I'm thinking about how the Target value is truncated to and later compared to the SHA-256hash of the Block Header.
Given that
The maximum target (lowest possible difficulty) is
0x00000000FFFF0000000000000000000000000000000000000000000000000000
and the current target is
0x00000000000004985C0000000000000000000000000000000000000000000000
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ (4bits * 32 = 128)
This part is always zero due to truncation
Question:
- Is it simply there to make coding and evaluation easier?
- Does this make the mining operation faster (more efficient comparisons)?
- Are the efficiencies of the previous bullet endian-specific?
- Are there statistical benefits or drawbacks to making it non-zero? (fractionally easier to find a block?)
I haven't seen a thoughtful explanation of why the decision was made, and I don't want to infer anything on my own.