CBC Mode error during encryption In CBC mode I know that if i get a corrupted Ci* block than Pi and Pi+1 will be wrong during the decryption, but i believe that this happens only when Ci* is not envolved during the feedback stage at the encryption, my question is what happens if the corrupted block is a feedback at the encryption stage, will it have the same impact as before?
As you can see in the Wikipedia article a ciphertext block is used for two different things during decryption:


as input to the block cipher to calculate the output, which is then XOR'ed with the previous ciphertext block to create the corresponding plaintext block at the same position;
as input to the XOR to calculate the next plaintext block, together with the next decrypted block.


You can see it clearly in the picture, by following the arrows of one of the ciphertext blocks:



(XOR is denoted by $\oplus$ )

As the block cipher is a pseudo random permutation the plaintext block at the same position is indistinguishable from random even if only one bit is changed. The fact that it is XOR'ed with the value of the previous ciphertext block does not change this fact. There is only one restriction: the result is not identical to the plaintext block that would be calculated without the change.

Things are different with the next plaintext block. Only those bits that have flipped in the ciphertext block will be flipped in the next plaintext block. So the change is localized and not random.

Next or previous blocks are not affected as the affected ciphertext block, intermediate values or plaintext blocks are not used as input anywhere else.