C++實(shí)現(xiàn)區(qū)塊鏈的源碼

更新時(shí)間：2021年01月20日 09:48:19 作者：程序小黑

這篇文章主要介紹了C++實(shí)現(xiàn)區(qū)塊鏈的源碼,本文通過實(shí)例代碼給大家介紹的非常詳細(xì)，對(duì)大家的學(xué)習(xí)或工作具有一定的參考借鑒價(jià)值，需要的朋友可以參考下

看了上面的算法，相信大家基本可以猜到，相對(duì)于比特幣的限量的性質(zhì)，對(duì)于本算法來說，難解程度的根本原因即為向量環(huán)路的迭代次數(shù)。迭代次數(shù)越多，則算法越難解，從而導(dǎo)致解題需要花費(fèi)更多的時(shí)候，再基于這點(diǎn)，在數(shù)學(xué)上，當(dāng)解題次數(shù)足夠大時(shí)，效率會(huì)無限小，從而導(dǎo)致了解題時(shí)間無限長(zhǎng)最后導(dǎo)致加密貨幣的發(fā)放無限小。
創(chuàng)世區(qū)塊創(chuàng)建（部分大媽在前面有實(shí)現(xiàn)，而區(qū)塊這一部分將會(huì)詳細(xì)解答）

void Make_First_Block()
{
Getpublickey();
blo.data = circle;
blo.pre_hash = 0;
blo.this_hash = (blo.pre_hash+public_Key) * (a+b);
Block.push_back(blo);
}

由于在區(qū)塊鏈中，本區(qū)快的數(shù)字簽名是基于上一區(qū)塊的數(shù)字簽名和區(qū)塊本身的DATA決定，所以，在這里我們采用了上一區(qū)塊的數(shù)字簽名加上難解的PublicKey乘上長(zhǎng)軸和短軸的和實(shí)現(xiàn)本區(qū)塊的數(shù)字簽名的算法。
添加區(qū)塊（當(dāng)當(dāng)前區(qū)塊被算出時(shí)，添加新區(qū)塊，檢查簽名正確性。）

void Append_Block()
{
pre_blo = blo;
bool flag = true;
auto temp = public_Key;
circle = circle + 1;
Getpublickey();
blo.data = circle;
blo.pre_hash = blo.this_hash;
blo.this_hash = (blo.pre_hash + public_Key) * (a + b);
for(list::iterator itor = Block.begin(); itor != Block.end(); itor++)
{
if ((*itor).this_hash != (*itor).pre_hash + temp * (a + b))
{
flag = false;
break;
}
}
if (flag) { Block.push_back(blo); };
}

這個(gè)迭代其實(shí)可以不用的，因?yàn)槲以谕獠窟€定義了一個(gè)block類型的全局變量Pre_block和blo。Pre_block存儲(chǔ)了上一個(gè)區(qū)塊的信息。而本區(qū)塊的信息則存儲(chǔ)在Blo中。只有當(dāng)用戶解出當(dāng)前區(qū)塊后，才可以得到新區(qū)塊。而data參數(shù)，為了方便僅存儲(chǔ)了當(dāng)前區(qū)塊所在的位置。
區(qū)塊的計(jì)算（用類實(shí)現(xiàn)）

class Get_Block :Create_Block {
public:
int diffcult;
int number = 1;
Get_Block():Create_Block(“OK”){
}
void calc()
{
double start = clock();
while (true){
for (unsigned long long z = 1; z < ULLONG_MAX; z++){
for (unsigned long long j = 1; j < 65535; j = j + 1) {
for (unsigned long long i = 1; i < 65535; i = i + 1) {
Cryptography *person = new Cryptography(i,j,z);
person->Getpublickey();
block bloc;
bloc.data = circle;
bloc.pre_hash = pre_blo.this_hash;
bloc.this_hash = (blo.pre_hash + person->public_Key) * (i + j);
if (blo.data == bloc.data &&blo.pre_hash== bloc.pre_hash && blo.this_hash == bloc.this_hash)
{
double end = clock();
cout << “歷時(shí)”<<end-start<<“毫秒獲得的第” << number++ <<“個(gè)區(qū)塊信息為：” << endl;
cout << “data:” << bloc.data << endl;
cout << “this_hash:” << bloc.this_hash << endl;
cout << “pre_hash:” << bloc.pre_hash << endl << “=======================” << endl << endl;
this->Append_Block();
start = clock();
}
delete []person;
}
}
}
}
}
};

