Скачать или смотреть Understanding Why Builtins Can't Become Bound Methods in Python

Discover the reasons behind the behavior of built-in functions when they are assigned as methods in Python. Learn how to make a lambda function behave like a built-in function using static methods.
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Understanding Why Builtins Can't Become Bound Methods in Python

In the world of Python programming, built-in functions and methods behave in certain, sometimes unexpected, ways. A common point of confusion arises around the question: Why can't builtins become bound methods? Let's unpack this issue with a clear understanding of the underlying concepts and provide a practical solution.

The Problem Explained

Consider the following class structure and function:

[[See Video to Reveal this Text or Code Snippet]]

When we assign the lambda function f as a method of the class Something:

[[See Video to Reveal this Text or Code Snippet]]

You might expect it to work seamlessly, but it results in an error:

[[See Video to Reveal this Text or Code Snippet]]

What Happened Here?

When you assign f to Something.open, Python treats open as a bound method. This means that when you call g.open(), Python automatically passes the instance g as the first argument, which is referred to as self. Since f is defined to take no arguments, the mismatch leads to an error.

A Comparison with Built-in Functions

Now, let's consider built-in functions. If you assign the built-in open function to Something.open:

[[See Video to Reveal this Text or Code Snippet]]

In this case, you get a different error:

[[See Video to Reveal this Text or Code Snippet]]

Here, the open function does not get bound to an instance, and thus it requires an actual file path (or other arguments), not an instance of the class. This showcases that not all functions are meant to work the same way when assigned as methods.

Why Is There a Difference?

The key difference between a custom function (like f) and a built-in function (like open) lies in how Python handles method binding:

For Lambdas (Custom Functions): They become bound methods and expect self because Python assumes you're trying to create instance-specific behavior.

For Built-in Functions: They maintain their standalone behaviors and expect their original parameters but do not accept an instance.

The Solution

If you want your lambda function f to behave like a built-in function (i.e., to not receive a self parameter), you can use a staticmethod. Here's how you can do it:

[[See Video to Reveal this Text or Code Snippet]]

What This Does:

The staticmethod decorator allows your method to be accessed without affecting or requiring an instance (like g).

As a result, calling g.open() will not pass the instance, and everything will work correctly.

Conclusion

Understanding the distinction between bound methods and standalone functions in Python can help you navigate potential pitfalls. The use of staticmethod allows flexibility, making it possible to define functions that behave as if they're built-in, removing the notion of requiring an instance. So, while built-ins don't naturally become bound methods, there is a straightforward approach to achieve similar behavior using static methods.

By mastering these nuances, you can write more effective and error-free Python code!