JS Addition

杜绝0.1 + 0.2 =0.30000000000000004 问题；

function add(num1, num2) {const num1Digits = (num1.toString().split('.')[1] || '').length;const num2Digits = (num2.toString().split('.')[1] || '').length;const baseNum = Math.pow(10, Math.max(num1Digits, num2Digits));return (num1 * baseNum + num2 * baseNum) / baseNum;
}

异步求和函数

function asyncAdd(a, b, callback) {setTimeout(function () {callback(null, a + b);}, 1000);
}

简化：两数之和

function sumT(a, b) {return await new Promise((resolve, reject) => {asyncAdd(a, b, (err, res) => {if(!err) {resolve(res)}reject(err)})})}const test = await sumT(1, 2)
console.log(test) // 3

加深：多数之和

function sum(...args) {return new Promise((resolve) => {args.reduce((acc, cur) => acc.then((total) => sumT(total, cur)), Promise.resolve(0)).then(resolve);});
}console.time('sum');await sum(1, 2, 3, 4, 5); // 15console.timeEnd('sum');

优化：使用 Promise.all

async function sum(...args) {console.log(args); // 用于考察每次迭代的过程if (args.length === 1) return args[0];  // 如果仅有一个，直接返回let result = [];// 两两一组，如果有剩余一个，直接进入for (let i = 0; i < args.length - 1; i += 2) {result.push(sumT(args[i], args[i + 1]));}if (args.length % 2) result.push(args[args.length - 1]);return sum(...(await Promise.all(result))); // Promise.all 组内求和
} test = await sum(1, 2, 3, 4, 5);

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