如何为我的Task实现实现Promise.all的等效功能？

时间：2019-03-16 17:05:52

标签： javascript functional-programming continuations continuation-passing

这是我的Task实现（即一种Promise，但符合monad法律且可取消）。它的工作原理坚如磐石：

const Task = k =>
  ({runTask: (res, rej) => k(res, rej)});

const tAp = tf => tk =>
  Task((res, rej) => tf.runTask(f => tk.runTask(x => res(f(x)), rej), rej));

const tOf = x => Task((res, rej) => res(x));

const tMap = f => tk =>
  Task((res, rej) => tk.runTask(x => res(f(x)), rej));

const tChain = fm => mx =>
  Task((res, rej) => mx.runTask(x => fm(x).runTask(res, rej), rej));

const log = x => console.log(x);
const elog = e => console.error(e);

const fetchName = (id, cb) => {
  const r = setTimeout(id_ => {
    const m = new Map([[1, "Beau"], [2, "Dev"], [3, "Liz"]]);

    if (m.has(id_))
      return cb(null, m.get(id_));

    else
      return cb("unknown id", null);
  }, 0, id);

  return () => clearTimeout(r);
};

const fetchNameAsync = id =>
  Task((res, rej) =>
    fetchName(id, (err, data) =>
      err === null
        ? res(data)
        : rej(err)));

const a = tAp(tMap(x => y => x.length + y.length)
  (fetchNameAsync(1)))
    (fetchNameAsync(3));

const b = tAp(tMap(x => y => x.length + y.length)
  (fetchNameAsync(1)))
    (fetchNameAsync(5));

a.runTask(log, elog); // 7
b.runTask(log, elog); // Error: "unknown id"

但是，我不知道如何实现awaitAll，它应该具有以下特征：

  • 它要么以单个Tasks的一系列结果来解析
  • 或者它立即拒绝并出现第一个错误，并取消所有其他Tasks
  • 它以“并行”方式执行Tasks 
const awaitAll = ms =>
  Task((res, rej) => ms.map(mx => mx.runTask(...?)));

任何提示都值得赞赏！

3 个答案:

答案 0 :(得分：3)

这是从其他答案以及相关的民间故事/任务中汲取灵感的另一种方法。与其实现一个复杂的tAll来处理迭代和组合任务的列表，我们将关注点分离为单个功能。

这是简化的tAnd-

const tAnd = (t1, t2) =>
{ const acc = []

  const guard = (res, i) => x =>
    ( acc[i] = x
    , acc[0] !== undefined && acc[1] !== undefined
        ? res (acc)
        : void 0
    )

  return Task
    ( (res, rej) =>
        ( t1 .runTask (guard (res, 0), rej) // rej could be called twice!
        , t2 .runTask (guard (res, 1), rej) // we'll fix this below
        )
    )
}

它是这样的-

tAnd
  ( delay (2000, 'a')
  , delay (500, 'b')
  )
  .runTask (console.log, console.error)

// ~2 seconds later
// [ 'a', 'b' ]

现在tAll轻而易举地实现-

const tAll = (t, ...ts) =>
  t === undefined
    ? tOf ([])
    : tAnd (t, tAll (...ts))

哦，别忘了沿途变平-

const tAll = (t, ...ts) =>
  t === undefined
    ? tOf ([])
    : tMap
        ( ([ x, xs ]) => [ x, ...xs ]
        , tAnd (t, tAll(...ts))
        )

它是这样的-

tAll
  ( delay (2000, 'a')
  , delay (500, 'b')
  , delay (900, 'c')
  , delay (1500, 'd')
  , delay (1800, 'e')
  , delay (300, 'f')
  , delay (2000, 'g')
  )
  .runTask (console.log, console.error)

// ~2 seconds later
// [ 'a', 'b', 'c', 'd', 'e', 'f', 'g' ]

tAll也可以正确处理错误-

tAll
  ( delay (100, 'test failed')
  , Task ((_, rej) => rej ('test passed'))
  )
  .runTask (console.log, console.error)

// test passed

正确设置tAnd十分困难，尽管与原始tAll相比，我们已经限制了程序的范围。合并后的任务只能解决一次，或只能拒绝一次-不能同时解决。这意味着也应避免双重决定/拒绝。强制执行这些约束需要更多代码-

const tAnd = (t1, t2) =>
{ let resolved = false
  let rejected = false

  const result = []

  const pending = ([ a, b ] = result) =>
    a === undefined || b === undefined

  const guard = (res, rej, i) =>
    [ x =>
        ( result[i] = x
        , resolved || rejected || pending ()
            ? void 0
            : ( resolved = true
              , res (result)
              )
        )
    , e =>
        resolved || rejected
          ? void 0
          : ( rejected = true
            , rej (e)
            )
    ]

  return Task
    ( (res, rej) =>
        ( t1 .runTask (...guard (res, rej, 0))
        , t2 .runTask (...guard (res, rej, 1))
        )
    )
}

