test: Add difficulty to boundary conversion method #191
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@@ Coverage Diff @@
## master #191 +/- ##
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+ Coverage 98.52% 98.58% +0.06%
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Files 22 24 +2
Lines 1765 1840 +75
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+ Hits 1739 1814 +75
Misses 26 26
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test/unittests/difficulty.cpp
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| uint32_t d[8]; | ||
| for (int i = 0; i < n; ++i) | ||
| d[i] = ethash::be::uint32(difficulty->word32s[8 - 1 - i]); |
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This is a bit hard to read: not immediate to understand you're swapping also words positions.
| uint32_t d[8]; | |
| for (int i = 0; i < n; ++i) | |
| d[i] = ethash::be::uint32(difficulty->word32s[8 - 1 - i]); | |
| int src{7 /*8-1*/}; | |
| for (int tgt = 0; tgt < n; ++tgt, --src) { | |
| d[tgt] = ethash::be::uint32(difficulty->word32s[src]); | |
| } |
test/unittests/difficulty.cpp
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| // For difficulty of 0 (division by 0) or 1 (256-bit overflow) return max boundary value. | ||
| if (n == 0 || (n == 1 && d[0] == 1)) | ||
| return max_boundary; |
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Maybe this check can be moved before the swapping loop ?
| // For difficulty of 0 (division by 0) or 1 (256-bit overflow) return max boundary value. | |
| if (n == 0 || (n == 1 && d[0] == 1)) | |
| return max_boundary; | |
| // For difficulty of 0 (division by 0) or 1 (256-bit overflow) return max boundary value. | |
| static const be_one = ethash::be::(1u); | |
| if (n == 0 || (n == 1 && difficulty->word32s[7] == be_one) | |
| return max_boundary; |
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I did that at first, but the check is simpler if we have the native order. And where the check is does not matter very much as these cases are only for completeness but do not happen in practice.
test/unittests/difficulty.cpp
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| ethash_hash256 boundary{}; | ||
| for (int i = 0; i < 8; ++i) | ||
| boundary.word32s[i] = ethash::be::uint32(q[8 - 1 - i]); |
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As a personal preference I'd use the suggestion above for better reading
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I moved these to separate function
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