在运行时附加__call__不起作用

Question

我正在尝试在运行时定义__call__ dunder方法，但没有成功。代码如下：

class Struct:
    pass

result=Struct()
dictionary={'a':5,'b':7}
for k,v in dictionary.items():
    setattr(result,k,v)

result.__call__=lambda self: 2

但是，解释器返回错误：

Traceback (most recent call last):
 File "<input>", line 1, in <module>
TypeError: 'Struct' object is not callable

但是，如果我从一开始就添加dunder方法，那么所有神奇的工作都会起作用：

class Foo():
   def __call__(self):
      return 42

foo=Foo()
foo() #returns 42

我在Windows 64位计算机上使用Python 3.4。

我在哪里做错了？

Answer 1

<强>被修改

您可以通过将var ajax = function (arg) { if (typeof arg.method !== "undefined" && typeof arg.url !== "undefined" && typeof arg.async !== "undefined" && typeof arg.success !== "undefined" && typeof arg.data !== "undefined") { var xmlhttp, i = 0, versions = [ "MSXML2.XmlHttp.6.0", "MSXML2.XmlHttp.5.0", "MSXML2.XmlHttp.4.0", "MSXML2.XmlHttp.3.0", "MSXML2.XmlHttp.2.0", "Microsoft.XmlHttp" ]; if (window.XMLHttpRequest) { xmlhttp = new XMLHttpRequest(); } else { for ( ; i < versions.length; i++) { try { xmlhttp = new ActiveXObject(versions[i]); break; } catch (e) { } } } xmlhttp.onreadystatechange = function () { if (xmlhttp.readyState == XMLHttpRequest.DONE) { if (xmlhttp.status == 200) { arg.success(xmlhttp.responseText); } else if (xmlhttp.status == 400) { console.log("There was an error 400"); } else { console.log("UNSUCCESSFUL"); } } } xmlhttp.open(arg.method, arg.url, arg.async); xmlhttp.send(arg.data); console.log("Method: " + arg.method + "\nURL: " + arg.url + "\nAsync: " + arg.async + "\nData: " + arg.data + "\n"); } }; 添加到类对象来附加__call__：

Struct.__call__ = lambda self: 2

但是如果你想为每个实例获得不同的值，你应该：

class Struct:
    def __call__(self):
        return self._call_ret

result=Struct()
dictionary={'a':5,'b':7}
for k,v in dictionary.items():
    setattr(result,k,v)
    result._call_ret = 2

print(result())

@Blckknght谢谢。