Python 如何将参数传递到构造函数中?

Python 如何将参数传递到构造函数中?,python,Python,此代码: class Net(nn.Module): def __init__(self): super(Net, self).__init__() self.conv1 = nn.Conv2d(1,6,5) self.conv2 = nn.Conv2d(6, 16, 5) self.fc1 = nn.Linear(16*5*5, 120) self.fc2 = nn.Linear(120, 84)

此代码:

class Net(nn.Module):
    def __init__(self):
        super(Net, self).__init__()

        self.conv1 = nn.Conv2d(1,6,5)
        self.conv2 = nn.Conv2d(6, 16, 5)
        self.fc1 = nn.Linear(16*5*5, 120)
        self.fc2 = nn.Linear(120, 84)
        self.fc3 = nn.Linear(84, 10)

     def forward(self, x):
            x = F.max_pool2d(F.relu(self.conv1(x)), (2,2))
            x = F.max_pool2d(F.relu(self.conv2(x)), 2)
            x = x.view(-1, self.num_flat_features(x))
            x = F.relu(self.fc1(x))
            x = F.relu(self.fc2(x))
            x = self.fc3(x)

            return x

    def num_flat_features(self, x):
        size = x.size()[1:]
        num_features = 1
        for s in size:
            num_features *= s

        return num_features

net = Net()

input = torch.randn(1,1,32,32)
out = net(input)

print(out)
我正在学习python并试图理解这个构造函数是如何工作的。我的问题有两行:

input = torch.randn(1,1,32,32)
out = net(input)
在init初始化中,我看不到如何使用“input”进行初始化

net = Net()
在不带参数的情况下调用_init__方法

out = net(input)
以输入为参数调用_call__方法。 由于Net没有实现这一点,它必须在基类nn.Module中实现


您可以找到nn.Module的源,并且有一个用输入作为参数定义的调用。

您不是在向类传递参数,而是在向对象传递参数。有区别

下面的示例显示了如何实现这一点。您需要实现调用方法

查看“调用”方法在哪里搜索呼叫__
class CallableClass:
    def __init__(self):
        pass


    def __call__(self, *args, **kwargs):
        print(args)


class Net(CallableClass):
    def __init__(self):
        super(Net, self).__init__()
        pass

net = Net()
net(100)