Merkle root with one transaction

Question

I am trying to build a merkle root which contains only the coinbase transaction from bitcoind.

After creating the coinbase transaction, do I need to convert it to little endian, hash it twice and then put it in the blockheader as it is, or do I have to it reconvert it again before I put it in the blockheader?

I mean:

step 1: convert coinbase transaction to little endian
step 2: double SHA256
step 3: reconvert to little endian or let it as it's?
step 4: put it on the blockheader

I am a little confused

Thank you for help

Murch · Accepted Answer · 2021-09-27T15:14:28.243

2

Whenever data is serialized to bytes in Bitcoin, the bytes are represented in little endian. (Data that is already in bytes does not need to be serialized or otherwise converted.)

When there is only a single element in the merkle tree, the root of the tree is the hash of that element. Therefore, you would only need to serialize the coinbase transaction and hash the bytes resulting from the serialization. The resulting hash is the coinbase's txid as well as the merkle root.

edited Sep 27 '21 at 15:14

answered Jul 24 '20 at 18:01

Murch

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ah, thank you, i thought we do this when it's an odd number of transactions, and according to the documentation, it says when it contains only the coinbase transaction, the merkle root will be the same – Hamita Jul 24 '20 at 18:14

score 0 · Answer 2 · answered Jul 24 '20 at 18:48

0

I put the answer for my question here, if someone else need it, the right answer is: Convert before hashing and after hashing

answered Jul 24 '20 at 18:48

Hamita

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Merkle root with one transaction

2 Answers2

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