最近在学习制作高精度计算器，前辈给了我一段参考代码，但是我对其中有一部分还是不太懂。
【计算器要求】
实现两个高精度大整数（200位以内的整数）的四则运算，输出这两个大整数的和、差、积、商及余数。
· 请大神如果有空，帮我对几个运算符号代码（主程序不需要）进行注释！越详细越好！十分感谢！
· 如果寻找到可优化的部分或者bug也请进行修改并标注，谢谢！
#include <iostream>
#include <string>
using namespace std;
inline int compare(string str1, string str2) {
if(str1.size() > str2.size()) //长度长的整数大于长度小的整数
return 1;
else if(str1.size() < str2.size())
return -1;
else
return str1.compare(str2); //若长度相等，从头到尾按位比较，compare函数：相等返回0，大于返回1，小于返回－1
}
string ADD_INT(string str1, string str2) {//高精度加法
string SUB_INT(string str1, string str2);
int sign = 1; //sign 为符号位
string str;
if(str1[0] == '-') {
if(str2[0] == '-') {
sign = -1;
str = ADD_INT(str1.erase(0, 1), str2.erase(0, 1));
} else {
str = SUB_INT(str2, str1.erase(0, 1));
}
} else {
if(str2[0] == '-')
str = SUB_INT(str1, str2.erase(0, 1));
else {
//把两个整数对齐，短整数前面加0补齐
string::size_type l1, l2;
int i;
l1 = str1.size(); l2 = str2.size();
if(l1 < l2) {
for(i = 1; i <= l2 - l1; i++)
str1 = "0" + str1;
} else {
for(i = 1; i <= l1 - l2; i++)
str2 = "0" + str2;
}
int int1 = 0, int2 = 0; //int2 记录进位
for(i = str1.size() - 1; i >= 0; i--) {
int1 = (int(str1[i]) - '0' + int(str2[i]) - '0' + int2) % 10;
int2 = (int(str1[i]) - '0' + int(str2[i]) - '0' +int2) / 10;
str = char(int1 + '0') + str;
}
if(int2 != 0) str = char(int2 + '0') + str;
}
}
//运算后处理符号位
if((sign == -1) && (str[0] != '0'))
str = "-" + str;
return str;
}
string SUB_INT(string str1, string str2) {//高精度减法
string MUL_INT(string str1, string str2);
int sign = 1; //sign 为符号位
string str;
int i;
if(str2[0] == '-')
str = ADD_INT(str1, str2.erase(0, 1));
else {
int res = compare(str1, str2);
if(res == 0) return "0";
if(res < 0) {
sign = -1;
string temp = str1;
str1 = str2;
str2 = temp;
}
string::size_type tempint;
tempint = str1.size() - str2.size();
for(i = str2.size() - 1; i >= 0; i--) {
if(str1[i + tempint] < str2[i]) {
str1[i + tempint - 1] = char(int(str1[i + tempint - 1]) - 1);
str = char(str1[i + tempint] - str2[i] + ':') + str;
} else
str = char(str1[i + tempint] - str2[i] + '0') + str;
}
for(i = tempint - 1; i >= 0; i--)
str = str1[i] + str;
}
//去除结果中多余的前导0
str.erase(0, str.find_first_not_of('0'));
if(str.empty()) str = "0";
if((sign == -1) && (str[0] != '0'))
str = "-" + str;
return str;
}
string MUL_INT(string str1, string str2) {//高精度乘法
int sign = 1; //sign 为符号位
string str;
if(str1[0] == '-') {
sign *= -1;
str1 = str1.erase(0, 1);
}
if(str2[0] == '-') {
sign *= -1;
str2 = str2.erase(0, 1);
}
int i, j;
string::size_type l1, l2;
l1 = str1.size(); l2 = str2.size();
for(i = l2 - 1; i >= 0; i --) { //实现手工乘法
string tempstr;
int int1 = 0, int2 = 0, int3 = int(str2[i]) - '0';
if(int3 != 0) {
for(j = 1; j <= (int)(l2 - 1 - i); j++)
tempstr = "0" + tempstr;
for(j = l1 - 1; j >= 0; j--) {
int1 = (int3 * (int(str1[j]) - '0') + int2) % 10;
int2 = (int3 * (int(str1[j]) - '0') + int2) / 10;
tempstr = char(int1 + '0') + tempstr;
}
if(int2 != 0) tempstr = char(int2 + '0') + tempstr;
}
str = ADD_INT(str, tempstr);
}
//去除结果中的前导0
str.erase(0, str.find_first_not_of('0'));
if(str.empty()) str = "0";
if((sign == -1) && (str[0] != '0'))
str = "-" + str;
return str;
}
string DIVIDE_INT(string str1, string str2, int flag) {//高精度除法
//flag = 1时,返回商; flag = 0时,返回余数
string quotient, residue; //定义商和余数
int sign1 = 1, sign2 = 1;
if(str2 == "0") { //判断除数是否为0
quotient = "ERROR!";
residue = "ERROR!";
if(flag == 1) return quotient;
