Do I understand the work of Promise.all correctly?

Question

Node.js

Option 1 Option 2
[1,2,3,4,5].map(num => heavyIOfunction);

Promise.all([1,2,3,4,5].map(num => heavyIOfunction));

Do I understand correctly that there is no difference between the two options and IO operations start to be executed in parallel from the heavyIOfunction call itself, and not from Promise.all? That is, Promise.all does not start running IO. Promise.all only registers a promise on the result of calling all IO operations.

Answer 1

Do I understand correctly that there is no difference between the two options and IO operations start to be executed in parallel from the heavyIOfunction call itself, and not from Promise.all?

That's right, but not in parallel, the process is one, to whom the answer will come faster (request, reading a file, etc.), that and slippers.

Promise.all only registers a promise on the result of calling all IO operations.

await Promise.all(arrOfPromises);
=

await arrOfPromises.reduce(async (acc, promise) => {
  await acc;
  return promise;
}, Promise.resolve());