Can the return of a promise be used as an input for the next function call in a Promise.all() execution?

Question

I've been looked on StackOverflow and haven't seen any direct examples of what I'm asking. I'm reading this article on memoization link here if you want to look.

It appears to me that you should be able to run them all together and use the return value from getSoupRecipe() as input for hireSoupChef()

async function makeSoupFromType(soupType) {
    
    let [ soupRecipe, soupPan, soupChef ] = await Promise.all([
        
        getSoupRecipe(soupType),
        buySoupPan(),
        hireSoupChef(soupRecipe.requiredSkills)
    ]);
    
    return await makeSoup(soupChef, soupRecipe, soupPan);
}

So the question is can all three async functions run at the same time and once getSoupRecipe returns I use it's variable name (which is soupRecipe) as input for hireSoupChef.

I would post all the other code here for you to see but I think it would probably make the question look too daunting, so the link is above. You don't necessarily have to look at it to understand I don't think because I think I've stated the question right, but nonetheless if you want to you can.

Answer 1

Not by itself, no. In your example the soupRecipe (and the other two variables) are only initialised after the Promise.all(…) has been awaited, so it can't be used in the expressions that are passed to Promise.all as an argument. There's no magic going on here, Promise.all is not part of the syntax but really just returning one promise that fulfills with an array once.

However, the approach outlined in this answer to How do I access previous promise results in a .then() chain? does enable the desired chaining of asynchronous actions:

async function makeSoupFromType(soupType) {
    const soupRecipePromise = getSoupRecipe(soupType);
    const [soupRecipe, soupPan, soupChef] = await Promise.all([
        soupRecipePromise,
        buySoupPan(),
        soupRecipePromise.then(({requiredSkills}) => hireSoupChef(requiredSkills))
    ]);
    
    return makeSoup(soupChef, soupRecipe, soupPan);
}

Alternatively, with plain async/await you could also use an IIFE to avoid the then() call:

async function makeSoupFromType(soupType) {
    const [soupPan, [soupRecipe, soupChef]] = await Promise.all([
        buySoupPan(),
        (async () => {
            const soupRecipe = await getSoupRecipe(soupType);
            const soupChef = await hireSoupChef(soupRecipe.requiredSkills);
            return [soupRecipe, soupChef];
        })(),
    ]);
    
    return makeSoup(soupChef, soupRecipe, soupPan);
}