java - Frege 中与 Java 的互操作性，尤其是在 IO Monad 中

Question

我想将这段 java 代码翻译成 Frege Haskell:

    PApplet pApplet = new PApplet();
    System.out.print(pApplet.toString());
    PApplet.runSketch(new String[]{"test"}, pApplet);

到目前为止我做了:

data PApplet = mutable native processing.core.PApplet
  where
    native new :: () -> IO PApplet
    native toString :: PApplet -> IO String

native runSketch processing.core.PApplet.runSketch
  :: ArrayOf RealWorld String -> PApplet -> IO ()

main _ = do p <- PApplet.new
            pStr <- p.toString
            putStrLn pStr
            args = JArray.fromList ["test"]
            runSketch args p

部分高达 main编译但随后出现这些错误:

E Process.fr:14: type error in expression fromList ("test":[])
    type is : STMutable t1 (JArray String)
    expected: ArrayOf RealWorld String
E Process.fr:15: type error in expression p
    type is : IO PApplet
    expected: PApplet
E Process.fr:12: type error in expression >>= p.toString (λpStr -> >> (putStrLn pStr) (runSketch (fromList ("test":[])) p))
    type is : IO ()
    expected: ()→t1
E Process.fr:11: type error in expression λp -> >>= p.toString (λpStr -> >> (putStrLn pStr) (runSketch (fromList ("test":[])) p))
    type is : IO ()
    expected: ()→t1
E Process.fr:11: type error in expression >>= new (λp -> >>= p.toString (λpStr -> >> (putStrLn pStr) (runSketch (fromList ("test":[])) p)))
    type is : ()→t1
    expected: IO ()
E Process.fr:11: type error in expression λ_ -> >>= new (λp -> >>= p.toString (λpStr -> >> (putStrLn pStr) (runSketch (fromList ("test":[])) p)))
    type is : ()→t1
    expected: IO ()
E Process.fr:12: can't find a type for  p.toString `toString`
    is neither an overloaded function nor a member of  IO PApplet

我正在努力满足编译器标准，但没有成功。经过无数次随机组合后，上面的这段代码对我来说似乎是最合理的。我需要在 do 中输入提示吗？堵塞？我不明白为什么 p <- PApplet.new评估为 IO PApplet ？以及如何制作JArray.fromList返回 ArrayOf RealWorld String ？弗雷格很棒，但互操作性相当令人生畏。是否有可能在 Frege github 上有更多关于它的例子？

Answer 1

你有

ST s X

你想要

X

你在 IO , 这只不过是 ST RealWorld

因此，最自然的解决方案是替换 =与 <-在行中

args = JArray.fromList ["test"]

一切就绪!

诚然，由于类型别名，整个故事有点困难:

type ArrayOf a x = Mutable a (JArray x)
type STMutable s a = ST s (Mutable s a)

是否选择了去锯齿进行翻译

ST s (Mutable s (JArray String))

回到

ST s (ArrayOf s String)

你可能已经看到了。