What is the difference between process.exitcode and process.exit() ? If I use process.exitode = 1 and process.exit(1) does this create any difference or its just a alternate way to do ?
2 Answers
Yes, there is a difference.
process.exitCodeonly sets the exit code that will be used when the process eventually exits. It does NOT tell the process to exit, only what code to use when it does.process.exit([code])will terminate the process with the given exit code, or with the value ofprocess.exitCodeif it has been set, or with the exit code 0 (success) by default.
The difference is that exit will exit as quickly as possible (after all 'exit' event listeners are called), even if there are pending async operations, including I/O operations. This can lead to surprising behaviour!
If you don't need to exit as soon as possible or if your code has a lot of async operations, it's safer to use exitCode and let the process exit gracefully when all operations have completed.
Comments
process.exitcode = 1 doesn't do anything consequential. Be careful of capitalisation.
From docs on process.exitCode:
A number which will be the process exit code, when the process either exits gracefully, or is exited via process.exit() without specifying a code.
Thus,
process.exitCode = 1;
// ...
process.exit();
is equivalent to
process.exit(1);
Notably,
process.exitCode = 1;
process.exitCode = 2;
process.exit();
is not equivalent to
process.exit(1);
process.exit(2);
Also,
process.exitCode = 1;
console.log("bye");
process.exit();
is not equivalent to
process.exit(1);
console.log("bye");
8 Comments
Vipin
it means if i do process.exitCode = 1, it will works with async task while the process.exit(1) don't ?
Amadan
I don't know what you mean by "it will works".
process.exitCode = 1 doesn't exit. They both work, but they do different things.Vipin
If I do
process.exit(1); setTimeout(() => { console.log("bye"); }, 2000); It exit the process without printing bye but if i change to process.exitCode = 1 it exit process by printing bye.Amadan
Yes, because
process.exitCode = 1 doesn't exit. It has nothing to do with asynchronicity. To exit, you need to either call process.exit, or you need to run out of script. Please read the quote again; it says nothing about exiting. Assignment to process.exitCode does not exit. No exiting is happening when you execute process.exitCode = 1.Amadan
You misunderstood what the documentation is saying. Reading the context around that quote, it says that it is better to set the exitCode and then allow the program to exit naturally (by either running out of things to do, or by throwing an exception that will cut through the remaining code) than using
process.exit, because process.exit is abrupt and doesn't allow for things to clean up nicely. Again, process.exitCode = 1 does not exit any more than let x = 3 does; it is a set-up step for the case when the program does end — whether naturally, or by process.exit with no arguments. |