[js] 第57天写一个方法，使得sum(x)(y)和sum(x,y)返回的结果相同 #238

haizhilin2013 · 2019-06-11T20:37:36Z

第57天写一个方法，使得sum(x)(y)和sum(x,y)返回的结果相同

AnsonZnl · 2019-06-12T08:34:04Z

写一个方法，使得sum(x)(y)和sum(x,y)返回的结果相同

var sum= function(x){
    if(arguments[1]){
        return x + arguments[1];
    }else{
        return function (y){
            return x + y
        }
    }
}
console.log(sum(1)(2))//3
console.log(sum(1,2))//3

群里的提议增加点难度使得sum(x)(y)(z)(...)(a)和sum(x,y,z,... a)返回的结果相同

function sum(x){
    if(arguments[1]){
        var arr = Array.prototype.slice.apply(arguments);
            x = arr.reduce((a, c)=> a+ c)
        return x;
    }else{
        function add(b) { 
            x = x + b; 
            return add;
        }
        add.toString = function() { 
            return x;
        }
        return add; 
    }
}
var res1 = sum(1)(2)(3)(4)(5)(6);
var res2 = sum(1,2,3,4,5,6);
console.log(res1)//21
console.log(res2)//21

在 JavaScript 高阶函数浅析获取的灵感

thisisandy · 2019-06-12T10:38:25Z

function sum(...args) {
  function curriedSum(...rest) {
    args.push(...rest);
    return curriedSum;
  }
  curriedSum.toString = () => args.reduce((prev, curr) => curr + prev);

  curriedSum.toNumber = () => args.reduce((prev, curr) => curr + prev);
  return curriedSum;
}

thisisandy · 2019-06-12T10:41:43Z

类似于python元编程，修改toString 和toNumber 方法，使其既可以当做数字也可以当做函数调用

SCLeoX · 2019-06-13T09:30:21Z

function sum(x){
if(arguments[1]){
var arr = Array.prototype.slice.apply(arguments);
x = arr.reduce((a, c)=> a+ c)
return x;
}else{
function add(b) {
x = x + b;
return add;
}
add.toString = function() {
return x;
}
return add;
}
}
var res1 = sum(1)(2)(3)(4)(5)(6);
var res2 = sum(1,2,3,4,5,6);
console.log(res1)//21
console.log(res2)//21

虽然我一时半会没想出什么解决方案但是你这个无限 curry 的不太行啊，除了显然的 typeof 结果不一样以外，还有这种，所以得到的值并不能放心用：

sum(1,2,3).toExponential(3);
// "6.000e+0"

sum(1)(2)(3).toExponential(3);
// VM1727:1 Uncaught TypeError: sum(...)(...)(...).toExponential is not a function
//    at <anonymous>:1:14

Gxmg · 2019-06-14T03:21:38Z

作者怎么也不最后给个答案呀

forever-z-133 · 2019-06-14T06:08:44Z

Function.prototype.currying = function() {
  var fn = this, _args = [];
  var _temp = function() {
    _args = _args.concat([].slice.call(arguments));
    return _temp;
  }
  _temp.toString = _temp.valueOf =function(){
    return fn.apply(fn, _args);
  }
  return _temp;
}
function adder() {
  return [].slice.call(arguments).reduce(function(a,b){return a+b});
}
var sum = adder.currying();
console.log(sum(2,4)(3));  // ƒ 9

haizhilin2013 · 2019-06-14T06:09:55Z

@Gxmg 没有最后的答案

hn-failte · 2019-06-25T06:27:45Z

此题考函数式编程：
const sum = (x, y) => (typeof y === "undefined") ? (y)=> x+y : x+y

console.log(sum(1,3));

console.log(sum(1)(3));

yuqingc · 2019-07-01T09:33:47Z

Lodash _.curry()
可以看人家实现

haizhilin2013 · 2019-07-01T09:34:47Z

@yuqingc 那你自己来实现一个呢？

chenyouf1996 · 2019-07-26T02:23:22Z

//函数柯理化
function sum1(x, y, z) {
  return x + y + z
}
function sum2() {
  var args = arguments
  return function () {
    var arr = []
    arr.push(...args, ...arguments)
    return arr.reduce((total, item) => total += item, 0)
  }
}
console.log(sum1(6, 9, 14)) //29
console.log(sum2(6)(9, 14)) //29

