diff options
author | Pieter Wuille <pieter.wuille@gmail.com> | 2012-06-16 13:36:00 +0200 |
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committer | Pieter Wuille <pieter.wuille@gmail.com> | 2012-10-20 23:08:56 +0200 |
commit | 0fa593d0fb3a6f731a31610e98ce6bfd6a50a96c (patch) | |
tree | e7c296456e910c22bf09dceee3291f6a30fa3c50 /src/main.cpp | |
parent | 69fc8047a9a96bcf5360f810c796049c27e16fcd (diff) |
Compact serialization for amounts
Special serializer/deserializer for amount values. It is optimized for
values which have few non-zero digits in decimal representation. Most
amounts currently in the txout set take only 1 or 2 bytes to
represent.
Diffstat (limited to 'src/main.cpp')
-rw-r--r-- | src/main.cpp | 54 |
1 files changed, 54 insertions, 0 deletions
diff --git a/src/main.cpp b/src/main.cpp index 43bb4b1793..ed677f31c8 100644 --- a/src/main.cpp +++ b/src/main.cpp @@ -4058,3 +4058,57 @@ void GenerateBitcoins(bool fGenerate, CWallet* pwallet) } } } + +// Amount compression: +// * If the amount is 0, output 0 +// * first, divide the amount (in base units) by the largest power of 10 possible; call the exponent e (e is max 9) +// * if e<9, the last digit of the resulting number cannot be 0; store it as d, and drop it (divide by 10) +// * call the result n +// * output 1 + 10*(9*n + d - 1) + e +// * if e==9, we only know the resulting number is not zero, so output 1 + 10*(n - 1) + 9 +// (this is decodable, as d is in [1-9] and e is in [0-9]) + +uint64 CTxOutCompressor::CompressAmount(uint64 n) +{ + if (n == 0) + return 0; + int e = 0; + while (((n % 10) == 0) && e < 9) { + n /= 10; + e++; + } + if (e < 9) { + int d = (n % 10); + assert(d >= 1 && d <= 9); + n /= 10; + return 1 + (n*9 + d - 1)*10 + e; + } else { + return 1 + (n - 1)*10 + 9; + } +} + +uint64 CTxOutCompressor::DecompressAmount(uint64 x) +{ + // x = 0 OR x = 1+10*(9*n + d - 1) + e OR x = 1+10*(n - 1) + 9 + if (x == 0) + return 0; + x--; + // x = 10*(9*n + d - 1) + e + int e = x % 10; + x /= 10; + uint64 n = 0; + if (e < 9) { + // x = 9*n + d - 1 + int d = (x % 9) + 1; + x /= 9; + // x = n + n = x*10 + d; + } else { + n = x+1; + } + while (e) { + n *= 10; + e--; + } + return n; +} |