Why is process.debugPort defined when not debugging? - node.js

Question

I've been exploring the various ways to detect whether a node process has a debugger attached or not. I came across the straightforward looking process.debugPort, which is documented as:
process.debugPort
Added in: v0.7.2
<number>
The port used by the Node.js debugger when enabled.
What do the docs mean by "when enabled"? Or perhaps more to the point, when is the debugger not enabled? A simple test program shows that the process.debugPort always seems to be defined.
'use strict'
const debugPort = process.debugPort
console.log(`debugPort: ${debugPort}`)
Here is some command line output showing its invocation in node v10.22.1 – I tested node v15.3.0 and observed the same behavior.
➜ node --version
v10.22.1
➜ node nodeDebug.js
debugPort: 9229
➜ node --inspect=0.0.0.0:9999 nodeDebug.js
Debugger listening on ws://0.0.0.0:9999/db5cdb75-4745-4ea1-9a87-3f81c811c563
For help, see: https://nodejs.org/en/docs/inspector
debugPort: 9999
So even when invoking node without a --inspect flag, the debugPort is defined (and set to the default of 9229).
As an aside, the best method I've discovered for detecting whether node is in debugging mode or not is inspector.url() which
Return the URL of the active inspector, or undefined if there is none.
I'm not sure if this is bulletproof, but it covers my use case.

Answer 1

Even when you start without the explicit --inspect flag, it's always possible to enable the inspector on an ad hoc basis by sending the process a SIGUSR1 signal like so:
kill -SIGUSR1 $myNodePID
As such, it makes sense that the debugPort property is always set to something, as it's possible that it might be shortly used.
See towards top of here for passing sentence confirming this.