第 15 题：实现 add(1)(2)(3) #21

lgwebdream · 2020-06-19T12:34:17Z

欢迎在下方发表您的优质见解

Genzhen · 2020-06-23T04:22:37Z

考点：函数柯里化

函数柯里化概念：柯里化（Currying）是把接受多个参数的函数转变为接受一个单一参数的函数，并且返回接受余下的参数且返回结果的新函数的技术。

1）粗暴版

function add (a) {
	return function (b) {
		return function (c) {
		    return a + b + c;
		}
	}
}
console.log(add(1)(2)(3)); // 6

2）柯里化解决方案

参数长度固定

const curry = (fn) =>
(judge = (...args) =>
    args.length === fn.length
    ? fn(...args)
    : (...arg) => judge(...args, ...arg));
const add = (a, b, c) => a + b + c;
const curryAdd = curry(add);
console.log(curryAdd(1)(2)(3)); // 6
console.log(curryAdd(1, 2)(3)); // 6
console.log(curryAdd(1)(2, 3)); // 6

参数长度不固定

function add (...args) {
    //求和
    return args.reduce((a, b) => a + b)
}

function currying (fn) {
    let args = []
    return function temp (...newArgs) {
        if (newArgs.length) {
            args = [
                ...args,
                ...newArgs
            ]
            return temp
        } else {
            let val = fn.apply(this, args)
            args = [] //保证再次调用时清空
            return val
        }
    }
}

let addCurry = currying(add)
console.log(addCurry(1)(2)(3)(4, 5)())  //15
console.log(addCurry(1)(2)(3, 4, 5)())  //15
console.log(addCurry(1)(2, 3, 4, 5)())  //15

H246802 · 2020-07-15T03:10:03Z

function currying(){
    let args = [...arguments]
    temp.getValue = ()=>{
        return args.reduce((a,b)=> a + b, 0)
    }
    function temp(...arg){
        if(arg.length){
            args = [
                ...args,
                ...arg
            ]
            return temp
        }
    }
    return temp
}

123456zzz · 2020-07-15T17:28:06Z

const add = (a: number, b: number, c: number) => a + b + c;

const adding = (...args: number[]) => args.reduce((pre, cur) => pre + cur, 0);

//参数确定
const curry = (fn: Function) => {
  let args = [];

  return function temp(...newArgs) {
    args.push(...newArgs);
    if (args.length === fn.length) {
      const val = fn.apply(this, args);
      args = [];
      return val;
    } else {
      return temp;
    }
  };
};

//参数不确定
const currying = (fn: Function) => {
  let args = [];

  return function temp(...newArgs) {
    if (newArgs.length) {
      args.push(...newArgs);
      return temp;
    } else {
      const val = fn.apply(this, args);
      args = [];
      return val;
    }
  };
};

const curryAdd = curry(add);
console.log(curryAdd(1)(2)(3)); // 6
console.log(curryAdd(1, 2)(3)); // 6
console.log(curryAdd(1)(2, 3)); // 6

let addCurry = currying(adding);
console.log(addCurry(1)(2)(3)(4, 5)()); //15
console.log(addCurry(1)(2)(3, 4, 5)()); //15
console.log(addCurry(1)(2, 3, 4, 5)()); //15

yaooooooooo · 2020-07-20T01:42:04Z

function curring(fn,arr = []){
  var length = fn.length;
  return (...args) => {
   const currentArr = [...arr.push(args)]
   if(arr.length < length){
        return curring(fn,currentArr];
    } else {
         fn(...currentArr )
    }
  }
}

GolderBrother · 2020-07-20T15:59:19Z

function currying(fn, args = []) {
    return function temp(...innerArgs) {
        if (innerArgs.length > 0) {
            // 收集后面传入的参数
            args = [...args, ...innerArgs];
            // 返回函数供后面可以继续调用
            return temp;
        } else {
            const val = fn.apply(this, args);
            // 清空参数数组，为了保证下次执行函数可以继续迭代
            args = [];
            return val;
        }
    }
}
// 求和函数
const add = (...args) => args.reduce((a, b) => a + b);
let addCurry = currying(add)
console.log(addCurry(1)(2)(3)(4, 5)())  //15
console.log(addCurry(1)(2)(3, 4, 5)())  //15
console.log(addCurry(1)(2, 3, 4, 5)())  //15

