为C ++中的作业分配创建一个IntegerNumber类，以将大整数合并为超出长范围的字符串 [英] Creating an IntegerNumber class for a homework assignment in C++ to sum large integers as strings exceeding long range
问题描述
所以我创建一个IntegerNumber类，需要能够输出一个大约26位数的整数的加法，例如：-12345678954688709764347890被存储到B是一个类型IntegerNumber。 A，B，C和D都是类型IntegerNumber。我没有使用operator =函数将每个的值彼此分配，如A = B或B = C的问题。稍后在主代码中，要求之一是能够输出数字的和，如D = A + B或者甚至比较A < B。
So I am creating an IntegerNumber class that needs to be able to output an addition of integers that are about 26 digits long, for example : -12345678954688709764347890 is stored into B which is a the type IntegerNumber. A,B,C, and D are all type IntegerNumber. I don't have a problem assigning the values of each to each other like A = B or B = C using an operator= function. Later in the main code, one of the requirements is to be able to output the sum of numbers like D = A + B or even do a comparison of A < B.
如果这些数字在数字的long或int范围内，我不会有麻烦。我有麻烦弄清楚当这些值是字符串时如何添加-12345678954688709764347890 + 5678954688709764347890。什么是最好的方式将它们转换为一个类型，它可以添加或甚至比较（A< B）？
I wouldn't have trouble doing this if these numbers were within the long or int range of numbers. I am having trouble figuring out how to do the addition of -12345678954688709764347890 + 5678954688709764347890 when these values are strings. What would be the best way to convert these into a type where it could be added or even compared ( A < B)?
这里是我到目前为止：
Here is what I have so far:
#include <iostream>
#include <cstring>
using namespace std;
class IntegerNumber
{
friend ostream& operator<<(ostream &, const IntegerNumber&);
friend IntegerNumber operator+(const IntegerNumber&, const IntegerNumber&);
friend bool operator<(const IntegerNumber&, const IntegerNumber&);
friend bool operator==(const IntegerNumber&, const IntegerNumber&);
friend bool operator!=(const IntegerNumber&, const IntegerNumber&);
private:
char *intnum;
public:
IntegerNumber(); //default constructor
IntegerNumber(const char *); //constructor with C-string argument
IntegerNumber(const IntegerNumber &); //copy constructor
~IntegerNumber(); //destructor
IntegerNumber& operator=(const IntegerNumber &rhsObject); //assignment operator
int Length(); //returns length of string
};
void main() {
IntegerNumber A; // IntegerNumber object is created and A contains the integer 0
IntegerNumber B("-12345678954688709764347890"); // IntegerNumber object B is created and B contains the negative number shown within the quotes " "
IntegerNumber C = "5678954688709764347890"; // IntegerNumber object C
//is created and C contains the positive number shown within the quotes " "
IntegerNumber D(B); // IntegerNumber object D is created and D contains
// the number that B contains
A = B; // assigns the value of A to that of B
cout << A << endl; // output to screen the integer in A
B = C; // assigns the value of B to that of C
cout << A << endl; // output to screen the integer in A
// value of A must be same as before.
cout << D << endl; // output to screen the integer in D
// value of D must be same as before.
