将内置方法分配为 class 属性的奇怪行为

Question

I encountered a strange behavior that I cannot explain when assigning builtin methods as attributes to a class in Python.

If I run the following python file:

class A:
    a = bin
    b = lambda x: bin(x)
    
print(A().a(2))
print(A().b(2))

The call to A().a(2) returns a byte string, but the call to A().b(2) raises:

TypeError: <lambda>() takes 1 positional argument but 2 were given

The signature of the builtin function bin is supposedly bin(number, /) , which seems identical to the signature provided by the lambda above. However, it appears as if A().a is treated as a static method, whereas A().b is treated like an "instance" method (with a self argument implicitly added to the provided lambda). There is an explanation of a similar issue here ( calling a function saved in a class attribute: different behavior with built-in function vs. normal function ), which claims that the reason these two are treated differently is because one is a builtin_function_or_method and the other is a function type.

However, there is inconsistent behavior even within builtins.

class B(int):
    c = pow
    d = round

print(B(1).d(2))
print(B(1).c(2))

In the case of pow and round , round is treated like an instance method while pow is treated as a static method. Both of these builtins are callables capable of taking two unnamed arguments.

This behavior exists across all the versions of Python 2.* and 3.* I've tried.

Answer 1

The answer you've referenced is correct. In the counter-example you gave the two built-in functions are indeed treated the same, that is no bound method object is created:

B(1).d(2) == round(2)  # not round(B(1), 2)
B(1).c(2) == pow(2)  # not pow(B(1), 2)

the issue arises from passing only one argument to pow which takes at least 2, as opposed to round which does only need one