JavaScript Promise then() ordering

Question

I\'m still learning JavaScript Promises, and I came across a behavior I don\'t understand.

Answer 1

The reason you see 6 early is because you didn't chain, you branched.

When you call p.then().then().then(), you've got a chain of promises that must execute in the correct order.
However, if you call p.then().then(); p.then(), you've got 2 promises attached to p - essentially creating a branch, and the 2nd branch will execute along with the first.

You can fix this by ensuring you chain them together p = p.then().then(); p.then();

FYI, you almost NEVER want to branch, unless you bring them back together (eg. Promise.all), or are intentionally creating a "fire and forget" branch.