return does not await async function

Question

In the following code, return does not return the awaited value. What can I so the the promise will be resolved before returning. Hence I want result to be SUCCESS instead of a promise.

const foo = async()=>{
    try{
        let a = await new Promise((resolve)=>{
            resolve('SUCCESS')
        })
        console.log("this is inside the try block");
        return a
    }catch{
        console.log('error')
    }
}
let result = foo();
console.log(result);

Answer 1

foo is an async function it will return a promise. To get the result of the promise, you will need to chain a then method to it:

const foo = async()=>{
    try{
        let a = await new Promise((resolve)=>{
            resolve('SUCCESS')
        })
        console.log("this is inside the try block");
        return a
    }catch{
        console.log('error')
    }
}

foo().then(result => console.log(result));

UPDATE:

To use the returned value, you can use it inside the then method or call another function with the result.

foo().then(result => {
  console.log(result);
  //do what you want with the result here
});

OR:

foo().then(result => {
  someFunction(result);
});

function someFunction(result) {
   console.log(result);
  //you can also do what you want here
}