else return residue;
}
if(str1 == "0") { //判断被除数是否为0
quotient = "0";
residue = "0";
}
if(str1[0] == '-') {
str1 = str1.erase(0, 1);
sign1 *= -1;
sign2 = -1;
}
if(str2[0] == '-') {
str2 = str2.erase(0, 1);
sign1 *= -1;
}
int res = compare(str1, str2);
if(res < 0) {
quotient = "0";
residue = str1;
} else if(res == 0) {
quotient = "1";
residue = "0";
} else {
string::size_type l1, l2;
l1 = str1.size(); l2 = str2.size();
string tempstr;
tempstr.append(str1, 0, l2 - 1);
//模拟手工除法
for(int i = l2 - 1; i < l1; i++) {
tempstr = tempstr + str1[i];
for(char ch = '9'; ch >= '0'; ch --) { //试商
string str;
str = str + ch;
if(compare(MUL_INT(str2, str), tempstr) <= 0) {
quotient = quotient + ch;
tempstr = SUB_INT(tempstr, MUL_INT(str2, str));
break;
}
}
}
residue = tempstr;
}
//去除结果中的前导0
quotient.erase(0, quotient.find_first_not_of('0'));
if(quotient.empty()) quotient = "0";
if((sign1 == -1) && (quotient[0] != '0'))
quotient = "-" + quotient;
if((sign2 == -1) && (residue[0] != '0'))
residue = "-" + residue;
if(flag == 1) return quotient;
else return residue;
}
string DIV_INT(string str1, string str2) {//高精度除法,返回商
return DIVIDE_INT(str1, str2, 1);
}
string MOD_INT(string str1, string str2) {//高精度除法,返回余数
return DIVIDE_INT(str1, str2, 0);
}
int main() {
char ch;
string s1, s2, res;
cout<<"请输入 丨 加法：“+ ” 丨 减法：“- ” 丨 乘法：“* ” 丨 除法：“/ ” 丨 余数：“% ”丨来确定计算类型！" <<endl;
while(cin >> ch)
{
cout<<"请输入要计算的数1:" <<endl;
cin >> s1;
cout<<"请输入要计算的数2:" <<endl;
cin >> s2;
switch(ch) {
case '+': res = "加法计算的结果是："+ADD_INT(s1, s2); break;
case '-': res = "减法计算的结果是："+SUB_INT(s1, s2); break;
case '*': res = "乘法计算的结果是："+MUL_INT(s1, s2); break;
case '/': res = "除法计算的结果是："+DIV_INT(s1, s2); break;
case '%': res = "除法计算后的余数是："+MOD_INT(s1, s2); break;
default : break;
}
cout << res << endl;
}
return(0);
}
我明白大概意思了，是两个字符串输入后不转换为数据，直接字符串里面按位进行计算的，我大概注释了一下，你看看：
inline int compare(string str1, string str2) {
if (str1.size() > str2.size()) //长度长的整数大于长度小的整数
return 1;
else if (str1.size() < str2.size())
return -1;
else
return str1.compare(str2); //若长度相等，从头到尾按位比较，compare函数：相等返回0，大于返回1，小于返回－1
}
string ADD_INT(string str1, string str2) {//高精度加法
string SUB_INT(string str1, string str2);
int sign = 1; //sign 为符号位
string str;//计算结果
if (str1[0] == '-') {
if (str2[0] == '-') {
sign = -1;
str = ADD_INT(str1.erase(0, 1), str2.erase(0, 1));// 两个数都是负数，sign负值-1表示结果为负
}
else {
str = SUB_INT(str2, str1.erase(0, 1));//1为负2为正，相减
}
}
else {
if (str2[0] == '-')
str = SUB_INT(str1, str2.erase(0, 1));//1为正2为负，相减
else {
//把两个整数对齐，短整数前面加0补齐
string::size_type l1, l2;
int i;
l1 = str1.size(); l2 = str2.size();//获取输入的两个数长度
if (l1 < l2) {//数1长度小于数2，在数1前面补零
for (i = 1; i <= l2 - l1; i++)
str1 = "0" + str1;
}
else {//数1长度大于等于数2，在数2前面补零，如果相等则不用补零
for (i = 1; i <= l1 - l2; i++)
str2 = "0" + str2;
}
int int1 = 0, int2 = 0; //int2 记录进位
for (i = str1.size() - 1; i >= 0; i--) {//从个位开始循环计算，int1是相加结果，int2是进位值
int1 = (int(str1[i]) - '0' + int(str2[i]) - '0' + int2) % 10;
int2 = (int(str1[i]) - '0' + int(str2[i]) - '0' + int2) / 10;
str = char(int1 + '0') + str;
}
if (int2 != 0) str = char(int2 + '0') + str;
}
}
//运算后处理符号位
if ((sign == -1) && (str[0] != '0'))//判断符合变量sign决定数字的正负
str = "-" + str;
return str;
}
string SUB_INT(string str1, string str2) {//高精度减法
string MUL_INT(string str1, string str2);
int sign = 1; //sign 为符号位
string str;
int i;
if (str2[0] == '-')
str = ADD_INT(str1, str2.erase(0, 1));//数2是负值时，将数2与数1做加法
