es6 还原 es5

Question

const compose = f => g => x => f(g(x));

const f = compose (x => x * 4) (x => x + 3);
f(2) // 20

求一个化简后的compose!!!!!!.......还有为什么f(2)等于8

Answer 1

"use strict";

var compose = function compose(f) {
  return function (g) {
    return function (x) {
      return f(g(x));
    };
  };
};
var f = compose(function (x) {
  return x * 4;
})(function (x) {
  return x + 3;
});
f(2);

地址

可能给compose的参数换个名字更阅读一些?

var compose = function(a) {
    return function(b) {
        return function(c) {
            return a(b(c));
        };
    };
};
var f = compose(function(x) {
    return x * 4;
})(function(x) {
    return x + 3;
});
console.log(f(2));

调用的话一共有三次
var f = compose(function(x) {
    return x * 4;
})(function(x) {
    return x + 3;
});
两次
f(2)
一次

三次调用再拆为

var f1=compose(function(x) {
    return x * 4;
})
var f2 = f1(function(x) {
    return x + 3;
})
f2(2);

1. compose(function(x) {return x * 4;})即参数a为函数function(x) {return x * 4;}内部a(b(c))相当于对a调用内部参数x为b(c),得到function(b){return function(c){return b(c)*4}}

2. 此时 f1 = function(b){return function(c){return b(c)*4}}同样对f1的调用f1(function(x) {return x + 3;})相当与参数b为function(x){return x+3},内部b(c)相当于将c作为参数传入b中,得到function(c){return (c+3)*4}

3. 此时f2=function(c){return (c+3)*4},f2(2)=20

有点绕，可以拿笔和纸写一下。

Answer 2

var compose= function compose(f){
    return function(g){
        return function (x){
            return f(g(x))
        }
    }
}
var f = compose(function(x){
    return x*4;
})(function(x){
    return x + 3;
});
f(2)//20

就像一楼说滴一样，就是这个东东