In ES6, what is the role of the 'package.json' file in module resolution?
- Defines dependencies and module paths
- Specifies the version of ECMAScript used
- Configures package settings
- Determines the module's entry point
The 'package.json' file in ES6 modules plays a crucial role by defining dependencies and specifying module paths. It helps in resolving modules and managing project dependencies effectively. Knowing how 'package.json' influences module resolution is essential for maintaining a well-structured project.
To handle both success and failure in an AJAX call using Promises, use .then() for success and _______ for errors.
- .error()
- .catch()
- .fail()
- .reject()
In Promises, .then() is used for success, and .catch() is used for handling errors. The .catch() method is specifically designed to handle the rejection of a promise, making it the correct choice in this scenario.
What are the implications of not calling super() in a derived class constructor?
- It has no implications
- It results in a syntax error
- It prevents the derived class from inheriting properties and methods
- It leads to a runtime error
Omitting 'super()' in a derived class constructor prevents proper initialization of the object and may lead to unexpected behavior, as the parent class constructor is not invoked. This can cause issues in inheriting properties and setting up the object's state correctly.
How does the temporal dead zone affect variables declared with let and const?
- Variables become undefined
- Variables are hoisted to the top of the scope
- Variables cannot be reassigned
- Variables retain their initial value
The temporal dead zone is a phase during which variables declared with let and const exist but cannot be accessed or assigned. Attempting to use such variables before their declaration results in a ReferenceError. However, the variables are hoisted, meaning their declarations are moved to the top of the scope, but their assignments remain in place. This affects the behavior of code and is crucial to understand for avoiding bugs.
Can you use the arguments object inside an arrow function?
- Yes, arrow functions have access to the arguments object
- No, arrow functions do not have access to the arguments object
- Only if the arrow function is declared with the 'arguments' keyword
- Yes, but only in strict mode
Arrow functions do not have their own 'arguments' object. They inherit it from the enclosing scope, which may lead to unexpected behavior. Option B is correct.
For side effects only, without importing any variables, use import _________ from 'module-name'.
- { sideEffect }
- sideEffect
- * as sideEffect
- sideEffect as default
In this scenario, the correct option is c) * as sideEffect. This syntax allows importing all exports from the 'module-name' without explicitly naming them.
In ES6, what is the behavior of using super in a method when extending a class?
- Calls the superclass constructor
- References the superclass prototype
- Creates a new instance of the superclass
- Invokes the superclass method
In ES6, when using super in a method of a subclass, it refers to the superclass, allowing access to its methods and properties. It is often used to call the superclass constructor using super() or invoke methods on the superclass. This facilitates extending the functionality of the superclass.
What is the base case in a recursive function?
- The initial condition
- The stopping condition
- The recursive condition
- The iterative condition
In a recursive function, the base case is the condition that prevents further recursion. It is crucial for stopping the recursion and ensuring that the function does not enter an infinite loop.
In ES6, mixins can be created by a function that takes a _________ as an argument and extends it.
- Class
- Function
- Object
- Module
In ES6, mixins are created by a function that takes an object as an argument and extends it. This approach allows for dynamic composition, enabling the combination of functionalities from different sources into a single object.
What are the implications of using default parameters on function length property?
- The length property reflects the number of parameters without default values
- The length property includes parameters with default values
- Default parameters do not affect the length property
- The length property is no longer relevant with default parameters
When default parameters are used in a function, they do not contribute to the length property. The length property only counts the number of parameters without default values. Therefore, the length might not accurately represent the total number of arguments the function can accept.