mirror of
https://github.com/dashpay/dash.git
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186 lines
5.0 KiB
C++
186 lines
5.0 KiB
C++
// Copyright (c) 2009-2010 Satoshi Nakamoto
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// Copyright (c) 2009-2014 The Bitcoin developers
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// Distributed under the MIT software license, see the accompanying
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// file COPYING or http://www.opensource.org/licenses/mit-license.php.
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#include "compressor.h"
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#include "hash.h"
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#include "pubkey.h"
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#include "script/standard.h"
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bool CScriptCompressor::IsToKeyID(CKeyID &hash) const
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{
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if (script.size() == 25 && script[0] == OP_DUP && script[1] == OP_HASH160
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&& script[2] == 20 && script[23] == OP_EQUALVERIFY
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&& script[24] == OP_CHECKSIG) {
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memcpy(&hash, &script[3], 20);
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return true;
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}
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return false;
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}
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bool CScriptCompressor::IsToScriptID(CScriptID &hash) const
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{
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if (script.size() == 23 && script[0] == OP_HASH160 && script[1] == 20
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&& script[22] == OP_EQUAL) {
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memcpy(&hash, &script[2], 20);
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return true;
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}
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return false;
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}
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bool CScriptCompressor::IsToPubKey(CPubKey &pubkey) const
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{
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if (script.size() == 35 && script[0] == 33 && script[34] == OP_CHECKSIG
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&& (script[1] == 0x02 || script[1] == 0x03)) {
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pubkey.Set(&script[1], &script[34]);
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return true;
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}
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if (script.size() == 67 && script[0] == 65 && script[66] == OP_CHECKSIG
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&& script[1] == 0x04) {
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pubkey.Set(&script[1], &script[66]);
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return pubkey.IsFullyValid(); // if not fully valid, a case that would not be compressible
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}
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return false;
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}
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bool CScriptCompressor::Compress(std::vector<unsigned char> &out) const
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{
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CKeyID keyID;
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if (IsToKeyID(keyID)) {
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out.resize(21);
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out[0] = 0x00;
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memcpy(&out[1], &keyID, 20);
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return true;
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}
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CScriptID scriptID;
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if (IsToScriptID(scriptID)) {
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out.resize(21);
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out[0] = 0x01;
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memcpy(&out[1], &scriptID, 20);
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return true;
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}
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CPubKey pubkey;
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if (IsToPubKey(pubkey)) {
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out.resize(33);
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memcpy(&out[1], &pubkey[1], 32);
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if (pubkey[0] == 0x02 || pubkey[0] == 0x03) {
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out[0] = pubkey[0];
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return true;
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} else if (pubkey[0] == 0x04) {
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out[0] = 0x04 | (pubkey[64] & 0x01);
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return true;
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}
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}
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return false;
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}
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unsigned int CScriptCompressor::GetSpecialSize(unsigned int nSize) const
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{
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if (nSize == 0 || nSize == 1)
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return 20;
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if (nSize == 2 || nSize == 3 || nSize == 4 || nSize == 5)
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return 32;
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return 0;
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}
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bool CScriptCompressor::Decompress(unsigned int nSize, const std::vector<unsigned char> &in)
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{
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switch(nSize) {
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case 0x00:
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script.resize(25);
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script[0] = OP_DUP;
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script[1] = OP_HASH160;
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script[2] = 20;
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memcpy(&script[3], &in[0], 20);
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script[23] = OP_EQUALVERIFY;
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script[24] = OP_CHECKSIG;
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return true;
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case 0x01:
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script.resize(23);
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script[0] = OP_HASH160;
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script[1] = 20;
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memcpy(&script[2], &in[0], 20);
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script[22] = OP_EQUAL;
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return true;
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case 0x02:
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case 0x03:
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script.resize(35);
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script[0] = 33;
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script[1] = nSize;
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memcpy(&script[2], &in[0], 32);
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script[34] = OP_CHECKSIG;
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return true;
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case 0x04:
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case 0x05:
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unsigned char vch[33] = {};
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vch[0] = nSize - 2;
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memcpy(&vch[1], &in[0], 32);
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CPubKey pubkey(&vch[0], &vch[33]);
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if (!pubkey.Decompress())
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return false;
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assert(pubkey.size() == 65);
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script.resize(67);
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script[0] = 65;
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memcpy(&script[1], pubkey.begin(), 65);
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script[66] = OP_CHECKSIG;
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return true;
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}
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return false;
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}
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// Amount compression:
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// * If the amount is 0, output 0
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// * first, divide the amount (in base units) by the largest power of 10 possible; call the exponent e (e is max 9)
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// * if e<9, the last digit of the resulting number cannot be 0; store it as d, and drop it (divide by 10)
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// * call the result n
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// * output 1 + 10*(9*n + d - 1) + e
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// * if e==9, we only know the resulting number is not zero, so output 1 + 10*(n - 1) + 9
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// (this is decodable, as d is in [1-9] and e is in [0-9])
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uint64_t CTxOutCompressor::CompressAmount(uint64_t n)
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{
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if (n == 0)
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return 0;
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int e = 0;
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while (((n % 10) == 0) && e < 9) {
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n /= 10;
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e++;
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}
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if (e < 9) {
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int d = (n % 10);
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assert(d >= 1 && d <= 9);
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n /= 10;
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return 1 + (n*9 + d - 1)*10 + e;
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} else {
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return 1 + (n - 1)*10 + 9;
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}
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}
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uint64_t CTxOutCompressor::DecompressAmount(uint64_t x)
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{
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// x = 0 OR x = 1+10*(9*n + d - 1) + e OR x = 1+10*(n - 1) + 9
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if (x == 0)
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return 0;
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x--;
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// x = 10*(9*n + d - 1) + e
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int e = x % 10;
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x /= 10;
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uint64_t n = 0;
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if (e < 9) {
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// x = 9*n + d - 1
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int d = (x % 9) + 1;
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x /= 9;
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// x = n
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n = x*10 + d;
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} else {
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n = x+1;
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}
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while (e) {
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n *= 10;
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e--;
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}
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return n;
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}
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