3

How do you generate an integrated address using a subaddress? I'm getting

Error: usage: integrated_address [payment ID]

enter image description here

1 Answer 1

5

It was decided that there should not be such a thing as an integrated subaddress. Only your main Monero address can become an integrated address (this functionality is retained for backwards compatibility).

The reason for this is that integrated addresses stand out on the blockchain. It is possible to see the difference between a payment sent to a standard Monero address vs an integrated address. This is because when an integrated address is used, an encrypted payment id appears in the "tx_extra" part of the transaction, and that makes the transaction stand out.

Since you can create as many subaddresses as you want, you can create an entire subaddress just for a particular sender. This means that you don't need to rely on a special payment id to identify the sender, because your wallet is able to correlate funds received to the subaddresses that you have given out.

2
  • 1
    So your saying payment ids will be obsolete by subaddresses then? Meaning exchanges will migrate to this? Commented Mar 29, 2018 at 18:53
  • 4
    @PatoshiパトシThe backwards compatibility means that exchanges could just ignore subaddresses and keep their current method of using integrated address variants of their main Monero address. But now they have options. A huge advantage of subaddresses for ordinary people is that they can give out different subaddresses to different people, and those people can't tell that the addresses are linked. If you gave out integrated addresses, however, people could easily see that they've just been given an integrated address variant of the same underlying Monero main wallet address.
    – knaccc
    Commented Mar 29, 2018 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.