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「関数型プログラミングの基礎」サンプルコード

License: MIT License

JavaScript 97.13% CoffeeScript 0.05% Ruby 0.36% Scala 1.28% Haskell 0.20% Shell 0.07% HTML 0.18% Dockerfile 0.74%

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Wrong flatMap definition? (without world version)

You define IO.flatMap as below. (https://github.com/akimichi/functionaljs/blob/master/bin/io_without_world.js#L48-L53)

flatMap: (instanceA) => {
  var self = this;
  return (actionAB) => { // actionAB:: a -> IO b
    return self.unit(self.run(actionAB(self.run(instanceA))));
  };
}

But this definition evaluates self.run(actionAB(self.run(instanceA)) eagerly.
For example, the code below displays hello monad before executing IO.run(e).

// putStr :: String => IO[Unit]
const putStr = (s) => IO.flatMap(IO.unit(s))((s) => {
  console.log(s);
  return IO.unit();
});
const e = putStr('hello monad'); // display 'hello monad' at this point
IO.run(e); // shoud be displayed at this point

So I think the definition should be as below.

flatMap: (instanceA) => {
  const self = this;
  return (actionAB) => { // actionAB:: a -> IO b
    return () => self.run(actionAB(self.run(instanceA)))
  };
}

https://github.com/akimichi/functionaljs/blob/master/bin/io_without_world.js#L48-L53
にて、IO.flatMapは以下のように定義されています。

flatMap: (instanceA) => {
  var self = this;
  return (actionAB) => { // actionAB:: a -> IO b
    return self.unit(self.run(actionAB(self.run(instanceA))));
  };
}

しかし、この定義ではself.run(actionAB(self.run(instanceA))を正格評価してしまうと思います。
(正格評価という表現が適切か分かりませんが、IO.flatMap(instanceA)(actionAB)の時点で評価されます。)
例えば、以下のコードではIO.run(e)の実行前にhello monadが表示されます。

// putStr :: String => IO[Unit]
const putStr = (s) => IO.flatMap(IO.unit(s))((s) => {
  console.log(s);
  return IO.unit();
});
const e = putStr('hello monad'); // この時点で'hello monad'が表示されます
IO.run(e); // 実際には、この時点で表示されるべきかと思います

そのため、flatMapの定義としては以下のほうが適切かと思いますが如何でしょうか。

flatMap: (instanceA) => {
  const self = this;
  return (actionAB) => { // actionAB:: a -> IO b
    return () => self.run(actionAB(self.run(instanceA)))
  };
}

よろしくお願いします。

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