如何以函数方式有条件地分配变量？

Question

我需要根据条件为变量分配一个值。我想在考虑功能性编程范例的情况下进行此操作，因此无法在外部范围中对其进行声明然后重新分配。

// let foo = undefined // no-let!
if(condition) {
  const foo = 1
} else {
  const foo = 0
}
do_some_stuff(foo) // Uncaught ReferenceError: foo is not defined

我知道，我可以在这里使用三元表达式，例如
const foo = condition ? 1 : 0
但是，如果我在自己的条件内还有其他例程可以做，例如：

if(condition) {
  const foo = 1
  do_stuff()
} else {
  const foo = 0
  do_other_stuff()
}
do_third_stuff(foo)

Answer 1

没有什么可以阻止您分裂两者：

const foo = condition ? 1 : 0;

if(condition) {
    do_stuff();
} else {
    do_other_stuff();
}

do_third_stuff(foo);

如果condition是昂贵的执行，只需在多次使用之前将其分配给变量：

let isFoo =  expensiveIsFooMethod();

const foo = isFoo ? 1 : 0;

if(isFoo) {
    do_stuff();
} else {
    do_other_stuff();
}

do_third_stuff(foo);

您是对的，如果您不必重复该条件，它会更干净，但是您已经引入了此限制，因为您使用的是const变量，因此无法赋值到您的const的多个位置。

我建议在这里胜过这两个选择。对您来说更重要的是：更简洁的语法，还是确保您永远不会覆盖该值？

Answer 2

可能您会使用Either之类的代数数据类型（ADT）对案件进行编码。也就是说，您可能会涉及两个子情况： left 和 right 。

从// --> Solution starts here开始查看代码。先前的代码是使用香草JavaScript来使代码可运行的微型标准FP库。检查一下，享受吧！

// Mini standard library
// -------------------------------

// The identity combinator
// I :: a -> a
const I = x => x

// Either ADT
const Either = (() => {
   // Creates an instance of Either.Right
   //
   // of :: b -> Either a b
   const of = x => ({ right: x })
   
   // Creates an instance of Either.Right
   //
   // Right :: b -> Either a b
   const Right = of
   
   // Creates an instance of Either.Left
   //
   // Left :: a -> Either a b
   const Left = x => ({ left: x })
   
   // Maps Either.Left or Either.Right in a single operation
   //
   // bimap :: (a -> c) -> (b -> d) -> Either a b -> Either c -> d
   const bimap = f => g => ({ left, right }) => left ? Left (f (left)) : Right (g (right))
   
   // Lifts a value to Either based on a condition, where false 
   // results in Left, and true is Right.
   //
   // tagBy :: (a -> Boolean) -> a -> Either a a
   const tagBy = f => x => f (x) ? Right (x) : Left (x)
   
   // Unwraps Either.Left or Either.Right with mapping functions
   //
   // either :: (a -> c) -> (b -> c) -> Either a b -> c
   const either = f => g => ({ left, right }) => left ? f (left) : g (right)
   
   // Unwraps Either.Left or Either.Right and outputs the raw value on them
   //
   // unwrap :: Either a b -> c
   const unwrap = either (I) (I)
   
   return { of, Right, Left, bimap, tagBy, either, unwrap }
}) ()

// --> Solution starts here

// Lifts to Either.Right if x is greater than 3, 
// otherwise, x is encoded as Left.
//
// tagGt3 :: Number -> Either Number Number
const tagGt3 = Either.tagBy (x => x > 3)

// doStuff :: Number -> Number
const doStuff = x => x + 1

// doStuff2 :: Number -> Number
const doStuff2 = x => x * 4

// doStuff3 :: Either Number Number -> Either Number Number
const doStuff3 = Either.bimap (doStuff) (doStuff2) // <-- here's the decision!

const eitherValue1 = doStuff3 (tagGt3 (2))
const eitherValue2 = doStuff3 (tagGt3 (30))


const output1 = Either.unwrap (eitherValue1)
const output2 = Either.unwrap (eitherValue2)

console.log ('output1: ', output1)
console.log ('output2: ', output2)

使用管道进行重构

现在，我介绍pipe来粘合一个或多个一元函数的组合，这使代码更加优雅：

// Mini standard library
// -------------------------------

// The identity combinator
// I :: a -> a
const I = x => x

// Pipes many unary functions
//
// pipe :: [a -> b] -> a -> c
const pipe = xs => x => xs.reduce ((o, f) => f (o), x)

// Either ADT
const Either = (() => {
   // Creates an instance of Either.Right
   //
   // of :: b -> Either a b
   const of = x => ({ right: x })
   
   // Creates an instance of Either.Right
   //
   // Right :: b -> Either a b
   const Right = of
   
   // Creates an instance of Either.Left
   //
   // Left :: a -> Either a b
   const Left = x => ({ left: x })
   
   // Maps Either.Left or Either.Right in a single operation
   //
   // bimap :: (a -> c) -> (b -> d) -> Either a b -> Either c -> d
   const bimap = f => g => ({ left, right }) => left ? Left (f (left)) : Right (g (right))
   
   // Lifts a value to Either based on a condition, where false 
   // results in Left, and true is Right.
   //
   // tagBy :: (a -> Boolean) -> a -> Either a a
   const tagBy = f => x => f (x) ? Right (x) : Left (x)
   
   // Unwraps Either.Left or Either.Right with mapping functions
   //
   // either :: (a -> c) -> (b -> c) -> Either a b -> c
   const either = f => g => ({ left, right }) => left ? f (left) : g (right)
   
   // Unwraps Either.Left or Either.Right and outputs the raw value on them
   //
   // unwrap :: Either a b -> c
   const unwrap = either (I) (I)
   
   return { of, Right, Left, bimap, tagBy, either, unwrap }
}) ()

// --> Solution starts here

// doStuff :: Number -> Number
const doStuff = x => x + 1

// doStuff2 :: Number -> Number
const doStuff2 = x => x * 4

const { tagBy, bimap, unwrap } = Either

// doStuff3 :: Number -> Number
const doStuff3 = pipe ([
   tagBy (x => x > 3),
   bimap (doStuff) (doStuff2), // <-- here's the decision!
   unwrap
])

const output1 = doStuff3 (2)
const output2 = doStuff3 (30)

console.log ('output1: ', output1)
console.log ('output2: ', output2)

Answer 3

因为您不想在外部声明 foo ，所以为什么不这样简单：

if(condition) {
  const foo = 1
  do_stuff()
  do_third_stuff(foo)
} else {
  const foo = 0
  do_other_stuff()
  do_third_stuff(foo)
}