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186 lines (175 loc) · 5.06 KB
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// Copyright (c) 2009-2010 Satoshi Nakamoto
// Copyright (c) 2009-2014 The Bitcoin Core developers
// Copyright (c) 2018-2022 The Flux Developers
// Distributed under the MIT software license, see the accompanying
// file COPYING or https://www.opensource.org/licenses/mit-license.php.
#include "compressor.h"
#include "hash.h"
#include "pubkey.h"
#include "script/standard.h"
bool CScriptCompressor::IsToKeyID(CKeyID &hash) const
{
if (script.size() == 25 && script[0] == OP_DUP && script[1] == OP_HASH160
&& script[2] == 20 && script[23] == OP_EQUALVERIFY
&& script[24] == OP_CHECKSIG) {
memcpy(&hash, &script[3], 20);
return true;
}
return false;
}
bool CScriptCompressor::IsToScriptID(CScriptID &hash) const
{
if (script.size() == 23 && script[0] == OP_HASH160 && script[1] == 20
&& script[22] == OP_EQUAL) {
memcpy(&hash, &script[2], 20);
return true;
}
return false;
}
bool CScriptCompressor::IsToPubKey(CPubKey &pubkey) const
{
if (script.size() == 35 && script[0] == 33 && script[34] == OP_CHECKSIG
&& (script[1] == 0x02 || script[1] == 0x03)) {
pubkey.Set(&script[1], &script[34]);
return true;
}
if (script.size() == 67 && script[0] == 65 && script[66] == OP_CHECKSIG
&& script[1] == 0x04) {
pubkey.Set(&script[1], &script[66]);
return pubkey.IsFullyValid(); // if not fully valid, a case that would not be compressible
}
return false;
}
bool CScriptCompressor::Compress(std::vector<unsigned char> &out) const
{
CKeyID keyID;
if (IsToKeyID(keyID)) {
out.resize(21);
out[0] = 0x00;
memcpy(&out[1], &keyID, 20);
return true;
}
CScriptID scriptID;
if (IsToScriptID(scriptID)) {
out.resize(21);
out[0] = 0x01;
memcpy(&out[1], &scriptID, 20);
return true;
}
CPubKey pubkey;
if (IsToPubKey(pubkey)) {
out.resize(33);
memcpy(&out[1], &pubkey[1], 32);
if (pubkey[0] == 0x02 || pubkey[0] == 0x03) {
out[0] = pubkey[0];
return true;
} else if (pubkey[0] == 0x04) {
out[0] = 0x04 | (pubkey[64] & 0x01);
return true;
}
}
return false;
}
unsigned int CScriptCompressor::GetSpecialSize(unsigned int nSize) const
{
if (nSize == 0 || nSize == 1)
return 20;
if (nSize == 2 || nSize == 3 || nSize == 4 || nSize == 5)
return 32;
return 0;
}
bool CScriptCompressor::Decompress(unsigned int nSize, const std::vector<unsigned char> &in)
{
switch(nSize) {
case 0x00:
script.resize(25);
script[0] = OP_DUP;
script[1] = OP_HASH160;
script[2] = 20;
memcpy(&script[3], &in[0], 20);
script[23] = OP_EQUALVERIFY;
script[24] = OP_CHECKSIG;
return true;
case 0x01:
script.resize(23);
script[0] = OP_HASH160;
script[1] = 20;
memcpy(&script[2], &in[0], 20);
script[22] = OP_EQUAL;
return true;
case 0x02:
case 0x03:
script.resize(35);
script[0] = 33;
script[1] = nSize;
memcpy(&script[2], &in[0], 32);
script[34] = OP_CHECKSIG;
return true;
case 0x04:
case 0x05:
unsigned char vch[33] = {};
vch[0] = nSize - 2;
memcpy(&vch[1], &in[0], 32);
CPubKey pubkey(&vch[0], &vch[33]);
if (!pubkey.Decompress())
return false;
assert(pubkey.size() == 65);
script.resize(67);
script[0] = 65;
memcpy(&script[1], pubkey.begin(), 65);
script[66] = OP_CHECKSIG;
return true;
}
return false;
}
// 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_t CTxOutCompressor::CompressAmount(uint64_t 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_t CTxOutCompressor::DecompressAmount(uint64_t 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_t 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;
}