In this example network from pyTorch tutorial

import torch
import torch.nn as nn
import torch.nn.functional as F

class Net(nn.Module):

    def __init__(self):
        super(Net, self).__init__()
        # 1 input image channel, 6 output channels, 3x3 square convolution
        # kernel
        self.conv1 = nn.Conv2d(1, 6, 3)
        self.conv2 = nn.Conv2d(6, 16, 3)
        # an affine operation: y = Wx + b
        self.fc1 = nn.Linear(16 * 6 * 6, 120)  # 6*6 from image dimension
        self.fc2 = nn.Linear(120, 84)
        self.fc3 = nn.Linear(84, 10)

    def forward(self, x):
        # Max pooling over a (2, 2) window
        x = F.max_pool2d(F.relu(self.conv1(x)), (2, 2))
        # If the size is a square you can only specify a single number
        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:]  # all dimensions except the batch dimension
        num_features = 1
        for s in size:
            num_features *= s
        return num_features

net = Net()

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

Why is the method forward() not explicitely called? I mean how does just calling net(output) calls forward() ? (which is what happens as far as I understand) By the way I dont understand what this line means:

super(Net, self).__init__()

I can imagine super() is calling the constructor of a parent class but …?


1 Answer 1


If you look at the Module implementation of pyTorch, you'll see that forward is a method called in the special method __call__ :

class Module(object):
   def __call__(self, *input, **kwargs):
      result = self.forward(*input, **kwargs)

As you construct a Net class by inheriting from the Module class and you override the default behavior of the __init__ constructor, you also need to explicitly call the parent's one with super(Net, self).__init__().

  • $\begingroup$ Thanks, great answer. This is something they could have put in the comments. $\endgroup$
    – SheppLogan
    Commented Aug 30, 2019 at 12:02
  • $\begingroup$ But so the __call__ merhod is called by the constructor of the Module class if i understood right? $\endgroup$
    – SheppLogan
    Commented Aug 30, 2019 at 12:05
  • 2
    $\begingroup$ Nope, it is called when you call an instance of the class, so in the example : out = net(input). See : stackoverflow.com/questions/9663562/… $\endgroup$
    – Elliot
    Commented Aug 30, 2019 at 12:08
  • $\begingroup$ I dont see any explicit call to _\call_ if i can say so lol $\endgroup$
    – SheppLogan
    Commented Aug 30, 2019 at 12:09
  • 1
    $\begingroup$ great thanks for your fantastic answers! $\endgroup$
    – SheppLogan
    Commented Aug 30, 2019 at 13:21

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