完整代碼：

#include 
#include <stdio.h>
#include <windows.h>
#include 
#include 
#include 
#include 
#include <time.h>
using namespace std;

struct Moving_Point {
unsigned long long x;
unsigned long long y;
};

int circle = 1;

class Martix {
public:
static const int circle_s = 1; //假定向量環(huán)路為1;
static const int KEY =Martix::circle_s * 8;
private:
unsigned long long martix_4_2[Martix::KEY / 2][2]; //存儲(chǔ)向量矩陣
unsigned long long martix_8_8[Martix::KEY][Martix::KEY]; //存儲(chǔ)由向量矩陣得到的轉(zhuǎn)置矩陣
unsigned long long martix_complete[KEY * 2]; //存儲(chǔ)操作完成后的矩陣（一維）

public:
Martix(string a) {};
Martix(int a, int b,int circle)
{
int key = 8;
int cir = circle;
while (cir–)
{
martix_4_2[key / 2 - 4][0] = (-1)*b; martix_4_2[key / 2 - 4][1] = (-1)*a;
martix_4_2[key / 2 - 3][0] = b; martix_4_2[key / 2 - 3][1] = (-1)*a;
martix_4_2[key / 2 - 2][0] = b; martix_4_2[key / 2 - 2][1] = a;
martix_4_2[key / 2 - 1][0] = (-1)*b; martix_4_2[key / 2 - 1][1] = a;
key += 8;
}
}
void Change_New_Martix() {
for (int i = 0; i < 2; i++)
{
for (int j = 0; j < 2; j++)
{
martix_8_8[i][j] = 0;
}
}
for (int j = 2; j < KEY / 2 + 2; j++) {
martix_8_8[0][j] = martix_4_2[j - 2][0] * (-1);
martix_8_8[1][j] = martix_4_2[j - 2][1] * (-1);
}
for (int i = 2; i < KEY / 2 + 2; i++) {
martix_8_8[i][0] = martix_4_2[i - 2][0] * (-1);
martix_8_8[i][1] = martix_4_2[i - 2][1] * (-1);
}
for (int i = 2; i < KEY / 2 + 2; i++)
{
for (int j = 2; j < KEY / 2 + 2; j++)
{
martix_8_8[i][j] = 0;
}
}
}
public:
void Save_Martix()
{
int key = 0;
for (int i = 0; i < KEY / 2 + 2; i++)
{
for (int j = 0; j < KEY / 2 + 2; j++)
{
if (martix_8_8[i][j] != 0)
{
martix_complete[key++] = martix_8_8[i][j];
}
}
}
}
unsigned long long GetPublicKey()
{
unsigned long long public_key = martix_complete[0];
for (int i = 1; i < KEY * 2; i++)
{
if (i % 2 == 0)
{
public_key = public_key + martix_complete[i];
}
else {
public_key = public_key * martix_complete[i];
}
}
return public_key;
}
};
class Cryptography :Martix
{
public:
/作為私鑰，發(fā)送方保存內(nèi)容/
unsigned long long a; //橢圓長(zhǎng)軸的半軸長(zhǎng)度
unsigned long long b; //橢圓短軸的半軸長(zhǎng)度
/作為公鑰，接收方接受公鑰/
unsigned long long public_Key; //通過橢圓矩陣算法得到的公鑰G
Moving_Point p; //隨機(jī)選定的在橢圓上的點(diǎn)

