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c++ - 当我只重载它以接受 Rational 类型的参数时,我的 Rational Number 类如何使用 += 和 long long 参数?

转载 作者:行者123 更新时间:2023-11-27 23:56:02 25 4
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我已经在我的实现文件中实现了 operator+= (Rational),但是我偶然发现 Rational+= long long 有效,尽管我还没有实现该特定功能。

我的 main 的相关函数是当我使用 plusequals += num 时。

plusequals 声明为 Rational plusequals,num 声明为 long long num。两者都被初始化为包含来自用户输入的值。

什么给了?在我看来,这应该行不通,但确实如此。

这是我的头文件:

#ifndef _RATIONAL_H_
#define _RATIONAL_H_

#include<iostream>


using namespace std;

class Rational
{
long long _p;
long long _q;

void simplify();

public:
Rational();
Rational (long long p, long long Q = 1);
Rational (const Rational&);

Rational& operator= (const Rational&);
Rational& operator+= (const Rational&);
Rational& operator-= (const Rational&);
Rational& operator*= (const Rational&);
Rational& operator/= (const Rational&);

friend ostream& operator<< (ostream&, const Rational&);
friend istream& operator>> (istream&, Rational&);

Rational operator+ (const Rational&);
Rational operator+ (long long) const;
friend Rational operator+ (long long, const Rational&);
Rational operator- (const Rational&);
Rational operator- (long long) const;
friend Rational operator- (long long, const Rational&);
Rational operator* (const Rational&);
Rational operator* (long long) const;
friend Rational operator* (long long, const Rational&);
Rational operator/ (const Rational&);
Rational operator/ (long long) const;
friend Rational operator/ (long long, const Rational&);

bool operator== (const Rational&) const;
bool operator== (long long) const;
friend bool operator== (long long, const Rational&);
bool operator!= (const Rational&) const;
bool operator!= (long long) const;
friend bool operator!= (long long, const Rational&);
bool operator> (const Rational&) const;
bool operator> (long long) const;
friend bool operator> (long long, const Rational&);
bool operator< (const Rational&) const;
bool operator< (long long) const;
friend bool operator< (long long, const Rational&);
bool operator>= (const Rational&) const;
bool operator>= (long long) const;
friend bool operator>= (long long, const Rational&);
bool operator<= (const Rational&) const;
bool operator<= (long long) const;
friend bool operator<= (long long, const Rational&);

Rational operator++ (int);
Rational operator-- (int);
Rational& operator++ ();
Rational& operator-- ();
Rational operator- () const;
Rational operator+ () const;

Rational pow (unsigned exp) const;
Rational inverse() const;
};

#endif

下面是实现:

#include "Rational.h"
#include <iostream>

void validate (long long, long long);

int gcd (long long, long long);

Rational::Rational()
{
_p = 0;
_q = 1;
}

Rational::Rational (long long p, long long Q)
{
validate (p, Q);
_p = p;
_q = Q;
}

Rational::Rational (const Rational& rat)
{
this->_p = rat._p;
this->_q = rat._q;
}
void Rational::simplify()
{
// Fixes negative denominators.
if (_q < 0)
{
_p *= -1;
_q *= -1;
}

// Simplifies Rational Numbers.
int denom = gcd(_p, _q);
_p /= denom;
_q /= denom;

}

Rational& Rational::operator= (const Rational& rat)
{
_p = rat._p;
_q = rat._q;

return *this;
}

Rational& Rational::operator+= (const Rational& rat)
{
_p = ((_p * rat._q) + (_q * rat._p));
_q *= rat._q;

this->simplify();

return *this;
}

Rational& Rational::operator-= (const Rational& rat)
{
_p = ((_p * rat._q) - (_q * rat._p));
_q *= rat._q;

this->simplify();

return *this;
}

Rational& Rational::operator*= (const Rational& rat)
{
_p *= rat._p;
_q *= rat._q;

this->simplify();

return *this;
}

Rational& Rational::operator/= (const Rational& rat)
{
if (rat._p == 0)
{
throw "Division by zero not allowed";
}
_p *= rat._q;
_q *= rat._p;

this->simplify();

return *this;
}

ostream& operator<< (ostream& os, const Rational& rat)
{
os << rat._p << ":" << rat._q;

return os;
}

istream& operator>> (istream& is, Rational& rat)
{
long long p, q;

is >> p >> q;
validate(p, q);
rat._p = p;
rat._q = q;
rat.simplify();

return is;
}

Rational Rational::operator+ (const Rational& rat)
{
Rational result(*this);

result += rat;
result.simplify();

return result;
}

Rational Rational::operator+ (long long num) const
{
Rational result(*this);
Rational temp(num);

result += temp;
result.simplify();

return result;
}

Rational operator+ (long long num, const Rational& rat)
{
Rational result(num);
result += rat;
result.simplify();

return result;
}

Rational Rational::operator- (const Rational& rat)
{
Rational result(*this);

result -= rat;
result.simplify();

return result;
}

Rational Rational::operator- (long long num) const
{
Rational result(*this);
Rational temp(num);

