Promise.all.then Not all code paths return a value

Question

I have the following function that I want to return Promise< number[] >

async fetchCommentLines(commitDict: CommitDict): Promise < number[] > {
    if (GitCommentService.isLoggedIn()) {
        const commentLines = Object.values(commitDict).map(async commit => {
            // ...
            // do the job and return number[]
            // ...
            return lineNums;
        });

        Promise.all(commentLines)
            .then(commitLines => {
                return Array.prototype.concat.apply([], commitLines);
            });
    } else {
        return [] as number[];
    }
}

Firstly I got "function lacks ending return statement and return type does not include 'undefined'"

And then I added undefined (so return type becomes Promise< number[] | undefined >)

But I'm getting "not all code paths return a value" this time.

It seems I'm not considering a possible code path with below code

Promise.all(...)
.then(val => {return ...})

What I'm missing?

I also tried this

Promise.all(...)
.then(val => {return ...})
.catch(e => {return ...})

But it was not helpful

Note: My main purpose is to return Promise< number[] >, not Promise< number[] | undefined >

Answer 1

Your branch with Promise.all never issues a return (value) statement. The then callback does, but not the code outside the then callback.

You probably want to return the result of Promise.all().then().

    return Promise.all(commentLines)
//  ^^^^^^
        .then(commitLines => {
            return Array.prototype.concat.apply([], commitLines);
        });