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Author Topic: In mining are the rest of target bytes 0 or 255?  (Read 152 times)
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July 22, 2019, 07:08:18 AM
Merited by ABCbits (1)
 #1
If nBits is 0x1b0404cb the target according to wiki is calculated as
Code:
0x00000000000404CB000000000000000000000000000000000000000000000000

so does this mean if the block hash was the following it would be rejected?
Code:
0x00000000000404CB000000000000000000000000000000000000000000000001

or do we set the remaining bytes in target to 0xff and then compare it with hash result? (Which means the above block hash could be accepted)
Code:
0x00000000000404CBFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
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July 22, 2019, 07:18:56 AM
Merited by achow101 (2), vapourminer (1), ABCbits (1), Coding Enthusiast (1)
 #2
As shown in

https://github.com/bitcoin/bitcoin/blob/d0f81a96d9c158a9226dc946bdd61d48c4d42959/src/arith_uint256.cpp#L213-L214

the rest of the bits remain 0. So that one is indeed rejected.
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