展开以下代码段，以在您自己的浏览器中验证结果-

const Task = k =>
  ({ runTask: (res, rej) => k (res, rej) })

const tOf = v =>
  Task ((res, _) => res (v))

const tMap = (f, t) =>
  Task
    ( (res, rej) =>
        t.runTask
          ( x => res (f (x)) 
          , rej
          )
    )

const tAnd = (t1, t2) =>
{ let resolved = false
  let rejected = false
  
  const result = []

  const pending = ([ a, b ] = result) =>
    a === undefined || b === undefined

  const guard = (res, rej, i) =>
    [ x =>
        ( result[i] = x
        , resolved || rejected || pending ()
            ? void 0
            : ( resolved = true
              , res (result)
              )
        )
    , e =>
        resolved || rejected
          ? void 0
          : ( rejected = true
            , rej (e)
            )
    ]

  return Task
    ( (res, rej) =>
        ( t1 .runTask (...guard (res, rej, 0))
        , t2 .runTask (...guard (res, rej, 1))
        )
    )
}

const tAll = (t, ...ts) =>
  t === undefined
    ? tOf ([])
    : tMap
        ( ([ x, xs ]) => [ x, ...xs ]
        , tAnd (t, tAll (...ts))
        )

const delay = (ms, x) =>
  Task (r => setTimeout (r, ms, x))

tAnd
  ( delay (2000, 'a')
  , delay (500, 'b')
  )
  .runTask (console.log, console.error)

tAll
  ( delay (2000, 'a')
  , delay (500, 'b')
  , delay (900, 'c')
  , delay (1500, 'd')
  , delay (1800, 'e')
  , delay (300, 'f')
  , delay (2000, 'g')
  )
  .runTask (console.log, console.error)

// ~2 seconds later
// [ 'a', 'b' ]
// [ 'a', 'b', 'c', 'd', 'e', 'f', 'g' ]

tAll
  ( delay (100, 'test failed')
  , Task ((_, rej) => rej ('test passed'))
  )
  .runTask (console.log, console.error)

// Error: test passed

串行处理

最棘手的位在并行处理要求中。如果要求要求提供 serial 行为，则实现起来会非常容易-

const tAnd = (t1, t2) =>
  Task
    ( (res, rej) =>
        t1 .runTask
          ( a =>
              t2 .runTask
                ( b =>
                    res ([ a, b ])
                , rej
                )
          , rej
          )
    )

tAll的实施方式保持不变。请注意现在延迟的差异，因为任务现在按顺序运行-

tAnd
  ( delay (2000, 'a')
  , delay (500, 'b')
  )
  .runTask (console.log, console.error)

// ~2.5 seconds later
// [ 'a', 'b' ]

还有许多tAll的任务-

tAll
  ( delay (2000, 'a')
  , delay (500, 'b')
  , delay (900, 'c')
  , delay (1500, 'd')
  , delay (1800, 'e')
  , delay (300, 'f')
  , delay (2000, 'g')
  )
  .runTask (console.log, console.error)

// ~ 9 seconds later
// [ 'a', 'b', 'c', 'd', 'e', 'f', 'g' ]

展开以下代码段，以在您自己的浏览器中验证结果-

const Task = k =>
  ({ runTask: (res, rej) => k (res, rej) })

const tOf = v =>
  Task ((res, _) => res (v))

const tMap = (f, t) =>
  Task
    ( (res, rej) =>
        t.runTask
          ( x => res (f (x)) 
          , rej
          )
    )

const tAnd = (t1, t2) =>
  Task
    ( (res, rej) =>
        t1 .runTask
          ( a =>
              t2 .runTask
                ( b =>
                    res ([ a, b ])
                , rej
                )
          , rej
          )
    )

const tAll = (t, ...ts) =>
  t === undefined
    ? tOf ([])
    : tMap
        ( ([ x, xs ]) => [ x, ...xs ]
        , tAnd (t, tAll (...ts))
        )

const delay = (ms, x) =>
  Task (r => setTimeout (r, ms, x))

tAnd
  ( delay (2000, 'a')
  , delay (500, 'b')
  )
  .runTask (console.log, console.error)

// ~2.5 seconds later
// [ 'a', 'b' ]

tAll
  ( delay (2000, 'a')
  , delay (500, 'b')
  , delay (900, 'c')
  , delay (1500, 'd')
  , delay (1800, 'e')
  , delay (300, 'f')
  , delay (2000, 'g')
  )
  .runTask (console.log, console.error)

