I writed a net about seq2seq .It have two class, one is encoder,the other is decoder
The loss backpropagation Two times , the net is ok, but third time,the loss backpropagation occur a error

This is the error:
RuntimeError: Trying to backward through the graph a second time, but the buffers have already been freed. Specify retain_graph=True when calling backward the first time.

This is my code:
for epoch in range(20):
for k in range(6250):

if (k+1)%5==0:
en_input=to_cuda(total_in[:,k-4:k+1,:])
target_out=to_cuda(total_out[:,k-4:k+1,:])
#==========encoder============================
output,hidden_out=encoder(en_input) #### this is my encoder
de_input=output[:,-1,:]
#==========decoder============================
for i in range(5):
de_input=to_cuda((de_input.view(3,1,25)))
output,hidden_out=decoder(de_input)##### This is my decoder
out=torch.cat((out,output),1)
out=out[:,1:6,:]
#==========loss backward======================
loss=criterion(out,target_out)
optimizer_en.zero_grad()
optimizer_de.zero_grad()
print(loss) ## first time:loss 0.4317 second time loss 0.4834
loss.backward()
optimizer_en.step()
optimizer_de.step()

If I write this code ,it will occur that error
optimizer.zero_grad()
loss.backward()
loss.backward()
optimizer.step()
I think ,it use the last time loss to backpropagation ,so this state happen.but I writed the code is this:
loss=criterion(out,target_out)
optimizer_en.zero_grad()
optimizer_de.zero_grad()
print(‘loss %.4f’%(loss))
loss.backward()
optimizer_en.step()
optimizer_de.step()

This error will occur if any part of your computational graph is shared with the previous iteration.
For example here, if out is not resetted between iterations, you will backpropagate twice in this part of the graph.

Don’t do that
You should not share part of the graph like that in general.
If in your particular case, you actually want to do it, use the retain_graph=True flag when you call backward.