在instanciation时没有调用metaclass`__call__`方法

Question

我试图理解元类。

我已阅读json_decode，我尝试使用这样的自定义__call__：

#!/usr/bin/env python3                                                                                               

class MetaTest(type):
    def __call__(cls, *args, **kwargs):
        print("MetaTest.__call__ called.")
        return super().__call__(cls, *args, **kwargs)


class Test:
    __metaclass__ = MetaTest


if __name__ == '__main__':
    Test()

我希望在实例时调用Test.__class__.__call__，Test.__class__成为MetaTest的实例，但事实并非如此，因为我没有运行此脚本时的输出。

那么为什么MetaTest.__call__在Test实施时被func createFirstRowView() { let b = SKSpriteNode(imageNamed: "obj_bg_card1") b.xScale = 0.8 b.yScale = 0.8 r = SKShapeNode(rectOf: CGSize(width: b.size.width*4, height: b.size.height)) r.position = CGPoint(x: frame.midX, y: frame.midY) //r.strokeColor = UIColor(red: 0, green: 0, blue: 0, alpha: 0) self.addChild(r) for i in 0..<4{ let bg = SKSpriteNode(imageNamed: "obj_bg_card1") bg.xScale = 0.8 bg.yScale = 0.8 let lel = 0.5 + (Double(i)*1) bg.position = CGPoint(x: -bg.size.width*2 + bg.size.width*CGFloat(lel), y: 0) r.addChild(bg) } r.position = CGPoint(x: frame.midX, y: frame.midY*1.4) let q = SKSpriteNode(imageNamed: "obj_bg_card1") q.xScale = 0.8 q.yScale = 0.8 t = SKShapeNode(rectOf: CGSize(width: q.size.width*4, height: q.size.height)) t.position = CGPoint(x: frame.midX, y: frame.midY) //t.strokeColor = UIColor(red: 0, green: 0, blue: 0, alpha: 0) self.addChild(t) for k in 0..<4{ let bg = SKSpriteNode(imageNamed: "obj_bg_card1") bg.xScale = 0.8 bg.yScale = 0.8 let lel = 0.5 + (Double(k)*1) bg.position = CGPoint(x: -bg.size.width*2 + bg.size.width*CGFloat(lel), y: 0) t.addChild(bg) } t.position = CGPoint(x: frame.midX, y: frame.midY*0.6) } 调用？

Answer 1

您正在为元类使用Python 2语法 - 但您的shebang行和标记表明您在Python 3上运行它。

只需更改您的类声明，即可使用Python＆＃39;声明元类。 3路：

#!/usr/bin/env python3                                                                                               

class MetaTest(type):
    def __call__(cls, *args, **kwargs):
        print("MetaTest.__call__ called.")
        return super().__call__(cls, *args, **kwargs)


class Test(metaclass=MetaTest):
    pass

if __name__ == '__main__':
    Test()

在类体上声明__metaclass__属性在Python 3-中没有任何效果，但由于它只是一个属性声明，因此它也不会引发任何错误。