answer:<tool_call>[{name: load_file_content, arguments: {file_path: /Users/username/Desktop/file1.txt}}, {name: load_file_content, arguments: {file_path: /Users/username/Documents/file2.txt}}]</tool_call>

answer:In TypeScript, you can use inheritance with classes and interfaces through the `extends` and `implements` keywords. Here's a basic example: ```typescript // Parent class class Animal { name: string; constructor(name: string) { this.name = name; } sound(): void { console.log(The animal makes a sound.); } } // Child class class Dog extends Animal { breed: string; constructor(name: string, breed: string) { super(name); // Call the parent class constructor this.breed = breed; } sound(): void { console.log(The dog barks.); } } // Create an instance of the child class const dog = new Dog(Fido, Golden Retriever); console.log(dog.name); // Output: Fido console.log(dog.breed); // Output: Golden Retriever dog.sound(); // Output: The dog barks. ``` In this example, the `Dog` class inherits the `name` property and the `sound()` method from the `Animal` class using the `extends` keyword. The `super()` function is used to call the parent class constructor. You can also use interfaces with inheritance. Here's an example: ```typescript // Interface interface Printable { print(): void; } // Parent class class Document implements Printable { content: string; constructor(content: string) { this.content = content; } print(): void { console.log(this.content); } } // Child class class PDFDocument extends Document { filename: string; constructor(content: string, filename: string) { super(content); // Call the parent class constructor this.filename = filename; } print(): void { console.log(`Printing {this.filename}:`); super.print(); // Call the parent class print method } } // Create an instance of the child class const pdf = new PDFDocument(Hello, World!, example.pdf); pdf.print(); // Output: // Printing example.pdf: // Hello, World! ``` In this example, the `Document` class implements the `Printable` interface, which requires the `print()` method. The `PDFDocument` class extends the `Document` class and also implements the `print()` method, which calls the parent class `print()` method using `super.print()`.

answer:To implement an interface and extend a class at the same time in TypeScript, you can use the `extends` keyword to extend the class and the `implements` keyword to implement the interface. Here's how you can do it: ```typescript class Animal { name: string; constructor(name: string) { this.name = name; } sound(): void { console.log(The animal makes a sound.); } } interface IPrintable { print(): void; } class Dog extends Animal implements IPrintable { constructor(name: string) { super(name); } sound(): void { console.log(The dog barks.); } print(): void { console.log(`Printing dog's name: {this.name}`); } } const dog = new Dog(Fido); dog.sound(); // Output: The dog barks. dog.print(); // Output: Printing dog's name: Fido ``` In this example, the `Dog` class extends the `Animal` class and implements the `IPrintable` interface. The `Dog` class must provide an implementation for the `print()` method, which is required by the `IPrintable` interface. You can also implement multiple interfaces and extend a class at the same time: ```typescript interface IRunnable { run(): void; } class Dog extends Animal implements IPrintable, IRunnable { constructor(name: string) { super(name); } sound(): void { console.log(The dog barks.); } print(): void { console.log(`Printing dog's name: {this.name}`); } run(): void { console.log(`The dog {this.name} is running.`); } } const dog = new Dog(Fido); dog.sound(); // Output: The dog barks. dog.print(); // Output: Printing dog's name: Fido dog.run(); // Output: The dog Fido is running. ```

answer:To call a method from a parent class in TypeScript, you can use the `super` keyword. The `super` keyword is used to access and call functions and properties of a parent class. In your example, you've demonstrated how to call the `speak()` method from the parent class `Animal` using `super.speak()` in the child class `Dog`. Here are a few key points about using `super`: 1. **Calling parent class methods:** To call a method from a parent class, you can use the `super` keyword followed by the method name, as shown in your example: `super.speak()`. 2. **Calling parent class constructor:** When a child class has a constructor, it must call the parent class constructor using `super()` before accessing `this`. In your example, the `Dog` class constructor calls the `Animal` class constructor using `super(name)`. 3. **Overriding parent class methods:** When a child class provides a method with the same name as a parent class method, it overrides the parent class method. In your example, the `Dog` class overrides the `speak()` method of the `Animal` class. Here's a more detailed example: ```typescript class Animal { name: string; constructor(name: string) { this.name = name; } speak(): void { console.log(The animal makes a sound.); } eat(): void { console.log(The animal is eating.); } } class Dog extends Animal { constructor(name: string) { super(name); // Call the parent class constructor } speak(): void { super.speak(); // Call the parent class speak method console.log(The dog barks.); } play(): void { console.log(The dog is playing.); super.eat(); // Call the parent class eat method } } const dog = new Dog(Fido); dog.speak(); // Output: // The animal makes a sound. // The dog barks. dog.play(); // Output: // The dog is playing. // The animal is eating. ``` In this example, the `Dog` class calls the `speak()` and `eat()` methods from the parent class `Animal` using `super.speak()` and `super.eat()`, respectively. The `Dog` class also overrides the `speak()` method and adds a new method `play()`.