else {
int res = compare(str1, str2);//数2为正，比较两个数长度
if (res == 0) return "0";//两数相等返回0
if (res < 0) {//数1小于数2，设置符号为负，交换两个值
sign = -1;
string temp = str1;
str1 = str2;
str2 = temp;
}
string::size_type tempint;
tempint = str1.size() - str2.size();//两个数的位数差
for (i = str2.size() - 1; i >= 0; i--) {//逐位计算相减
if (str1[i + tempint] < str2[i]) {
str1[i + tempint - 1] = char(int(str1[i + tempint - 1]) - 1);//数1的当前位小于数2
str = char(str1[i + tempint] - str2[i] + ':') + str;
}
else
str = char(str1[i + tempint] - str2[i] + '0') + str;//数1的当前位大于数2
}
for (i = tempint - 1; i >= 0; i--)//将前面未进行计算的的位补上
str = str1[i] + str;
}
//去除结果中多余的前导0
str.erase(0, str.find_first_not_of('0'));
if (str.empty()) str = "0";
if ((sign == -1) && (str[0] != '0'))
str = "-" + str;
return str;
}
string MUL_INT(string str1, string str2) {//高精度乘法
int sign = 1; //sign 为符号位
string str;
if (str1[0] == '-') {//分别对数1数2进行判断，为负数则符号位修改一次
sign *= -1;
str1 = str1.erase(0, 1);
}
if (str2[0] == '-') {
sign *= -1;
str2 = str2.erase(0, 1);
}
int i, j;
string::size_type l1, l2;
l1 = str1.size(); l2 = str2.size();
for (i = l2 - 1; i >= 0; i--) { //实现手工乘法
string tempstr;
int int1 = 0, int2 = 0, int3 = int(str2[i]) - '0';//int1记录结果，int2记录进位
if (int3 != 0) {
for (j = 1; j <= (int)(l2 - 1 - i); j++)
tempstr = "0" + tempstr;
for (j = l1 - 1; j >= 0; j--) {
int1 = (int3 * (int(str1[j]) - '0') + int2) % 10;
int2 = (int3 * (int(str1[j]) - '0') + int2) / 10;
tempstr = char(int1 + '0') + tempstr;
}
if (int2 != 0) tempstr = char(int2 + '0') + tempstr;
}
str = ADD_INT(str, tempstr);//按位累计每次相乘的结果
}
//去除结果中的前导0
str.erase(0, str.find_first_not_of('0'));
if (str.empty()) str = "0";
if ((sign == -1) && (str[0] != '0'))
str = "-" + str;
return str;
}
string DIVIDE_INT(string str1, string str2, int flag) {//高精度除法
//flag = 1时,返回商; flag = 0时,返回余数
string quotient, residue; //定义商和余数
int sign1 = 1, sign2 = 1;
if (str2 == "0") { //判断除数是否为0
quotient = "ERROR!";
residue = "ERROR!";
if (flag == 1) return quotient;
else return residue;
}
if (str1 == "0") { //判断被除数是否为0
quotient = "0";
residue = "0";
}
if (str1[0] == '-') {//根据数1数2正负判断结果正负
str1 = str1.erase(0, 1);
sign1 *= -1;
sign2 = -1;
}
if (str2[0] == '-') {
str2 = str2.erase(0, 1);
sign1 *= -1;
}
int res = compare(str1, str2);
if (res < 0) {//数1小于数2，商为0，余数为数1
quotient = "0";
residue = str1;
}
else if (res == 0) {//相等，商为1，余数0
quotient = "1";
residue = "0";
}
else {//数1大于数2进行除法计算
string::size_type l1, l2;
l1 = str1.size(); l2 = str2.size();
string tempstr;
tempstr.append(str1, 0, l2 - 1);
//模拟手工除法
for (int i = l2 - 1; i < l1; i++) {//逐位计算
tempstr = tempstr + str1[i];
for (char ch = '9'; ch >= '0'; ch--) { //试商
string str;
str = str + ch;
if (compare(MUL_INT(str2, str), tempstr) <= 0) {//数2乘以多少小于等于数1
quotient = quotient + ch;//商累计
tempstr = SUB_INT(tempstr, MUL_INT(str2, str));//余数累计
break;
}
}
}
residue = tempstr;
}
//去除结果中的前导0
quotient.erase(0, quotient.find_first_not_of('0'));
if (quotient.empty()) quotient = "0";
if ((sign1 == -1) && (quotient[0] != '0'))
quotient = "-" + quotient;
if ((sign2 == -1) && (residue[0] != '0'))
residue = "-" + residue;
if (flag == 1) return quotient;
else return residue;
}
string DIV_INT(string str1, string str2) {//高精度除法,返回商
return DIVIDE_INT(str1, str2, 1);
}
string MOD_INT(string str1, string str2) {//高精度除法,返回余数
return DIVIDE_INT(str1, str2, 0);
}
http://www.pudn.com/downloads709/sourcecode/windows/detail2845853.html