What happens when the JavaScript engine encounters the 'debugger' statement?
function updateUser(userData) {
debugger;
// User update logic
const updatedUser = {...user, ...userData};
saveToDatabase(updatedUser);
return updatedUser;
}
The debugger statement triggers a powerful debugging mechanism: 1) When encountered, it pauses code execution if a debugging tool (like browser DevTools) is open, 2) It's equivalent to setting a manual breakpoint in the DevTools UI, 3) When execution pauses, the developer can inspect the current state of all variables, 4) The debugging interface allows stepping through code execution line by line, 5) If no debugging tool is open/connected, the statement has no effect and is simply ignored, 6) This provides an easy way to insert breakpoints directly in code without using the DevTools interface, 7) It's especially useful for debugging specific conditions that are hard to reproduce, 8) Unlike console.log(), no code cleanup is needed as the debugger statement has no effect in production when DevTools is closed.