请实现一个模块 math，支持链式调用math.add(2,4).minus(3).times(2);

[Sunny-117]

No description provided.

[weirong111]

class math {
  constructor(initValue = 0) {
    this.value = initValue;
  }
  add(...args) {
    this.value = args.reduce((pre, cur) => pre + cur, this.value);
    return this;
  }
  minus(...args) {
    this.value = this.value - args.reduce((pre, cur) => pre + cur);
    return this;
  }
  times(timer) {
    this.value = timer * this.value;
    return this;
  }
  getVal() {
    return this.value;
  }
}

[cscty]

class math {
constructor(value = 0) {
this.value = value;
}
add(...args) {
this.value = args.reduce(
(ljq, cur) => ljq + cur,
this.value
);
return this;
}
minus(...args) {
this.value =
this.value - args.reduce((ljq, cur) => ljq + cur);
return this;
}
times(...args) {
this.value *= args.reduce((ljq, cur) => ljq * cur);
return this;
}
}
let m = new math(2);
console.log(m.times(3, 4, 5));

[mafeiGitHub]

const math = {
    result: 0,
    add: function (...args) {
        this.result += args.reduce((pre, cur) => pre + cur, this.result);
        return this;
    },
    minus: function (value) {
        this.result -= value;
        return this;
    },
    times: function (value) {
        this.result *= value;
        return this;
    },
    getResult: function () {
        return this.result;
    }
};
// test code
const result = math.add(2, 4).minus(3).times(2).getResult();
console.log(result); // 输出: 6

[hexianga]

这个题有两种想法：

1. op 操作：将每一次函数调用视为一个 op 维护在一个数组中，当要通过一个方法获取之前操作的最后结果的时候，遍历所有 op，执行计算
2. 在每一次调用过程中，实时计算一个值，最后返回的时候也是这个值

[wmdmomo]

class Math {
  constructor(value = 0) {
    this.value = value;
  }
  add(...args) {
    this.value = args.reduce((acc, cur) => acc + cur, this.value);
    return this;
  }
  minus(...args) {
    this.value = args.reduce((acc, cur) => acc - cur, this.value);
    return this;
  }
  times(...args) {
    this.value = args.reduce((acc, cur) => acc * cur, this.value);
    return this;
  }
}
const math = new Math();
console.log(math.add(2, 4).minus(1).times(2, 4));