How can I create a class that implements a specific interface in Python?

To create a class that implements a specific interface in Python, you can use the following steps:

1. Define the interface as a protocol with the `@objc protocol` attribute. The protocol should define the methods that the implementing class must implement.

2. Define the class that will implement the interface.

3. Use the `@interface` attribute to declare the class and the `: ` syntax to specify that the class implements the interface.

4. Implement the methods defined in the interface in the class.

Here is an example:

@objc protocol MyProtocol

– (void)doSomething;

@end

@interface MyClass: NSObject

– (void)doSomething;

@end

@implementation MyClass

– (void)doSomething {

// Implement the method.

}

@end
​

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