// ~ 9 seconds later
// [ 'a', 'b', 'c', 'd', 'e', 'f', 'g' ]

tAll
  ( delay (100, 'test failed')
  , Task ((_, rej) => rej ('test passed'))
  )
  .runTask (console.log, console.error)

// Error: test passed

如何实施tOr和tRace

为完整起见，这里是tOr。请注意，这里的tOr相当于民间故事的Task.concat-

const tOr = (t1, t2) =>
{ let resolved = false
  let rejected = false

  const guard = (res, rej) =>
    [ x =>
        resolved || rejected
          ? void 0
          : ( resolved = true
            , res (x)
            )
    , e =>
        resolved || rejected
          ? void 0
          : ( rejected = true
            , rej (e)
            )
    ]

  return Task
    ( (res, rej) =>
        ( t1 .runTask (...guard (res, rej))
        , t2 .runTask (...guard (res, rej))
        )
    )
}

解决或拒绝两项任务中最先完成的任务-

tOr
  ( delay (2000, 'a')
  , delay (500, 'b')
  )
  .runTask (console.log, console.error)

// ~500 ms later
// 'b'

还有tRace-

const tRace = (t = tOf (undefined), ...ts) =>
  ts .reduce (tOr, t)

解决或拒绝许多任务中第一个完成的任务-

tRace
  ( delay (2000, 'a')
  , delay (500, 'b')
  , delay (900, 'c')
  , delay (1500, 'd')
  , delay (1800, 'e')
  , delay (300, 'f')
  , delay (2000, 'g')
  )
  .runTask (console.log, console.error)

// ~300 ms later
// 'f'

展开以下代码段，以在您自己的浏览器中验证结果-

const Task = k =>
  ({ runTask: (a, b) => k (a, b) })

const tOr = (t1, t2) =>
{ let resolved = false
  let rejected = false

  const guard = (res, rej) =>
    [ x =>
        resolved || rejected
          ? void 0
          : ( resolved = true
            , res (x)
            )
    , e =>
        resolved || rejected
          ? void 0
          : ( rejected = true
            , rej (e)
            )
    ]

  return Task
    ( (res, rej) =>
        ( t1 .runTask (...guard (res, rej))
        , t2 .runTask (...guard (res, rej))
        )
    )
}

const tRace = (t = tOf (undefined), ...ts) =>
  ts. reduce (tOr, t)

const delay = (ms, x) =>
  Task (r => setTimeout (r, ms, x))

tOr
  ( delay (2000, 'a')
  , delay (500, 'b')
  )
  .runTask (console.log, console.error)

// ~500 ms later
// 'b' 

tRace
  ( delay (2000, 'a')
  , delay (500, 'b')
  , delay (900, 'c')
  , delay (1500, 'd')
  , delay (1800, 'e')
  , delay (300, 'f')
  , delay (2000, 'g')
  )
  .runTask (console.log, console.error)

// ~300 ms later
// note `f` appears in the output first because this tRace demo finishes before the tOr demo above
// 'f'

tRace
  ( delay (100, 'test failed')
  , Task ((_, rej) => rej ('test passed'))
  )
  .runTask (console.log, console.error)

// Error: test passed

如何实施tAp

在评论中，我们谈论的是适用性tAp。我认为tAll使实现起来相当容易-

const tAp = (f, ...ts) =>
  tMap
    ( ([ f, ...xs ]) => f (...xs)
    , tAll (f, ...ts)
    )

tAp接受一个任务包装的函数和任意数量的任务包装的值，并返回一个新任务-

const sum = (v, ...vs) =>
  vs.length === 0
    ? v
    : v + sum (...vs)

tAp
  ( delay (2000, sum)
  , delay (500, 1)
  , delay (900, 2)
  , delay (1500, 3)
  , delay (1800, 4)
  , delay (300, 5)
  )
  .runTask (console.log, console.error)

// ~2 seconds later
// 15

除非任务有副作用，否则我无法看到tAp的“并行”实现违反适用法律的原因。

展开以下代码段，以在您自己的浏览器中验证结果-

const Task = k =>
  ({ runTask: (res, rej) => k (res, rej) })

const tOf = v =>
  Task ((res, _) => res (v))

const tMap = (f, t) =>
  Task
    ( (res, rej) =>
        t.runTask
          ( x => res (f (x)) 
          , rej
          )
    )

const tAp = (f, ...ts) =>
  tMap
    ( ([ f, ...xs ]) => f (...xs)
    , tAll (f, ...ts)
    )

const tAnd = (t1, t2) =>
{ let resolved = false
  let rejected = false
  
  const result = []