result -= temp;
result.simplify();

return result;
}

Rational operator- (long long num, const Rational& rat)
{
Rational result(num);
result -= rat;
result.simplify();

return result;
}

Rational Rational::operator* (const Rational& rat)
{
Rational result(*this);

result *= rat;
result.simplify();

return result;
}

Rational Rational::operator* (long long num) const
{
Rational result(*this);
Rational temp(num);
result *= temp;
result.simplify();

return result;
}

Rational operator* (long long num, const Rational& rat)
{
Rational result(num);
result *= rat;
result.simplify();

return result;
}

Rational Rational::operator/ (const Rational& rat)
{
Rational result(*this);

result /= rat;
result.simplify();

return result;
}

Rational Rational::operator/ (long long num) const
{
Rational result(*this);
Rational temp(num);

result /= temp;
result.simplify();

return result;
}

Rational operator/ (long long num, const Rational& rat)
{
Rational result(num);
result /= rat;
result.simplify();

return result;
}

bool Rational::operator== (const Rational& rat) const
{
bool result;

if ((this->_p == rat._p) && (this->_q == rat._q))
{
result = true;
}
else
{
result = false;
}

return result;
}

bool Rational::operator== (long long num) const
{
bool result;
Rational temp(num);

result = (*this == temp);

return result;
}

bool operator== (long long num, const Rational& rat)
{
bool result;

result = (rat == num);

return result;
}

bool Rational::operator!= (const Rational& rat) const
{
return !(*this == rat);
}

bool Rational::operator!= (long long num) const
{
return !(*this == num);
}

bool operator!= (long long num, const Rational& rat)
{
return !(num == rat);
}

bool Rational::operator> (const Rational& rat) const
{
bool result;

if ((this->_p / this->_q) > (rat._p / rat._q))
{
result = true;
}
else
{
result = false;
}

return result;
}

bool Rational::operator> (long long num) const
{
bool result;
Rational temp(num);

result = (*this > temp);

return result;
}

bool operator> (long long num, const Rational& rat)
{
bool result;

result = (rat < num);

return result;
}

bool Rational::operator< (const Rational& rat) const
{
bool result;

if (!(*this > rat) && !(*this == rat))
{
result = true;
}
else
{
result = false;
}

return result;
}

bool Rational::operator< (long long num) const
{
bool result;
Rational temp(num);

result = (*this < temp);

return result;
}

bool operator< (long long num, const Rational& rat)
{
bool result;

result = (rat > num);

return result;
}

bool Rational::operator>= (const Rational& rat) const
{
bool result;

if (!(*this < rat))
{
result = true;
}
else
{
result = false;
}

return result;
}

bool Rational::operator>= (long long num) const
{
bool result;
Rational temp(num);

result = (*this >= temp);

return result;
}

bool operator>= (long long num, const Rational& rat)
{
bool result;

result = (rat <= num);

return result;
}

bool Rational::operator<= (const Rational& rat) const
{
bool result;

if (!(*this > rat))
{
result = true;
}
else
{
result = false;
}

return result;
}

bool Rational::operator<= (long long num) const
{
bool result;
Rational temp(num);

result = (*this <= temp);

return result;
}

bool operator<= (long long num, const Rational& rat)
{
bool result;

result = (rat >= num);

return result;
}

Rational Rational::operator++ (int) // Postfix
{
Rational temp(*this);

this->_p++;
this->_q++;

return temp;
}

Rational Rational::operator-- (int) // Postfix
{
Rational temp(*this);

this->_p--;
this->_q--;

return temp;
}

Rational& Rational::operator++()
{
this->_p++;
this->_q++;

return *this;
}

Rational& Rational::operator--()
{
this->_p--;
this->_q--;

return *this;
}

Rational Rational::operator-() const
{
Rational temp(-(this->_p), (this->_q));

return temp;
}

Rational Rational::operator+() const
{
Rational temp(+(this->_p), +(this->_q));

return temp;
}

Rational Rational::pow (unsigned exp) const
{
Rational result(*this);
Rational temp(*this);

if (exp == 0)
{
result = 1;
}
else
{
for (unsigned i = 1; i < exp; i++)
{
result *= temp;
}
}

return result;
}

Rational Rational::inverse() const
{
Rational temp(this->_q, this->_p);

return temp;
}

void validate(long long p, long long q)
{
p++; // Supress error for unused value. Decided to keep value in parameter list to maintain clarity.
if (q == 0)
{
throw "Zero Denominator";
}
}

int gcd(long long p, long long q)
{
// Euclid's Algorithm
if (q == 0)
{
return p;
}
return gcd (q, p%q);
}

为了更好地衡量,我正在使用的测试 main():

#include <string>
#include <iostream>
#include "Rational.h"

int main()
{
while (true)
{
Rational rat1;
Rational rat2;
Rational rat3;
long long num;
unsigned exp;
Rational power(rat1);
string hello = "hello";

