[英]Node.js: async-await with/without Promise
當我為我的項目設計一些需要實現的邏輯原型時,我注意到 async-await 和 Promise 的一些有趣行為。
// Notice this one returns a Promise
var callMe = function(i) {
return new Promise((resolve, reject) => {
setTimeout(() => {
console.log(i)
resolve(`${i} is called :)`)
}, (i+1)*1000)
})
}
// But this one doesn't
var callYou = function(i) {
setTimeout(() => {
console.log(i)
}, (i+1)*1000)
}
async function run() {
console.log("Start")
for(let i = 0; i < 3; i++) {
let val = await callYou(i)
# also try with callMe()
#let val = await callMe(i)
console.log(val)
}
console.log("End")
}
run()
使用let val = await callYou(i)
,結果看起來像這樣
Start
callYou()
callYou()
callYou()
End
0
1
2
而使用let val = await callMe(i)
,結果看起來像這樣
Start
0
0 is called :)
1
1 is called :)
2
2 is called :)
End
我期待兩個函數的行為相似,因為異步 function 基本上返回 promise。 有人可以解釋為什么會這樣嗎?
async
函數返回承諾,但callYou
或callMe
是async
函數,即使它們在哪里,當回調傳遞到setTimeout
function 時,當 function 返回時callYou
將被實現。
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