由于闭包和 Python 在对象自省上的灵活性，我们可以提炼出 operator 函数的创建 。
# 一个创建闭包的函数，实现了二元运算的逻辑_MISSING = object()def _create_binary_op(name: str, operator: str) -> Any: """Create a binary operation function. The `name` parameter specifies the name of the special method used for the binary operation (e.g. `sub` for `__sub__`). The `operator` name is the token representing the binary operation (e.g. `-` for subtraction). """ lhs_method_name = f"__{name}__" def binary_op(lhs: Any, rhs: Any, /) -> Any: """A closure implementing a binary operation in Python.""" rhs_method_name = f"__r{name}__" # lhs.__*__ lhs_type = type(lhs) try: lhs_method = debuiltins._mro_getattr(lhs_type, lhs_method_name) except AttributeError: lhs_method = _MISSING # lhs.__r*__ (for knowing if rhs.__r*__ should be called first) try: lhs_rmethod = debuiltins._mro_getattr(lhs_type, rhs_method_name) except AttributeError: lhs_rmethod = _MISSING # rhs.__r*__ rhs_type = type(rhs) try: rhs_method = debuiltins._mro_getattr(rhs_type, rhs_method_name) except AttributeError: rhs_method = _MISSING call_lhs = lhs, lhs_method, rhs call_rhs = rhs, rhs_method, lhs if ( rhs_type is not _MISSING # Do we care? and rhs_type is not lhs_type # Could RHS be a subclass? and issubclass(rhs_type, lhs_type) # RHS is a subclass! and lhs_rmethod is not rhs_method # Is __r*__ actually different? ): calls = call_rhs, call_lhs elif lhs_type is not rhs_type: calls = call_lhs, call_rhs else: calls = (call_lhs,) for first_obj, meth, second_obj in calls: if meth is _MISSING: continue value = meth(first_obj, second_obj) if value is not NotImplemented: return value else: exc = TypeError( f"unsupported operand type(s) for {operator}: {lhs_type!r} and {rhs_type!r}" ) exc._binary_op = operator raise exc
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