cout << B << endl; // output to screen the integer in B
// value of B must be same as that of C
D = A + B;
cout << D << endl; // output the sum of the numbers A and B
if ( A < B ) {
C = A + B;
cout << C << endl; // output the sum of A and B
}
else {
A = B + C;
cout << A << endl; // output the sum of B and C
}
if (A == B || C != D)
cout << A << " " << D << endl; // output values of A and D
}
IntegerNumber::IntegerNumber() {
intnum = new char[2];
intnum = "0";
}
IntegerNumber::IntegerNumber(const char *str) {
intnum = new char[strlen(str) +1];
strcpy(intnum, str);
}
IntegerNumber::IntegerNumber(const IntegerNumber &ob) {
intnum = new char[strlen(ob.intnum) +1];
strcpy(intnum, ob.intnum);
}
IntegerNumber::~IntegerNumber() {
delete [] intnum;
}
IntegerNumber& IntegerNumber::operator=(const IntegerNumber &ob) {
if (this != &ob) {
delete [] intnum;
intnum = new char[strlen(ob.intnum) +1];
strcpy(intnum, ob.intnum);
}
return *this;
}
int IntegerNumber::Length() {
return strlen(intnum);
}
ostream& operator<<(ostream &out, const IntegerNumber &ob) {
out << ob.intnum;
return out;
}
IntegerNumber operator+(const IntegerNumber &lhs, const IntegerNumber &rhs) {
int strLength = strlen(lhs.intnum) + strlen(rhs.intnum) +1;
char *tmpStr = new char[strLength];
strcpy(tmpStr, lhs.intnum);
strcat(tmpStr, rhs.intnum);
IntegerNumber retStr(tmpStr);
delete [] tmpStr;
return retStr;
}
bool operator==(const IntegerNumber& lhs, const IntegerNumber& rhs) {
return (strcmp(lhs.intnum, rhs.intnum) == 0);
}
bool operator!=(const IntegerNumber& lhs, const IntegerNumber& rhs) {
return (strcmp(lhs.intnum, rhs.intnum) != 0);
}
bool operator<(const IntegerNumber& lhs, const IntegerNumber& rhs) {
return (strcmp(lhs.intnum, rhs.intnum) < 0);
}
由于某些原因，我有strcpy的警告： 警告4警告C4996：'strcpy'：此函数或变量可能不安全。请考虑使用strcpy_s。要禁用弃用，请使用_CRT_SECURE_NO_WARNINGS。有关详细信息，请参阅联机帮助。 c：\users\danny\documents\visual studio 2010\projects\hw6\hw6\hw6.cpp 106 1 HW6
For some reason, I'm having warnings for strcpy: Warning 4 warning C4996: 'strcpy': This function or variable may be unsafe. Consider using strcpy_s instead. To disable deprecation, use _CRT_SECURE_NO_WARNINGS. See online help for details. c:\users\danny\documents\visual studio 2010\projects\hw6\hw6\hw6.cpp 106 1 HW6
还有strcat有相同的错误，我试图更改为strcpy_s和strcat_s但我得到一个错误说： 6 IntelliSense：没有重载函数的实例strcpy_s匹配参数列表c： \users\danny\documents\visual studio 2010 \projects\hw6\hw6\hw6.cpp 89 3 HW6
And also strcat with the same error, I tried changing to strcpy_s and strcat_s but I get an error saying: 6 IntelliSense: no instance of overloaded function "strcpy_s" matches the argument list c:\users\danny\documents\visual studio 2010\projects\hw6\hw6\hw6.cpp 89 3 HW6
推荐答案
在类中有一个类型为 std :: vector< char> 的字段，数字，然后你可以对 operator +（）（就像你在学校中做的）向量的相应数字求和并返回结果。
Have a field of type std::vector<char> in your class, and store all the digits of the big number, in it and then you can sum the corresponding digits of the vectors in operator+() (just like you did in school) and return the result.
class IntegerNumber
{
//make sure that m_digits contains only digit: digit means, 0 to 9.
//when you add 9 plus 4, it becomes 14, but you don't put in into m_digits,
//rather you just put 3 (unit digit of 13), the 1 goes in the second round of sum!
std::vector<char> m_digits;
public:
IntegerNumber();
IntegerNumber(const std::string &number)
{
//parse the string 'number' and populate the m_digits;
}
IntegerNumber operator+(const IntegerNumber & number);
{
IntegerNumber result;
//sum all the corresponding digits of number.m_digits and this->m_digits
//and store in result.m_digits;
return result;
}
//...
};
编辑：
顺便说一句，以下是开始的样子： http://www.ideone.com/Yb5Nn
By the way, here is how the start should look like : http://www.ideone.com/Yb5Nn
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