Building React App

Question 1

We can integrate reactJS in any of the existing web app

Accepted Answer

True

Answer

ReactJS is designed to be highly modular and can be integrated into existing web applications incrementally. Developers don't need to rewrite their entire frontend; they can add React components to specific parts of an application, allowing for gradual adoption and coexistence with other frameworks or plain JavaScript.

Question 2

ReactJS does not need HTML or any of the HTML preprocessor to make UI elements

Accepted Answer

False

Answer

ReactJS heavily relies on JSX (JavaScript XML), which is a syntax extension that allows you to write HTML-like code directly within your JavaScript. While it's technically possible to create UI elements using `React.createElement()` without JSX, JSX is the recommended and most common way to define UI structures in React, making it a crucial part of the development workflow.

Question 3

Props can be modified in child component directly like state

Accepted Answer

False

Answer

Props (short for properties) are used to pass data from a parent component to a child component in React. They are read-only within the child component, meaning a child component cannot directly modify the props it receives. This ensures a unidirectional data flow and helps maintain predictable component behavior, unlike state, which is mutable and managed within a component.

Question 4

About hooks it is incorrect to say:

Accepted Answer

Hooks don't work inside classes and allow you to use React without classes.

Answer

Hooks are a feature introduced in React 16.8 that allow functional components to 'hook into' React state and lifecycle features. A key characteristic of Hooks is that they *do not* work inside class components; they are specifically designed for functional components, enabling developers to write React applications entirely with functions and avoid class-based patterns.

Question 5

What is the result of the expression

const [add, setSum] = useState(0)
setSum(add + 4)
setSum(adding - (1x2))

Accepted Answer

2

Answer

Let's trace the state changes: `useState(0)` initializes `add` to 0. The first `setSum(add + 4)` updates the state to `0 + 4 = 4`. Assuming 'adding' is a typo for 'add', the second `setSum(add - (1*2))` then updates the state to `4 - 2 = 2`. Therefore, the final value of the state is 2.

React Practice Test

React Practice Test

Building React App