终极答案
function add() {
let args = [...arguments]
let fn = function () {
args.push(...arguments)
return fn
}
fn.toString = function () {
return args.reduce((x,y) => x+y)
}
return fn
}
toString和valueOf的类型转换
//案例1
//sum(2,3)和sum(2)(3)均能输出outPut5，这个问题较为简单，只要返回一个函数即可，本例中使用了es6语法
function sum(x = 0, y) {
if ([...arguments][1] == undefined) {
let befor = [...arguments][0];
return function suum(afte) {
let sum2 = befor + afte;
console.log(`outPut${sum2}`);
}
} else {
let sum1 = x + y;
console.log(`outPut${sum1}`);
}
}
sum(2, 3); //输出outPut5
sum(2)(3); //输出outPut5
//案例二、add(2,3,4...)和add(2)(3)(4)...都输出相同结果的解决方案
function add(x) {
var sum = x;
var tmp = function (y) {
sum = sum + y;
return tmp;
};
tmp.toString = function () {
return sum;
};
return tmp;
}
console.log(add(1)(2)(3)); //6
console.log(add(1)(2)(3)(4)); //10
console.log(add(1)(2)(3)(4)(5)); //15
console.log(add(1)(2)(3)(4)(5)(6)); //21
//扩展案例3、此函数使用es6实现输入sum(),sum(2,3,4...)多个值相加，还有sum(2)(3)(4)...等多个数值分别相加的值相等，此案例也是最完整的解决方案
function sum(...args) {
if ([...args].length == 1) {
let sum2 = [...args][0];
var suum = function (y) {
sum2 += y;
console.log(`output ${sum2}`)
return suum;
}
suum.valueOf = function () {
return sum2;
}
return suum;
} else {
let sum1 = 0;
for (var i = 0; i < [...args].length; i++) {
sum1 += [...args][i];
}
return sum1
}
}
//sum(1); //outPut1 0
//sum(2, 3, 4); //outPut1 9
//sum(2)(3)(4)(5); //outPut1 5//outPut1 9//outPut1 14
//每加一次就会输出一次Sum。如果不想输出三次则可以在函数中加一个valueOf方法或者toString方法。然后将Summ函数中的console去掉。
//然后console.log(sum(2)(3)(4)(5));即可
console.log(sum(2)(3)(4)(5))
console.log(sum(2))
