带有ostream运算符的C ++类与VS一起使用,无法使用gcc进行编译

时间:2018-04-18 03:11:43

标签: c++ visual-studio unix gcc ostream

我正在为C ++中的CS2工作。我已经完成了我的代码,它在Visual Studio中按预期运行。但是要提交代码,我们必须将其复制到我们的unix服务器并确保它仍然运行。

我无法让它运行所以我想我肯定错过了VS正在纠正的错误?

我向我的教授展示了我的代码,他同意看起来一切都是正确的。

有人可以帮忙吗?

rational.h

#ifndef  RATIONAL_H
#define RATIONAL_H
#include <iostream>
#include <ostream>
#include <cmath>
#include <cstdlib>
using namespace std;


class rational
{

    friend ostream& operator<<(ostream &, rational&);
    friend istream& operator>>(istream &, rational&);
public:

    rational operator+(const rational &)const;
    rational operator-(const rational &)const;

    bool rational::operator>(const rational &r2);

    rational(int n = 0, int d = 1);

    rational add(const rational &r2) const;

    void add(const rational &r1, const rational &r2);

    rational subtract(const rational &r2) const;

    void subtract(const rational &r1, const rational &r2);


    rational multiply(const rational &r2) const;


    rational divide(const rational &r2) const;

    int compare(const rational &r2) const;

private:
    int num;    // numerator
    int denom;  // denominator
};
#endif

rational.cpp

#include "rational.h"
using namespace std;

ostream& operator<<(ostream &out, rational &robj)
{
    out << robj.num << "/" << robj.denom;
    return out;
}
istream& operator>>(istream &in, rational &obj)
{
    cout << "Enter values for the numerator and denominator of a rational number: ";
    in >> obj.num >> obj.denom;

    return in;
}

rational::rational(int n, int d)
{
    num = n;
    denom = d;
}

rational rational::operator+(const rational &r2) const
{
    rational sum;
    sum.denom = (*this).denom * r2.denom;
    sum.num = ((*this).num * r2.denom) + (r2.num * (*this).denom);
    return sum;

}

rational rational::operator-(const rational &r2) const
{
    rational diff;
    diff.denom = (*this).denom * r2.denom;
    diff.num = ((*this).num * r2.denom) - (r2.num * (*this).denom);
    return diff;
}

bool rational::operator>(const rational &r2)
{
    double x1, x2;

    x1 = double((*this).num) / double((*this).denom);
    x2 = double(r2.num) / double(r2.denom);

    return (x1 > x2);
}

rational rational::add(const rational &r2) const
{
    rational sum;
    sum.denom = (*this).denom * r2.denom;
    sum.num = ((*this).num * r2.denom) + (r2.num * (*this).denom);
    return sum;
}

void rational::add(const rational &r1, const rational &r2)
{
    (*this).denom = r1.denom * r2.denom;
    (*this).num = (r1.num * r2.denom) + (r2.num * r1.denom);

}
rational rational::subtract(const rational &r2) const
{
    rational diff;
    diff.denom = (*this).denom * r2.denom;
    diff.num = ((*this).num * r2.denom) - (r2.num * (*this).denom);
    return diff;
}
void rational::subtract(const rational &r1, const rational &r2)
{
    (*this).denom = r1.denom * r2.denom;
    (*this).num = (r1.num * r2.denom) - (r2.num * r1.denom);

}
rational rational::multiply(const rational &r2) const
{
    rational multi;
    multi.denom = (*this).denom * r2.denom;
    multi.num = (*this).num * r2.num;
    return multi;

}
rational rational::divide(const rational &r2) const
{
    rational div;
    div.denom = (*this).denom * r2.num;
    div.num = (*this).num * r2.denom;
    return div;

}
int rational::compare(const rational &r2) const
{
    if ((*this).denom == r2.denom && (*this).num == r2.num)
        return 0;
    else if (double((*this).num) / double((*this).denom) > double(r2.num) / double(r2.denom))
        return 1;
    else
        return -1;
}

的main.cpp

#include "rational.h"

using namespace std;

int main()
{
    rational r1(1,4), r2(1,3),r3;

    cout << "r1 is initialized by the 2nd constructor: r1 = " << r1 << endl;
    cout << "r2 is initialized by the 2nd constructor: r2 = " << r2 << endl << endl;

    cout << "Testing the comapre() member function, found:" << endl << "\t";
    int compare = r1.compare(r2);

    switch (compare){
        case 0:
            cout << r1 << " is equal to " << r2;
            break;
        case 1:
            cout << r1 << " is greater than " << r2;
            break;
        case -1:
            cout << r1 << " is less than " << r2;
            break;
    }
    cout << endl << endl;

    cout << "Testing the four arithmetic member functions:" << endl;
    r3.add(r1, r2);
    cout << "\tr1 + r2 = " << r1 << " + " << r2 << " = " << r3 << endl;
    r3.subtract(r1, r2);
    cout << "\tr1 - r2 = " << r1 << " - " << r2 << " = " << r3 << endl;
    cout << "\tr1 * r2 = " << r1 << " * " << r2 << " = " << r1.multiply(r2) << endl;
    cout << "\tr1 / r2 = " << r1 << " / " << r2 << " = " << r1.divide(r2) << endl;