JL003 · 2020-07-24T02:49:04Z

除了typeof为function 正常运算都没问题
function add(num) {
const add2 = (num2) => {
num = num + num2;
add2.toString = () => num;
return add2;
}
return add2;
}
add(1)(2)(3) + 5 // 11
add(1)(2)(3)(4) // ƒ 10

ABoyCDog · 2020-08-11T06:07:01Z

有个疑问，就是参数长度不固定时，addCurry(1)(2)(3)(4, 5)()，最后的需要加函数执行的（），但是题目不是没有吗，也就是只要addCurry(1)(2)(3)(4, 5)这样

huzedong2015 · 2020-09-01T03:48:41Z

function curry(fn, ...args) {
    if (args.length >= fn.length) {
        return fn(...args);
    }

    return (...args2) => curry(fn, ...args, ...args2);
}

function sum(a, b, c) {
    return a + b + c;
}

let add = curry(sum);

console.log(add(1)(2)(3));

mapoio · 2020-10-17T07:36:47Z

function add (...params) {
    let result = params.reduceRight((a, b) => a + b)
    const tmp = (...paramsInit) => {
        result = [...paramsInit, result].reduceRight((a, b) => a + b)
        return tmp
    }
    tmp.toString = () => `${result}`
    tmp.valueOf = () => result
    return tmp
}

可以这样实现
@ABoyCDog

m7yue · 2021-02-08T07:25:10Z

const curryAdd = (...args) => {
  const Add = (...args2) => {
    return curry(...args, ...args2)
  }
  Add.toString = () => args.reduce((t, v) => t+v);
  
  return Add
}

console.log(curryAdd(1)(2)(3,4))

wjiantao · 2021-07-11T14:37:46Z

function curry(fn, args) {
  var length = fn.length;
  args = args || [];
  return function () {
    newArgs = args.conct(Array.prototype.slice.call(arguments));
    return newArgs.length < length
      ? curry.call(this, fn, newArgs)
      : fn.apply(this, newArgs);
  };
}

function multiFn(a, b, c) {
  return a * b * c;
}

var multi = curry(multiFn);

console.log(multi(2)(3)(4));

geekftz · 2021-07-13T23:51:39Z

const sum = (arr) => {
return arr.reduce((pre, cur, index) => {
return pre + cur
}, 0)
}

function add() {
const outArgs = Array.from(arguments)
let res = sum(outArgs);

function newAdd() {
if (arguments.length > 0) {
const innerArgs = Array.from(arguments);
res += sum(innerArgs)
return newAdd
} else {
return res;
}
}

newAdd.valueOf = function () {
return res
}

newAdd.toString = function () {
return res + ''
}

return newAdd
}

var result = add(1, 2)(3)(4)(5); // 参数可能为多个的情况下

console.log(result());

xcs-go · 2021-07-24T08:52:49Z

function curry(fn, args) {
    const { length } = fn;
    const params = args || [];
    return function() {
        const args = params.concat([].slice.call(arguments));
        if (args.length < length) {
            return curry.apply(this, fn, args); 
       } else {
           return fn.apply(this, args);
      }
   }
}

Evllis · 2021-08-11T10:02:22Z

const curryAdd = (...args) => {
  const Add = (...args2) => {
    return curry(...args, ...args2)
  }
  Add.toString = () => args.reduce((t, v) => t+v);
  
  return Add
}

console.log(curryAdd(1)(2)(3,4))

