[Solved] Given this Angular component code, which part will be removed by tree shaking if unused?import { Component } from '@angular/core'; @Component({ selector: 'app-test', template:... | Angular

Angular - Performance Optimization

Given this Angular component code, which part will be removed by tree shaking if unused?

import { Component } from '@angular/core';

@Component({ selector: 'app-test', template: 'Test' })
export class TestComponent {
  usedMethod() { return 'used'; }
  unusedMethod() { return 'unused'; }
}

AusedMethod()

BBoth methods

CNeither method

DunusedMethod()

Step-by-Step Solution

Solution:

Step 1: Identify which methods are called
Only usedMethod() is called or referenced; unusedMethod() is not.
Step 2: Understand tree shaking effect
Unused methods like unusedMethod() get removed to reduce bundle size.
Final Answer:
unusedMethod() -> Option D
Quick Check:
Unused code removed = unusedMethod() [OK]

Quick Trick: Unused methods get removed by tree shaking [OK]

Common Mistakes:

Assuming all methods stay regardless of usage
Thinking usedMethod() is removed
Believing tree shaking removes entire classes

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More Angular Quizzes

Given this Angular component code, which part will be removed by tree shaking if unused?

Step 1: Identify which methods are called

Step 2: Understand tree shaking effect

Final Answer:

Quick Check:

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