  const pending = ([ a, b ] = result) =>
    a === undefined || b === undefined

  const guard = (res, rej, i) =>
    [ x =>
        ( result[i] = x
        , resolved || rejected || pending ()
            ? void 0
            : ( resolved = true
              , res (result)
              )
        )
    , e =>
        resolved || rejected
          ? void 0
          : ( rejected = true
            , rej (e)
            )
    ]

  return Task
    ( (res, rej) =>
        ( t1 .runTask (...guard (res, rej, 0))
        , t2 .runTask (...guard (res, rej, 1))
        )
    )
}

const tAll = (t, ...ts) =>
  t === undefined
    ? tOf ([])
    : tMap
        ( ([ x, xs ]) => [ x, ...xs ]
        , tAnd (t, tAll (...ts))
        )

const delay = (ms, x) =>
  Task (r => setTimeout (r, ms, x))

const sum = (v, ...vs) =>
  vs.length === 0
    ? v
    : v + sum (...vs)

tAp
  ( delay (2000, sum)
  , delay (500, 1)
  , delay (900, 2)
  , delay (1500, 3)
  , delay (1800, 4)
  , delay (300, 5)
  )
  .runTask (console.log, console.error)

// ~2 seconds later
// 15

答案 1 :(得分：2)

这是使用计数器和包装在另一个Task中的循环的一种可能方法。使用计数器是因为任务可以按任何顺序完成，否则很难知道外部Task何时可以最终解决-

const assign = (o = {}, [ k, v ]) =>
  Object .assign (o, { [k]: v })

const tAll = (ts = []) =>
{ let resolved = 0
  const acc = []
  const run = (res, rej) =>
  { for (const [ i, t ] of ts .entries ())
      t .runTask
        ( x =>
            ++resolved === ts.length
              ? res (assign (acc, [ i, x ]))
              : assign (acc, [ i, x ])
        , rej
        )
  }
  return Task (run)
}

我们编写了一个简单的delay函数进行测试-

const delay = (ms, x) =>
  Task ((res, _) => setTimeout (res, ms, x))

const tasks =
  [ delay (200, 'a')
  , delay (300, 'b')
  , delay (100, 'c')
  ]

tAll (tasks) .runTask (console.log, console.error)
// ~300 ms later
// => [ 'a', 'b', 'c' ]

如果 any 任务失败，则外部任务将被拒绝-

const tasks =
  [ delay (200, 'a')
  , delay (300, 'b')
  , Task ((_, rej) => rej (Error('bad')))
  ]

tAll (tasks) .runTask (console.log, console.error)
// => Error: bad

展开以下代码段，以在您自己的浏览器中验证结果-

const assign = (o = {}, [ k, v ]) =>
  Object .assign (o, { [k]: v })

const Task = k =>
  ({runTask: (res, rej) => k(res, rej)});

const tAll = (ts = []) =>
{ let resolved = 0
  const acc = []
  const run = (res, rej) =>
  { for (const [ i, t ] of ts .entries ())
      t .runTask
        ( x =>
            ++resolved === ts.length
              ? res (assign (acc, [ i, x ]))
              : assign (acc, [ i, x ])
        , rej
        )
  }
  return Task (run)
}

const delay = (ms, x) =>
  Task ((res, _) => setTimeout (res, ms, x))

const tasks =
  [ delay (200, 'a')
  , delay (300, 'b')
  , delay (100, 'c')
  ]

tAll (tasks) .runTask (console.log, console.error)
// ~300 ms later
// => [ 'a', 'b', 'c' ]

这是tAll的另一种实现，它用for换成forEach，并删除了另一个命令式块{ ... }-

const tAll = (ts = []) =>
{ let resolved = 0
  const acc = []
  const run = (res, rej) => (t, i) =>
    t .runTask
      ( x =>
          ++resolved === ts.length
            ? res (assign (acc, [ i, x ]))
            : assign (acc, [ i, x ])
      , rej
      )
  return Task ((res, rej) => ts .forEach (run (res, rej)))
}

答案 2 :(得分：2)

另一种使用递归和2 Task基本案例的解决方案，然后允许仅使用两个变量来管理状态：

  const tAll = ([first, second, ...rest]) =>
   !second
     ? first
     : rest.length 
        ? tMap(
            results => results.flat()
          )(tAll([ tAll([first, second]), tAll(rest) ]))
        : Task((res, rej, a, b, done) => (
            first.runTask(
               value => !done && b ? (res([value, b.value]), done = true) : (a = { value }),
               err => !done && (rej(err), done = true)
            ),
            second.runTask(
               value => !done && a ? (res([a.value, value]), done = true) : (b = { value }),
              err => !done && (rej(err), done = true)
            ) 
         ));
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