Node.js stacktraces不包含用户代码

Question

When debugging Node.js code I often encounter call stacks that do not include my program's code in them, only node_modules/non-user code, despite the current line of execution being at a location in my code. This defeats the purpose of following the call stack to see the execution path through my application code.

Why are my source files not showing in the call stack?

Answer 1

It appears that you're looking at an asynchronous stack trace where your code is not in the stack except for your callback because your code unwound/finished and THEN the async callback was called.

All .then() handlers for all promises are called asynchronously with a clean stack. That's per the promise specification. So, promises always let the current thread of execution finish and unwind and then they fire .then() handlers with no user code on the stack when the callback is called. What you are describing is how synchronous code would work, not asynchronous code. We could talk a lot more specifically instead of theoretically if you showed actual code and described where you're looking at the call stack.

Async progress often has to be tracked with logging because you can't easily step through it and you can't just break and look at stack traces either.

As a little simpler example to look at:

function foo() {
   setTimeout(() => {
       console.log("timer");    // set breakpoint here
   }, 100);
}

foo();

The function foo() has finished executing and returned before the callback is called and thus the stack trace will not have any of your code (other than just the callback) on it.

While .then() handlers use a slightly different scheduler than setTimeout() , the principle is the same.