected:

import Foreign.Ptr
foreign import ccall "wrapper" foo :: IO (FunPtr ())

The error is:

test.hs:2:1:
    Unacceptable type in foreign declaration: IO (FunPtr ())
    When checking declaration:
      foreign import ccall safe "wrapper" foo :: IO (FunPtr ())

However, my reading of the 2010 Report suggests this should be acceptable. Specifically:

* Prelude.IO t is a marshallable foreign result type when t is a marshallable foreign type,
* all basic foreign types are marshallable foreign types, and
* FunPtr? a is a basic foreign type for all a.

(Political note: I include this chain of reasoning not because I think others too stupid to recreate it, but because I think it likely that I am not reading the Report correctly, and want to make it easy to detect and correct any misconceptions I have.)

[igloo]

The problem is:

Dynamic wrapper.
    The type of a wrapper stub has to be of the form ft -> IO (FunPtr ft), where ft may be any foreign type.

e.g. this is accepted:

import Foreign.Ptr
foreign import ccall "wrapper" foo :: () -> IO (FunPtr ())

I'll leave the ticket open, though, as I think we should give the expected pattern in the error message.

[simonpj]

See also #5610.