Vi-jay · 2019-07-31T03:42:14Z

function currying(fn) {
  let args = [];
  return function closureFn() {
    args = args.concat(Array.from(arguments));
    if (args.length < fn.length) return closureFn;
    const res = fn.apply(this, args);
    args = [];
    return res;
  }
}

const sum = currying((a, b) => a + b);
console.log(sum(1, 2) === sum(1)(2)); // true

lizheng1991 · 2019-08-23T03:34:34Z

综合一下大家的写法，再自由发挥一下~话说怎么让代码高亮啊？？？

function sumfn(){
  var result=0;
  function fn(...arg){
    result = result + arg.reduce((a,b)=>a+b,0);
    return fn;
  }
  fn.toString=function(){
    var re=result;
    result=0;
    return re;
  }
  return fn;
}

var sum=sumfn();
sum(1,2,3);
sum(1)(2)(3);
sum(1)(2)(3,4,5);

YiChongXie · 2019-09-29T00:41:00Z

function add(){
var args = [...arguments]
var fn = function (){
args.push(...arguments)
return fn
}
fn.toString = function(){
return args.reduce((x,y) => x + y)
}
return fn
}

geng130127 · 2019-11-20T10:20:25Z

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

rni-l · 2020-01-19T08:46:02Z

function curry(...args) {
  if (args.length > 1) return args2.reduce((acc, cur) => acc + cur, 0)
  let val = 0
  function add(...args2) {
    val = args2.reduce((acc, cur) => acc + cur, val)
    return add
  }
  add.toString = function() { return val }
  val = args[0]
  return add
}

// curry(1,6)(2,5)(3,4)(7,8,9) -> 45
// curry(1) -> 1

276378532 · 2020-02-24T15:01:50Z

 function Curry(fn, length) {
            var length = length || fn.length;
            return function (...arg) {
                if (arg.length < length) {
                    return Curry(fn.bind(this, ...arg), length - arg.length); 
                } else {
                    return fn(...arg)
                }
            }
        }
函数式编程 柯里化

Alex-Li2018 · 2020-08-12T01:53:24Z

function sum(x,y) {
    return x + y;
}
sum(2,3)

const sum = x => y => x + y
function sum (x) {
     return function(y) {
        return x + y
    }
}
sum(2)(3)

laboonly · 2020-09-10T15:10:15Z


function add() {
  // 第一次执行时，定义一个数组专门用来存储所有的参数
  var _args = [].slice.call(arguments);

  // 在内部声明一个函数，利用闭包的特性保存_args并收集所有的参数值
  var adder = function () {
      var _adder = function() {
          // [].push.apply(_args, [].slice.call(arguments));
          _args.push(...arguments);
          return _adder;
      };

      // 利用隐式转换的特性，当最后执行时隐式转换，并计算最终的值返回
      _adder.toString = function () {
          return _args.reduce(function (a, b) {
              return a + b;
          });
      }

      return _adder;
  }
  // return adder.apply(null, _args);
  return adder(..._args);
}

MrZ2019 · 2020-12-07T07:33:32Z

写一个方法，使得sum(x)(y)和sum(x,y)返回的结果相同

var sum= function(x){
    if(arguments[1]){
        return x + arguments[1];
    }else{
        return function (y){
            return x + y
        }
    }
}
console.log(sum(1)(2))//3
console.log(sum(1,2))//3

群里的提议增加点难度使得sum(x)(y)(z)(...)(a)和sum(x,y,z,... a)返回的结果相同

function sum(x){
    if(arguments[1]){
        var arr = Array.prototype.slice.apply(arguments);
            x = arr.reduce((a, c)=> a+ c)
        return x;
    }else{
        function add(b) { 
            x = x + b; 
            return add;
        }
        add.toString = function() { 
            return x;
        }
        return add; 
    }
}
var res1 = sum(1)(2)(3)(4)(5)(6);
var res2 = sum(1,2,3,4,5,6);
console.log(res1)//21
console.log(res2)//21

