gpt4 book ai didi

c++:覆盖运算符时的多个定义

转载 作者:太空狗 更新时间:2023-10-29 20:22:55 25 4
gpt4 key购买 nike

我正在为有理数实现一个类,这当然涉及重写“==”和“!=”等常见运算符。我很确定我遗漏了一个愚蠢的错误,不要犹豫,索要我没有提供的任何文件。谢谢!

理性.hpp:

#ifndef RATIONAL_HPP
#define RATIONAL_HPP

#include "test.hpp"

#include <cstdlib>
#include <iosfwd>
#include <iostream>
#include <assert.h>

// Mathematical helper functions.
//
// NOTE: These are defined in rational.cpp.
int gcd(int, int);
int lcm(int, int);


// Represents a rational number. The rational numbers are the set of
// numbers that can be represented as the quotient of two integers.
struct Rational
{
// TODO: Define the following:
// 1. A default constructor
int n;
int d;
Rational()
:n(0), d(1) {}

// 2. A constructor that takes an integer value
Rational(int num)
:n(num), d(1){}
// 3. A constructor that takes a pair of values
Rational(int numer, int denom)
:n(numer), d(denom) {
assert( d != 0);
int gcdnum;
if ((numer % denom) != 0){
//do nothing
}else{
gcdnum = gcd(numer, denom);
numer /= gcdnum;
denom /= gcdnum;
Rational(numer, denom);
}
}
// Returns the numerator.
int num() const {
return n;
}

// Returns the denominator
int den() const {
return d;
}

};


bool operator==(Rational a, Rational b){
return (a.n == b.n && a.d == b.d);
}
bool operator!=(Rational a, Rational b){
return (a.n != b.n && a.d != b.d);
}
bool operator < (Rational a, Rational b){
int lcdNum = lcm(a.d, b.d);
int newAN, newBN; //allows for comparisons without altering actuial value
newAN = a.n * lcdNum;
newBN = b.n * lcdNum;
return newAN < newBN;
}
bool operator > (Rational a, Rational b){
int lcdNum = lcm(a.d, b.d);
int newAN, newBN; //allows for comparisons without altering actuial value
newAN = a.n * lcdNum;
newBN = b.n * lcdNum;
return newAN > newBN;
}
bool operator <= (Rational a, Rational b){
int lcdNum = lcm(a.d, b.d);
int newAN, newBN; //allows for comparisons without altering actuial value
newAN = a.n * lcdNum;
newBN = b.n * lcdNum;
return newAN <= newBN;
}

bool operator >= (Rational a, Rational b){
int lcdNum = lcm(a.d, b.d);
int newAN, newBN; //allows for comparisons without altering actuial value
newAN = a.n * lcdNum;
newBN = b.n * lcdNum;
return newAN >= newBN;
}

// 3. The standard arithmetic operators
// - r1 + r2
// - r1 - r2
// - r1 * r2
// - r1 / r2
// - r1 / r2
Rational operator + (Rational a, Rational b){
int lcdNum = lcm(a.d, b.d);
int newAN, newBN; //allows for comparisons without altering actuial value
newAN = a.n * lcdNum;
newBN = b.n * lcdNum;
Rational c((newAN + newBN), (a.d * lcdNum));
return c;
}
Rational operator - (Rational a, Rational b){
int lcdNum = lcm(a.d, b.d);
int newAN, newBN; //allows for comparisons without altering actuial value
newAN = a.n * lcdNum;
newBN = b.n * lcdNum;
Rational c((newAN + newBN), (a.d * lcdNum));
return c;
}
Rational operator * (Rational a, Rational b){
Rational c((a.n * b.n), (a.d * b.d));
return c;
}
Rational operator / (Rational a, Rational b){
Rational c((a.n * b.d), (a.d * b.n)); //multiplies by the reciprocal
return c;
}


std::ostream& operator<<(std::ostream&, Rational);
std::istream& operator>>(std::istream&, Rational&);


#endif

理性.cpp:

//
// rational.hpp: Definition of rational class and its interace.

#include "rational.hpp"

#include <iostream>


// -------------------------------------------------------------------------- //
// Helper functions

// Compute the GCD of two integer values using Euclid's algorithm.
int
gcd(int a, int b)
{
while (b != 0) {
int t = b;
b = a % b;
a = t;
}
return a;
}


// Compute the LCM of two integer values.
int
lcm(int a, int b)
{
return (std::abs(a) / gcd(a, b)) * std::abs(b);
}


// -------------------------------------------------------------------------- //
// Rational implementation


// TODO: Make this print integers when the denominator is 1.
std::ostream&
operator<<(std::ostream& os, Rational r)
{
if(r.den() == 1){
return os << r.num();
}else{
return os << r.num() << '/' << r.den();
}
}


// TODO: Make this read integer values if no '/' is given as a separator.
// You may assume that there is no space between the numerator and the
// slash. Hint, find and read the reference documentation for istream::peek().
std::istream&
operator>>(std::istream& is, Rational& r)
{
int p, q;
char c;
is >> p;
c = is.peek();
if (c == '/'){
is >> c >> q;
if (!is)
return is;

// Require that the divider to be a '/'.
if (c != '/') {
is.setstate(std::ios::failbit);
return is;
}

// Make sure that we didn't read p/0.
if (q == 0) {
is.setstate(std::ios::failbit);
return is;
}

r = Rational(p, q);
return is;
}else{
is.setstate(std::ios::failbit);
}
}

rc.cpp:

// main.cpp: rational number test suite

#include "rational.hpp"

#include <iostream>
#include <iomanip>

#include <unistd.h>


int
main()
{
// Determine if input is coming from a terminal.
bool term = isatty(0);

// This will continue reading until it reaches the end-of-input.
// If you are using this interactivly, type crtl-d to send the
// end of input character to the terminal.
while (std::cin) {
Rational r1;
Rational r2;
std::string op;

if (term)
std::cout << "> ";

std::cin >> r1 >> op >> r2;
if (!std::cin)
break;

// FIXME: Add all of the other overlaoded operators by adding
// cases for each of them.
if (op == "==")


std::cout << std::boolalpha << (r1 == r2) << '\n';
else if (op == "!=")
std::cout << std::boolalpha << (r1 != r2) << '\n';
else if (op == "<")
std::cout << std::boolalpha << (r1 < r2) << '\n';
else if (op == ">")
std::cout << std::boolalpha << (r1 > r2) << '\n';
else if (op == "<=")
std::cout << std::boolalpha << (r1 <= r2) << '\n';
else if (op == ">=")
std::cout << std::boolalpha << (r1 >= r2) << '\n';
else if (op == "+")
std::cout << (r1 + r2) << '\n';
else if (op == "-")
std::cout << (r1 - r2) << '\n';
else if (op == "*")
std::cout << (r1 * r2) << '\n';
else if (op == "/")
std::cout << (r1 / r2) << '\n';
else
std::cerr << "invalid operator: " << op << '\n';

}

// If we got to the end of the file without fatal errors,
// return success.
if (std::cin.eof())
return 0;

// Otherwise, diagnose errors in input and exit with an error
// code.
if (std::cin.fail()) {
std::cerr << "input error\n";
return 1;
}

return 0;
}

最佳答案

#includerational.hpp 的每个翻译单元都将获得比较运算符函数的定义,这肯定会导致链接时的重复定义.

尝试在它们前面添加一个“内联”关键字。

关于c++:覆盖运算符时的多个定义,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35591871/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com