I have the following code.
self.bn1 = nn.BatchNorm2d(2)
self.conv1 = nn.Conv2d(2, 20, kernel_size=1)
self.bn2 = nn.BatchNorm2d(20)
self.conv2 = nn.Conv2d(20, 40, kernel_size=3)
self.active = True
def forward(self, x):
    if not self.active:
        self.eval()
        out = self.conv1(F.relu(self.bn1(x)))
        out = self.conv2(F.relu(self.bn2(out)))

        out.detach()

Does  out.detach() mean that, for next iteration, the forward pass will not be computed?
I know detach will prevent the back propagation computation.