我的代码高尔夫同事想出了这个（ES6），包容的：

(s,f)=>[...Array(f-s+1)].map((e,i)=>i+s)

非包容性：

(s,f)=>[...Array(f-s)].map((e,i)=>i+s)

已经给出了非常好的答案，但我没有看到ES6迭代器的完整使用来实现范围的完整实现，所以这里是：

/**
 * inspired by Python's built-in range utility function
 * implemented using ES6 Iterable, Iterator  protolols (interfaces)
 */
class Range {
  constructor(...args) {
    this.start = args.length <= 1 ? 0 : args[0];
    this.end = args.length <= 1 ? args[0] || 0 : args[1];
    this.step = args.length <= 2 ? 1 : args[2];
  }

  [Symbol.iterator]() {
    return this;
  }

  next() {
    if (this.end > this.start) {
      const result = { done: false, value: this.start };
      this.start = this.start + this.step;
      return result;
    } else return { done: true, value: undefined };
  }
}

/**
 * Symbol.iterator is part of the couple of inbuilt symbols
 * available in JavaScript. It allows for hooking into the
 * inbuilt for of iteration mechanism. This is why the
 * Symbol.iterator comes into play when talking about
 * iterables and iterators in JavaScript.
 */

function range(...args) {
  return new Range(...args);
}

console.log([...range(4)]);        // [0, 1, 2, 3]
console.log([...range(2, 5)]);     // [2, 3, 4]
console.log([...range(1, 10, 3)]); // [1, 4, 7]

您还可以执行以下操作：

const range = Array.from(Array(size)).map((el, idx) => idx+1).slice(begin, end);

一个定义了硬整数的范围有很多答案，但是如果你不知道这一步，而你想在这一步之间多走几步呢？

我写这段代码就是为了做到这一点。这很不言自明。

const stepScale = (min, max, numberOfSteps) => {
  const _numberOfSteps = numberOfSteps - 1
  const scaleBy = (max - min) / _numberOfSteps

  const arr = []
  for (let i = 0; i <= _numberOfSteps; i += 1) {
    arr.push(min + scaleBy * i)
  }
  return arr
}

export default stepScale

stepScale(5, 10, 4)
// [5, 6.666666666666667, 8.333333333333334, 10]

对于npmhttps://npm.im/@通用标准/步长

我想补充一点，我认为这是一个非常可调的版本，速度非常快。

const range = (start, end) => {
    let all = [];
    if (typeof start === "string" && typeof end === "string") {
        // Return the range of characters using utf-8 least to greatest
        const s = start.charCodeAt(0);
        const e = end.charCodeAt(0);
        for (let i = s; i <= e; i++) {
            all.push(String.fromCharCode(i));
        }
    } else if (typeof start === "number" && typeof end === "number") {
        // Return the range of numbers from least to greatest
        for(let i = end; i >= start; i--) {
            all.push(i);
        }
    } else {
        throw new Error("Did not supply matching types number or string.");
    }
    return all;
}
// usage
const aTod = range("a", "d");

如果您愿意，也可以使用打字机

const range = (start: string | number, end: string | number): string[] | number[] => {
    const all: string[] | number[] = [];
    if (typeof start === "string" && typeof end === "string") {
        const s: number = start.charCodeAt(0);
        const e: number = end.charCodeAt(0);
        for (let i = s; i <= e; i++) {
            all.push(String.fromCharCode(i));
        }
    } else if (typeof start === "number" && typeof end === "number") {
        for (let i = end; i >= start; i--) {
            all.push(i);
        }
    } else {
        throw new Error("Did not supply matching types number or string.");
    }
    return all;
}
// Usage
const negTenToten: number[] = range(-10, 10) as number[];

受到其他答案的影响。用户已离开。

您可以使用以下一行代码使事情简短明了

var启动=4；var端=20；console.log（数组（end-start+1）.fill（start）.map（（x，y）=>x+y））；