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.
Topic: In mining are the rest of target bytes 0 or 255? (Read 176 times)
If nBits is 0x1b0404cb the target according to wiki is calculated as
so does this mean if the block hash was the following it would be rejected?
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:
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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