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第272天 用不同的方法改装下面的代码,使它的结果输出为0-4[代码]
我也要出题
for (var i = 0; i < 5; i++){ setTimeout(() => { console.log(i); }, 1000) }
for (let i = 0; i < 5; i++){ setTimeout(() => { console.log(i); }, 1000) }
使用 setTimeout 的第三个入参(有兼容问题),使用自运行函数,抽离成新方法
存个数组延时后使用 shift 获取
补充两个
for (var i = 0; i < 5; i++) { setTimeout(console.log(i), 1000) } for (var i = 0; i < 5; i++) { setTimeout(eval('console.log(i)'), 1000) }
for(var i = 0; i < 5; i++){ ((j) => { setTimeout(() => { console.log(j) },1000) })(i); }
let output = (i) => { return new Promise(resolve => { setTimeout(() => { console.log(i) resolve() },1000) }) } (async function () { for (var i = 0; i < 5;i++) { await output(i) } })()
let task = [] let output = (i) => { return new Promise(resolve => { setTimeout(() => { console.log(i) },1000*i) }) } for (var i = 0; i < 5;i++) { task.push(output(i)) } Promise.all(task)
for(var i=0;i<5;i++) { (function(n){ setTimeout( ()=> console.log(n), 1000) }(i)) }
for(let i=0;i<5;i++) { setTimeout( ()=> console.log(i), 1000) }
Activity
HCLQ commentedon Jan 13, 2020
for (let i = 0; i < 5; i++){ setTimeout(() => { console.log(i); }, 1000) }
forever-z-133 commentedon Jan 13, 2020
使用 setTimeout 的第三个入参(有兼容问题),使用自运行函数,抽离成新方法
forever-z-133 commentedon Jan 13, 2020
存个数组延时后使用 shift 获取
Jingce-lu commentedon Jan 14, 2020
补充两个
yini0223 commentedon Jan 15, 2020
for(var i = 0; i < 5; i++){
((j) => {
setTimeout(() => {
console.log(j)
},1000)
})(i);
}
renqi1996 commentedon Feb 20, 2020
xiaoqiangz commentedon Aug 29, 2022
for(var i=0;i<5;i++) {
(function(n){
setTimeout( ()=> console.log(n), 1000)
}(i))
}
for(let i=0;i<5;i++) {
setTimeout( ()=> console.log(i), 1000)
}