fork vs spawn, nodejs

child.js

var i = 0;

setInterval(function() {
  if (i++ % 2) {
    console.log('I\'m the child!');
  } else {
    console.error('I\'m the child!');
  }
}, 1500);

fork.js

var fork = require('child_process').fork;

var child = fork('child.js');

child.on('close', function (code) {
  if (code !== 0) {
    console.log('child process exited with code ' + code);
  }
});

setTimeout(function() {
  throw new Error('ZEPELE');
}, 5000);

spawn.js

var spawn = require('child_process').spawn;

var child = spawn('node', ['child.js']);

child.stdout.on('data', function (data) {
  console.log('stdin:', data.toString());
});

child.stderr.on('data', function (data) {
  console.log('child stderr: ', data.toString());
});

child.on('close', function (code) {
  if (code !== 0) {
    console.log('child process exited with code ' + code);
  }
});

setTimeout(function() {
  throw new Error('ZEPELE');
}, 5000);

depends on the use case, fork create a copy of the v8 engine instance with copy of node_modules, and spawn contains fork and create an entire new instance of V8 engine.