Javascript 异步 function 而不是新的 promise

Question

I just learnt " async " that " async " ensures that the function returns promise then i tried to make a promise with writing " async " instead of writing " new Promise " and it didn't work out so i want to know is it wrong syntax or it will never work out and i will have to write only new promise ?

---

// This is a code

async function gets(){
    let gets1 = async()=>{
        return 45;
    }
    gets1.then((value)=>{
        console.log(value, "inside gets1")
    })
}
gets()

Answer 1

As already pointed out, you need to call a function fn().then( , not just reference it fn.then( .
Also if you don't use the await , you also don't need the wrapping await

 const gets = () => { const gets1 = async() => { return 45; } gets1().then((value) => { console.log(value, "inside gets1") }) }; gets();

Or use async with await to get rid of Promise.then() :

 const gets = async () => { const gets1 = async() => { return 45; }; const value = await gets1(); console.log(value, "inside gets1") }; gets();

Make sure to read top to bottom: async Function

Answer 2

You forgot to call the function

It should be gets1().then() and not gets1.then()

 async function gets(){ let gets1 = async()=>{ return 45; } gets1().then((value)=>{ console.log(value, "inside gets1") }) } gets()

Answer 3

The whole thing will work without the outer async function too:

 let gets1 = async()=>"msg from gets1"; gets1().then(console.log)