Haskell FFI：顶级 FunPtr 到顶级函数？
似乎最好只为顶级函数创建一次 FunPtr，而不是在需要时创建一个新函数（针对同一函数）并处理其释放。
我是否忽略了除 foreign import ccall "wrapper" 之外的获取 FunPtr 的其他方法？如果没有，我的解决方法将如下面的代码所示。这样安全吗？
type SomeCallback = CInt -> IO ()
foreign import ccall "wrapper" mkSomeCallback :: SomeCallback -> IO (FunPtr SomeCallback)
f :: SomeCallback
f i = putStrLn ("It is: "++show i)
{-# NOINLINE f_FunPtr #-}
f_FunPtr :: FunPtr SomeCallback
f_FunPtr = unsafePerformIO (mkSomeCallback f)
编辑：已验证“每次都创建一个新的”变体（main =forever (mkSomeCallback f)）实际上会泄漏内存，如果不freeHaskellFunPtr它的话。
It seems desirable to create a FunPtr to a top-level function just once instead of creating a new one (to the same function) whenever it's needed and dealing with its deallocation.
Am I overlooking some way to obtain the FunPtr other than foreign import ccall "wrapper"? If not, my workaround would be as in the code below. Is that safe?
type SomeCallback = CInt -> IO ()
foreign import ccall "wrapper" mkSomeCallback :: SomeCallback -> IO (FunPtr SomeCallback)
f :: SomeCallback
f i = putStrLn ("It is: "++show i)
{-# NOINLINE f_FunPtr #-}
f_FunPtr :: FunPtr SomeCallback
f_FunPtr = unsafePerformIO (mkSomeCallback f)
Edit: Verified that the "creating a new one every time" variant (main = forever (mkSomeCallback f)) does in fact leak memory if one doesn't freeHaskellFunPtr it.
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原则上，这应该是安全的 - GHC 内部代码使用类似的模式来初始化单例，例如 IO 监视句柄队列。请记住，您无法控制 mkSomeCallback 何时运行，并且不要忘记
NOINLINE。This should, in principle, be safe - GHC internal code uses a similar pattern to initialize singletons such as the IO watched-handles queues. Just keep in mind that you have no control over when mkSomeCallback runs, and don't forget the
NOINLINE.