public:
Cryptography(string a) :Martix(“OK”) {};
Cryptography(unsigned long long in_a, unsigned long long in_b,int diffcult) :Martix(in_a, in_b,diffcult)
{
this->a = in_a;
this->b = in_b;
p.x = 0;
p.y = 0;
public_Key = Getpublickey();
}
unsigned long long Getpublickey()
{
Get_Public_Key();
return public_Key;
}
Moving_Point GetPoint()
{
Get_Point();
return p;
}
public:
void PrintPrivateKey() {
cout << “#############私鑰:#############” << endl;
cout << “長(zhǎng)軸：” << 2this->a << “\t\t”;
cout << “短軸：” << 2this->b << endl;
}
private:
void Get_Point()
{
if (p.x == 0 && p.y == 0)
{
while (!Is_Moving_Point())
{
Get_Moving_Point_P();
}
}
}
void Get_Public_Key()
{
this->Change_New_Martix();
this->Save_Martix();
this->public_Key = this->GetPublicKey();
}
void Get_Moving_Point_P() //得到一個(gè)隨機(jī)的在橢圓上的點(diǎn)的坐標(biāo)
{
for (int i = 0; i < this->a; i++)
{
for (int j = 0; j < this->b; j++)
{
p.x = i;
p.y = j;
}
}
}
bool Is_Moving_Point() {
if (pow(b, 2)*pow(p.y, 2) + pow(a, 2)*pow(p.x, 2) == pow(a, 2)*pow(b, 2) && p.y <= a && p.x <= b)
return true;
else
return false;
}
};
struct block {
unsigned long long this_hash;
unsigned long long pre_hash;
unsigned long long data;
};

block blo;
block pre_blo = {0,0,0};
class Create_Block:public Cryptography {
public:
list Block;
public:
Create_Block(string a):Cryptography(“OK”) {};
Create_Block(int x = rand()*2, int y = rand(), int diffcult = 1):Cryptography(x,y,diffcult){
}
void Make_First_Block()
{
Getpublickey();
blo.data = circle;
blo.pre_hash = 0;
blo.this_hash = (blo.pre_hash+public_Key) * (a+b);
Block.push_back(blo);
}
void Append_Block()
{
pre_blo = blo;
bool flag = true;
auto temp = public_Key;
circle = circle + 1;
Getpublickey();
blo.data = circle;
blo.pre_hash = blo.this_hash;
blo.this_hash = (blo.pre_hash + public_Key) * (a + b);
for(list::iterator itor = Block.begin(); itor != Block.end(); itor++)
{
if ((*itor).this_hash != (*itor).pre_hash + temp * (a + b))
{
flag = false;
break;
}
}
if (flag) { Block.push_back(blo); };
}
};

class Get_Block :Create_Block {
public:
int diffcult;
int number = 1;
Get_Block():Create_Block(“OK”){
}
void calc()
{
double start = clock();
while (true){
for (unsigned long long z = 1; z < ULLONG_MAX; z++){
for (unsigned long long j = 1; j < 65535; j = j + 1) {
for (unsigned long long i = 1; i < 65535; i = i + 1) {
Cryptography *person = new Cryptography(i,j,z);
person->Getpublickey();
block bloc;
bloc.data = circle;
bloc.pre_hash = pre_blo.this_hash;
bloc.this_hash = (blo.pre_hash + person->public_Key) * (i + j);
if (blo.data == bloc.data &&blo.pre_hash== bloc.pre_hash && blo.this_hash == bloc.this_hash)
{
double end = clock();
cout << “歷時(shí)”<<end-start<<“毫秒獲得的第” << number++ <<“個(gè)區(qū)塊信息為：” << endl;
cout << “data:” << bloc.data << endl;
cout << “this_hash:” << bloc.this_hash << endl;
cout << “pre_hash:” << bloc.pre_hash << endl << “=======================” << endl << endl;
this->Append_Block();
start = clock();
}
delete []person;
}
}
}
}
}
};

int main()
{
Create_Block * one = new Create_Block();
one->Make_First_Block();
Get_Block* two = new Get_Block();
two->calc();
return 0;
}

不得不說第一個(gè)區(qū)塊的挖掘永遠(yuǎn)是最快的。第二個(gè)區(qū)塊確實(shí)要等好久才可以得出。以上即為C/C++實(shí)現(xiàn)區(qū)塊鏈的全部源碼。僅用于學(xué)習(xí)交流，不得用于商業(yè)用途，轉(zhuǎn)載必究。

作者：程序小黑
來源：CSDN
原文：https://blog.csdn.net/qq_27180763/article/details/82588305
版權(quán)聲明：本文為博主原創(chuàng)文章，轉(zhuǎn)載請(qǐng)附上博文鏈接！

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