测一测吧，测好了再发出来。curry已经undefined了。

1uckyneo · 2021-09-28T17:34:10Z

const curry = (fn) => {
  const prevArgs = [];

  const rec = (...currArgs) => {
    prevArgs.push(...currArgs);
    
    if(fn.length - prevArgs.length <= 0) {
      return fn(...prevArgs);
    }
    
    return rec;
  }

  return rec;
}

const add = (a: number, b: number, c: number) => a + b + c;
const curryAdd = curry(add);

zizxzy · 2021-11-02T13:10:15Z

函数的柯里化的应用

const curry = (fn, args) => {
  const length = fn.length;
  let _args = args || [];
  return function () {
    let argArray = _args.slice(0);
    for (let i = 0; i < arguments.length; i++) {
      argArray.push(arguments[i]);
    }
    if (argArray.length < length) {
      return curry(fn, argArray);
    } else {
      return fn.apply(this, argArray);
    }
  }
}
const add = curry(function (a, b, c) {
  return a + b + c;
})
console.log(add(1)(2)(3));

ikozn · 2022-02-26T13:20:34Z

const add = a => b => c => a + b + c

wringY · 2022-03-28T09:35:03Z

function curry(fn) {
const length = fn.length
return function _curry(...args) {
if (args.length >= length) {
return fn.apply(this, args)
} else {
return _curry.bind(this, ...args)
}
}
}
// 测试
function add(x, y, z) {
return this.val + x + y + z
}
let fun1 = curry(add)
let obj = {
val: 10,
fun1
}
console.log(obj.fun1(1)(2)(3))

chinbor · 2023-08-02T06:58:03Z

function curry(...args1) {
  const fn = args1[0]
  const restArgs = args1.slice(1)

  function curried(...args2) {
    return curry.call(this, fn, ...[...restArgs, ...args2])
  }

  curried.toString = function() {
    return fn.call(this, ...restArgs)
  }

  return curried
}

function sum(...args) {
  let result = 0

  args.forEach(arg => {
    result += arg
  })

  return result
}

const add = curry(sum)

// toString 修改了但是不生效，需要自己利用 + 操作符进行转换
console.log(+add(1)(2)(3, 4, 5))

Kisthanny · 2024-03-20T13:44:37Z

暴力写法

function add(a) {
  return function (b) {
    return function (c) {
      return a + b + c;
    };
  };
}

console.log(add(1)(2)(3))

如果不限调用3次呢，比如add(1)(2)(3)(4)返回8

// 通过返回一个拥有获取值方法的函数
function add(x) {
  let sum = x;

  function innerAdd(y) {
    sum += y;
    return innerAdd;
  }

  innerAdd.getResult = function () {
    return sum;
  };

  return innerAdd;
}

console.log(add(1)(2)(3)(4).getResult()); // 输出 10
// 缺点：获取值时需要额外调用.getResult()

题目没说add(1)(2)(3)就是返回值了，有没有可能是类里的方法

class MyNumber {
  constructor(value) {
    this.value = value;
  }
  add(num) {
    this.value += num;
    return this.add.bind(this);
  }
}
const myNum = new MyNumber(0);
myNum.add(1)(2)(3);
console.log(myNum.value);

函数柯里化

// Function.length返回函数必传参数的长度
const curry = (fn) => {
  return function curried(...args) {
    if (args.length >= fn.length) {
      return fn(...args);
    } else {
      return function (...nextArgs) {
        return curried(...args, ...nextArgs);
      };
    }
  };
};
const baseAdd = (a, b, c) => a + b + c;
const add = curry(baseAdd);
console.log(add(1)(2)(3)); // 6
console.log(add(1, 2)(3)); // 6
console.log(add(1)(2, 3)); // 6
console.log(add(1, 2, 3)); // 6
console.log(add(1)(2)); // 返回方法
try {
  console.log(add(1)(2)(3)(4)); // 报错
} catch (error) {
  console.error(error);
}
// 缺点，baseAdd长度写死，比暴力好在修改baseAdd更简单