// Get input
try
{
cout << "Please enter a numerator and denominator separated by a space: ";
cin >> rat1;
cout << endl;
cout << "Please enter a second numerator and denomintator seperated by a space: ";
cin >> rat2;
cout << endl;
cout << "Please enter a numerator and denominator separated by a space for a third Rational number: ";
cin >> rat3;
cout << endl;
cout << "Enter a number to use for arithmetic operations: ";
cin >> num;
cout << endl;
cout << "Please enter a positive integer to use an exponent :";
cin >> exp;
cout << endl;
}
catch (char const* err)
{
cerr << err << " - Non-Zero denominators only ya big goon.\n";
}

cout << endl;

cout << "You put: " << rat1 << " and: " << rat2 << endl;

Rational plusequals (rat1);
Rational minusequals (rat1);
Rational timesequals (rat1);
Rational divequals (rat1);

plusequals += rat2;
minusequals -= rat2;
timesequals *= rat2;
try
{
divequals /= rat2;
}
catch (const char* msg)
{
cerr << msg << endl;
}

cout << "+= : " << plusequals << "\n-= : " << minusequals << "\n*= : " << timesequals << "\n/= : " << divequals << endl;

plusequals = rat1;
minusequals = rat1;
timesequals = rat1;
divequals = rat1;

plusequals += num;
minusequals -= num;
timesequals *= num;
try
{
divequals /= num;
}
catch (const char* msg)
{
cerr << msg << endl;
}

cout << "\nRational = " << rat1<< ", num = " << num << " :\n";
cout << rat1 << " += " << num << ": " << plusequals << endl << rat1 << " -= " << num << ": " << minusequals << endl << rat1 <<" *= " << num << ": " << timesequals << endl << rat1 << " /= " << num << ": " << divequals << endl;

plusequals = rat1;
minusequals = rat1;
timesequals = rat1;
divequals = rat1;

plusequals += rat3;
minusequals -= rat3;
timesequals *= rat3;
try
{
divequals /= rat3;
}
catch (const char* msg)
{
cerr << msg << endl;
}

cout << "\nRational = " << rat1<< ", Rational = " << rat3 << " :\n";
cout << rat1 << " += " << rat3 << ": " << plusequals << endl << rat1 << " -= " << rat3 << ": " << minusequals << endl << rat1 << " *= " << rat3 << ": " << timesequals << endl << rat1 << " /= " << rat3 << ": " << divequals << endl;


power = rat1.pow(exp);
cout << endl << rat1 << " raised to the power of " << exp << " : ";
cout << power << endl;

cout << "The multiplicative inverse of " << rat1 << " is " << rat1.inverse() << endl;

// Comparison
cout << endl << endl;
if (rat1 == rat2)
{
cout << rat1 << " = " << rat2 << endl;
}
if (rat1 != rat2)
{
cout << rat1 << " != " << rat2 << endl;
}
if (rat1 <= rat2)
{
cout << rat1 << " <= " << rat2 << endl;
}
if (rat1 >= rat2)
{
cout << rat1 << " >= " << rat2 << endl;
}
if (rat1 < rat2)
{
cout << rat1 << " < " << rat2 << endl;
}
if (rat1 > rat2)
{
cout << rat1 << " > " << rat2 << endl;
}
}
}

此外,我应该指出,我知道我不应该在我的头文件中使用 using namespace std;...但是我们需要直接从我们的教授那里复制头文件。他还坚持要我们使用我在这个特定项目中使用的括号样式,我认为这很丑陋,但事实就是如此。他很讲究。不过,如果您对作业有任何其他改进/见解,我们将不胜感激。

最佳答案

要回答关于 += 的问题:请看这个构造函数:

Rational (long long p, long long Q = 1);

构造函数没有用explicit关键字修饰,这意味着它可以用于隐式转换。

当你有一个 += 表达式,其操作数类型为 Rationallong long 时,编译器会搜索所有定义operator += 然后查看是否有任何一个可以接受操作数类型(即重载决策)。

如果没有直接匹配(在您的情况下它们不匹配),它将尝试查看是否有任何隐式转换可以使它们匹配。您的 += 两边都需要 Rational。左侧的操作已经是一个 Rational,那里不需要任何东西。右侧是 long long,但存在从 long longRational 的隐式转换(通过构造函数)。因此,编译器发出代码,从 long long 中创建一个临时的 Rational,然后将这个临时的 Rational 提供给您的 operator +=

您可以通过将构造函数标记为 explicit 来阻止这些隐式转换,但我认为您这样做是错误的。毕竟,写“1/2 + 1 = 3/2”有什么问题吗?我们不需要写成“1/2 + 1/1”。

数字之间的隐式转换很有用。由于从 intdouble 的隐式转换,3.14 + 1 是完全有效的 C++(并且完全合理)。


关于您的其他要求,

Any other improvements/insights into the assignment would be much appreciated, though.

这不是 Stack Overflow 的主题。假设您的代码有效,您可能会在 Code Review 上获得有趣的反馈.

关于c++ - 当我只重载它以接受 Rational 类型的参数时,我的 Rational Number 类如何使用 += 和 long long 参数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42580541/

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