}

编译命令g++ -c main.cpp rational.cpp

我收到错误

  no match for 'operator<<' in 'std::operator<< [with _Traits = std::char_traits<char>] 

另一整页错误。

1 个答案:

答案 0 :(得分:1)

使用Unix(Linux)上的gcc和clang / llvm编译器,我有关于两个运算符定义的rational::前缀的错误消息(对于现代g ++和clang ++,有颜色编码,红色表示错误):

$ gcc *.cpp -o a -w
In file included from main.cpp:1:0:
rational.h:20:10: error: extra qualification ‘rational::’ on member ‘operator>’ [-fpermissive]
     bool rational::operator>(const rational &r2);
          ^~~~~~~~

在类声明中声明的运算符不需要这些前缀,只需在类中使用bool operator>(const rational &r2);

其他错误会为r1.multiply(r2)r1.divide(r2)的两行生成大量文本,并且真正原因不容易找到。那是:

$ g++ main.cpp rational.cpp -w -o program
main.cpp: In function ‘int main()’:
main.cpp:33:58: error: no match for ‘operator<<’ (operand types are ‘std::basic_ostream<char>’ and ‘rational’)
     cout << "\tr1 * r2 = " << r1 << " * " << r2 << " = " << r1.multiply(r2) << endl;
     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~
....skip... no known conversion for ...
In file included from main.cpp:1:0:
rational.h:13:21: note: candidate: std::ostream& operator<<(std::ostream&, rational&) <near match>
     friend ostream& operator<<(ostream &, rational&);
                     ^~~~~~~~
rational.h:13:21: note:   conversion of argument 2 would be ill-formed:
main.cpp:33:72: error: invalid initialization of non-const reference of type ‘rational&’ from an rvalue of type ‘rational’
     cout << "\tr1 * r2 = " << r1 << " * " << r2 << " = " << r1.multiply(r2) << endl;
                                                             ~~~~~~~~~~~^~~~
main.cpp:34:58: error: no match for ‘operator<<’ (operand types are ‘std::basic_ostream<char>’ and ‘rational’)
     cout << "\tr1 / r2 = " << r1 << " / " << r2 << " = " << r1.divide(r2) << endl;
     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~

所以,编译器不能使用你的输出函数变体...为第二个参数ostream搜索“候选函数不可行:期望一个l值”我发现https://stackoverflow.com/a/23209236

  

你需要:

     

ostream& operator<< (ostream& out, Polynomial const& poly) {

     

这是因为临时对象不能绑定到非const引用。

因此,创建ostream& operator<<(ostream &out, rational &robj)方法的变体以接受const引用,如ostream& operator<<(ostream &out, rational const &robj),当您尝试在具有multipy和divide的行中输出临时有理对象时。

我的补丁:

diff -ur orig/rational.cpp new/rational.cpp
--- orig/rational.cpp   2018-04-18 06:30:14.734426770 +0300
+++ new/rational.cpp    2018-04-18 06:46:49.839818729 +0300
@@ -1,10 +1,15 @@
 #include "rational.h"
 using namespace std;
-
 ostream& operator<<(ostream &out, rational &robj)
 {
     out << robj.num << "/" << robj.denom;
     return out;
+}
+
+ostream& operator<<(ostream &out, rational const &robj)
+{
+    out << robj.num << "/" << robj.denom;
+    return out;
 }
 istream& operator>>(istream &in, rational &obj)
 {
diff -ur orig/rational.h new/rational.h
--- orig/rational.h     2018-04-18 06:20:14.431675001 +0300
+++ new/rational.h      2018-04-18 06:47:00.691799442 +0300
@@ -11,13 +11,14 @@
 {

     friend ostream& operator<<(ostream &, rational&);
+    friend ostream& operator<<(ostream &, rational const&);
     friend istream& operator>>(istream &, rational&);
 public:

     rational operator+(const rational &)const;
     rational operator-(const rational &)const;

-    bool rational::operator>(const rational &r2);
+    bool operator>(const rational &r2);

     rational(int n = 0, int d = 1);

输出:

$ g++ main.cpp rational.cpp -o program
$ ./program
r1 is initialized by the 2nd constructor: r1 = 1/4
r2 is initialized by the 2nd constructor: r2 = 1/3

Testing the comapre() member function, found:
        1/4 is less than 1/3

Testing the four arithmetic member functions:
        r1 + r2 = 1/4 + 1/3 = 7/12
        r1 - r2 = 1/4 - 1/3 = -1/12
        r1 * r2 = 1/4 * 1/3 = 1/12
        r1 / r2 = 1/4 / 1/3 = 3/4
$ clang++ main.cpp rational.cpp -o program
$ ./program
r1 is initialized by the 2nd constructor: r1 = 1/4
r2 is initialized by the 2nd constructor: r2 = 1/3

Testing the comapre() member function, found:
        1/4 is less than 1/3

Testing the four arithmetic member functions:
        r1 + r2 = 1/4 + 1/3 = 7/12
        r1 - r2 = 1/4 - 1/3 = -1/12
        r1 * r2 = 1/4 * 1/3 = 1/12
        r1 / r2 = 1/4 / 1/3 = 3/4