Solution:Step 1: Analyze the child process codeThe child waits 100ms then calls process.exit(2), so it exits with code 2.Step 2: Confirm exit event logs the codeThe exit event logs the exit code, which will be 2 after 100ms.Final Answer:2 -> Option DQuick Check:Delayed exit code is logged correctly

[Solved] Consider this code:const { spawn } = require('child_process');... child.on('exit', (code) => console.log(code));What will be printed after 100ms? | Node.js

Node.js - Child Processes

Consider this code:

const { spawn } = require('child_process');
const child = spawn('node', ['-e', 'setTimeout(() => process.exit(2), 100)']);
child.on('exit', (code) => console.log(code));

What will be printed after 100ms?

A0

Bnull

Cundefined

D2

Step-by-Step Solution

Solution:

Step 1: Analyze the child process code
The child waits 100ms then calls process.exit(2), so it exits with code 2.
Step 2: Confirm exit event logs the code
The exit event logs the exit code, which will be 2 after 100ms.
Final Answer:
2 -> Option D
Quick Check:
Delayed exit code is logged correctly [OK]

Quick Trick: Exit code logs after process.exit call [OK]

Common Mistakes:

Expecting immediate exit code 0
Thinking exit code is null or undefined
Ignoring asynchronous delay

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More Node.js Quizzes

Consider this code:

Step 1: Analyze the child process code

Step 2: Confirm exit event logs the code

Final Answer:

Quick Check:

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