在 JavaScript 高阶函数浅析获取的灵感

Huauauaa · 2021-05-07T01:10:28Z

export const sum = (...x) => {
  let result = x.reduce((a, b) => a + b);
  const add = (...y) => {
    if (y.length === 0) {
      return result;
    }
    result += y.reduce((a, b) => a + b);
    return add;
  };

  return add;
};

console.log(sum(1, 2, 3, 4)());  // 10
console.log(sum(1)(2)(3)(4)()); // 10
console.log(sum(1, 2)(3)(4)()); // 10
console.log(sum(1)(2, 3, 4)()); // 10

xiaoqiangz · 2022-06-09T03:54:35Z

柯里化函数
function sum() {
let args = [...arguments]
let fn = function() {
let fn_args = [...arguments]
return sum.apply(null, args.concat(fn_args))
}
fn.toString = function() {
return args.reduce((a, b) => a + b)
}
return fn
}
console.log(sum(1,2,3)+'')

dragon-sing · 2022-10-31T03:42:18Z

const sum = (a , b) => {
  if (b) {
    return a + b;
  } else {
    return (another) => {
      return a + another;
    }
  }
}

haizhilin2013 added the js JavaScript label Jun 18, 2019

haizhilin2013 mentioned this issue Jan 28, 2021

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[js] 第57天写一个方法，使得sum(x)(y)和sum(x,y)返回的结果相同 #238

[js] 第57天写一个方法，使得sum(x)(y)和sum(x,y)返回的结果相同 #238

haizhilin2013 commented Jun 11, 2019

AnsonZnl commented Jun 12, 2019 •

edited

thisisandy commented Jun 12, 2019 •

edited

thisisandy commented Jun 12, 2019

SCLeoX commented Jun 13, 2019

Gxmg commented Jun 14, 2019

forever-z-133 commented Jun 14, 2019 •

edited

haizhilin2013 commented Jun 14, 2019

hn-failte commented Jun 25, 2019 •

edited

yuqingc commented Jul 1, 2019

haizhilin2013 commented Jul 1, 2019

chenyouf1996 commented Jul 26, 2019

Vi-jay commented Jul 31, 2019

lizheng1991 commented Aug 23, 2019

YiChongXie commented Sep 29, 2019

geng130127 commented Nov 20, 2019

rni-l commented Jan 19, 2020 •

edited

276378532 commented Feb 24, 2020

Alex-Li2018 commented Aug 12, 2020

laboonly commented Sep 10, 2020

MrZ2019 commented Dec 7, 2020

Huauauaa commented May 7, 2021

xiaoqiangz commented Jun 9, 2022

dragon-sing commented Oct 31, 2022

[js] 第57天 写一个方法，使得sum(x)(y)和sum(x,y)返回的结果相同 #238

[js] 第57天 写一个方法，使得sum(x)(y)和sum(x,y)返回的结果相同 #238

Comments

haizhilin2013 commented Jun 11, 2019

AnsonZnl commented Jun 12, 2019 • edited

thisisandy commented Jun 12, 2019 • edited

thisisandy commented Jun 12, 2019

SCLeoX commented Jun 13, 2019

Gxmg commented Jun 14, 2019

forever-z-133 commented Jun 14, 2019 • edited

haizhilin2013 commented Jun 14, 2019

hn-failte commented Jun 25, 2019 • edited

yuqingc commented Jul 1, 2019

haizhilin2013 commented Jul 1, 2019

chenyouf1996 commented Jul 26, 2019

Vi-jay commented Jul 31, 2019

lizheng1991 commented Aug 23, 2019

YiChongXie commented Sep 29, 2019

geng130127 commented Nov 20, 2019

rni-l commented Jan 19, 2020 • edited

276378532 commented Feb 24, 2020

Alex-Li2018 commented Aug 12, 2020

laboonly commented Sep 10, 2020

MrZ2019 commented Dec 7, 2020

Huauauaa commented May 7, 2021

xiaoqiangz commented Jun 9, 2022

dragon-sing commented Oct 31, 2022

[js] 第57天写一个方法，使得sum(x)(y)和sum(x,y)返回的结果相同 #238

[js] 第57天写一个方法，使得sum(x)(y)和sum(x,y)返回的结果相同 #238

AnsonZnl commented Jun 12, 2019 •

edited

thisisandy commented Jun 12, 2019 •

edited

forever-z-133 commented Jun 14, 2019 •

edited

hn-failte commented Jun 25, 2019 •

edited

rni-l commented Jan 19, 2020 •

edited