不限长度的函数柯里化

const infiniteCurry = (fn) => {
  let lastArgsLength = 0;
  return function curried(...args) {
    if (args.length === lastArgsLength) {
      return fn(...args);
    } else {
      lastArgsLength = args.length;
      return function (...nextArgs) {
        return curried(...args, ...nextArgs);
      };
    }
  };
};
const infiniteAdd = (...args) => {
  return Array.from(args).reduce((a, b) => a + b);
};
const add = infiniteCurry(infiniteAdd);
console.log(add(1)(2)());
console.log(add(1)(2)(3)());
console.log(add(1)(2)(3)(4)());
console.log(add(1, 2, 3)(4, 5)());
console.log(add(1, 2, 3, 4, 5, 6)());
// 缺点：获取值时要()

lgwebdream added JavaScript teach_tag 滴滴 company labels Jun 19, 2020

Genzhen closed this as completed Jul 20, 2020

Genzhen reopened this Jul 29, 2020

QuocVi1994 mentioned this issue Apr 20, 2021

[手写源码] 函数变变变～ curry 化！！！ QuocVi1994/Blog#13

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第 15 题：实现 add(1)(2)(3) #21

第 15 题：实现 add(1)(2)(3) #21

lgwebdream commented Jun 19, 2020 •

edited by Genzhen

Genzhen commented Jun 23, 2020 •

edited

H246802 commented Jul 15, 2020

123456zzz commented Jul 15, 2020

yaooooooooo commented Jul 20, 2020 •

edited by Genzhen

GolderBrother commented Jul 20, 2020 •

edited by Genzhen

JL003 commented Jul 24, 2020 •

edited

ABoyCDog commented Aug 11, 2020

huzedong2015 commented Sep 1, 2020

mapoio commented Oct 17, 2020

m7yue commented Feb 8, 2021

wjiantao commented Jul 11, 2021

geekftz commented Jul 13, 2021

xcs-go commented Jul 24, 2021

Evllis commented Aug 11, 2021

1uckyneo commented Sep 28, 2021

zizxzy commented Nov 2, 2021

ikozn commented Feb 26, 2022 •

edited

wringY commented Mar 28, 2022 •

edited

chinbor commented Aug 2, 2023

Kisthanny commented Mar 20, 2024

第 15 题：实现 add(1)(2)(3) #21

第 15 题：实现 add(1)(2)(3) #21

Comments

lgwebdream commented Jun 19, 2020 • edited by Genzhen

Genzhen commented Jun 23, 2020 • edited

H246802 commented Jul 15, 2020

123456zzz commented Jul 15, 2020

yaooooooooo commented Jul 20, 2020 • edited by Genzhen

GolderBrother commented Jul 20, 2020 • edited by Genzhen

JL003 commented Jul 24, 2020 • edited

ABoyCDog commented Aug 11, 2020

huzedong2015 commented Sep 1, 2020

mapoio commented Oct 17, 2020

m7yue commented Feb 8, 2021

wjiantao commented Jul 11, 2021

geekftz commented Jul 13, 2021

xcs-go commented Jul 24, 2021

Evllis commented Aug 11, 2021

1uckyneo commented Sep 28, 2021

zizxzy commented Nov 2, 2021

ikozn commented Feb 26, 2022 • edited

wringY commented Mar 28, 2022 • edited

chinbor commented Aug 2, 2023

Kisthanny commented Mar 20, 2024

暴力写法

如果不限调用3次呢，比如add(1)(2)(3)(4)返回8

题目没说add(1)(2)(3)就是返回值了，有没有可能是类里的方法

函数柯里化

不限长度的函数柯里化

lgwebdream commented Jun 19, 2020 •

edited by Genzhen

Genzhen commented Jun 23, 2020 •

edited

yaooooooooo commented Jul 20, 2020 •

edited by Genzhen

GolderBrother commented Jul 20, 2020 •

edited by Genzhen

JL003 commented Jul 24, 2020 •

edited

ikozn commented Feb 26, 2022 •

edited

wringY commented Mar 28, 2022 •

edited