函数中的JavaScript函数 [英] JavaScript function in a function

问题描述

我试图得到这个q，但是没有成功...任务是在以下条件下构建函数:

I have tried to get this q but without any success... The task is to build a function by this conditions:

// b('m') -> 'bm'
// b()()('m') -> 'boom'
// b()()()()('m') -> 'boooom'

那是我的尝试:

var b = (a) => {
    var counter = 0;
    var times = counter += 1;
    var d = (a, o, times) => {
        var o = 'o'.repeat(times);
        return ('b' + o + a);
    };
    return d();
};

console.log(b('m'));

推荐答案

作为naomik和NinaScholz提出的另一种变体，您可以避免使用额外的可选参数，而使用 this 上下文(在ES6中)存储一个数字基元值，该值跟踪要产生多少'o'个字符:

As another variant of what naomik and NinaScholz proposed, you could avoid the extra optional argument, and use the this context (in ES6) to store a number primitive value that tracks how many 'o' characters to produce:

function b(a) {
    return a ? 'b' + 'o'.repeat(+this) + a : b.bind((+this||0)+1);
}

console.log(b('m'));
console.log(b()('m'));
console.log(b()()('m'));
console.log(b()()()('m'));

仅当使用(真实的)参数调用 b 时才组成字符串:该字符串是"b" 后跟许多"o"字符(由 this 确定)，最后是参数(即实际调用中的"m" ).如果未定义 this (或在非严格模式下为全局对象)，则将"o" 的计数视为 0 .当使用一个参数立即调用 b 时会发生这种情况(无链接).

The string is only composed when b is called with a (truthy) argument: that string is a "b" followed by a number of "o" characters (determined by this) and finally the argument (i.e. "m" in the actual calls). If this is not defined (or is the global object, in non-strict mode), the count of "o" is considered to be 0. This happens when b is called immediately with an argument (no chaining).

在不带参数的情况下调用 b 时，将返回函数本身，但绑定了 this 并已通过 1 进行了增强.注意:实际上它不是函数本身，而是函数的绑定变体.

When b is called without argument, the function itself is returned, but with a this bound to it that has been increased with 1. NB: so it is not actually the function itself, but a bound variant of it.

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