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搜索这个问题的标题给了我很多人引用相同的错误,但在不同的情况下,不幸的是,那里提供的答案是针对他们的情况的,我看不出他们能如何帮助我。

我正在尝试operator<<为模板类重载。下面是一个测试用例:

矢量.h:

#ifndef __INCL_VECTOR_H__
#define __INCL_VECTOR_H__

#include <array>

template < class T, unsigned int N >
class Vector
{
public:
    Vector();
    Vector( std::array< T, N > );

    template < class U, unsigned int M > friend Vector< U, M > operator+ ( const Vector< U, M >&, const Vector< U, M >& );

    template < class U, unsigned int M > friend std::ostream& operator<< ( std::ostream&, Vector< U, M >& );

    T& operator[] ( const unsigned int& );

protected:
    std::array< T, N > _values;
};

#include "Vector.hpp"

#endif

矢量.hpp:

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

template < class T, unsigned int N >
Vector< T, N >::Vector()
{
}

template < class T, unsigned int N >
Vector< T, N >::Vector( std::array< T, N > otherArray )
{
    _values = *( new std::array< T, N >( otherArray ) );
}

template < class U, unsigned int M >
Vector< U, M > operator+ ( const Vector< U, M > &lhVector, const Vector< U, M > &rhVector )
{
    Vector< U, M > sumVector;

    for( unsigned int i = 0; i < M; i++ )
        sumVector[i] = lhVector[i] + rhVector[i];

    return sumVector;
}

template < class U, unsigned int M >
std::ostream& operator<< ( std::ostream &out, Vector< U, M > &cVector )
{
    out << "< ";

    for( int i = M - 1; i >= 0; i-- )
    {
        out << cVector[i];
        if( i )
            out << ", ";
    }

    out << " >";

    return out;
}

template < class T, unsigned int N >
T& Vector< T, N >::operator[] ( const unsigned int &index )
{
    return _values[ index ];
}

矢量测试.cpp:

#include "Vector.h"

#include <iostream>
#include <array>

using namespace std;

int main( int argc, char* argv[] )
{
    Vector< int, 3 > u( array< int, 3 > {  1, 4,  2 } );
    Vector< int, 3 > v( array< int, 3 > { -2, 3, -1 } );

    cout << "u = " << u << endl;
    cout << "v = " << v << endl;
    cout << "u + v = " << u + v << endl;

    return 0;
}

导致错误的行是cout << "u + v = " << u + v << endl;; 前两行按预期工作。

报错信息如下(编译为g++ -std=c++11 Vector.h vectorTest.cpp):

vectorTest.cpp: In function ‘int main(int, char**)’:
vectorTest.cpp:15:31: error: cannot bind ‘std::basic_ostream<char>’ lvalue to ‘std::basic_ostream<char>&&’
In file included from /usr/include/c++/4.7/iostream:40:0,
                 from Vector.hpp:2,
                 from Vector.h:34:
/usr/include/c++/4.7/ostream:600:5: error:   initializing argument 1 of ‘std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&&, const _Tp&) [with _CharT = char; _Traits = std::char_traits<char>; _Tp = Vector<int, 3u>]’
In file included from Vector.h:34:0:
Vector.hpp: In instantiation of ‘Vector<U, M> operator+(const Vector<U, M>&, const Vector<U, M>&) [with U = int; unsigned int M = 3u]’:
vectorTest.cpp:15:31:   required from here
Vector.hpp:40:9: error: passing ‘const Vector<int, 3u>’ as ‘this’ argument of ‘T& Vector<T, N>::operator[](const unsigned int&) [with T = int; unsigned int N = 3u]’ discards qualifiers [-fpermissive]
Vector.hpp:40:9: error: passing ‘const Vector<int, 3u>’ as ‘this’ argument of ‘T& Vector<T, N>::operator[](const unsigned int&) [with T = int; unsigned int N = 3u]’ discards qualifiers [-fpermissive]

我无法理解这些错误消息告诉我什么。我会很感激任何帮助。

4

2 回答 2

6

第一个问题:

要使您的程序编译,只需使用左值引用const作为您的第二个参数operator <<(在friend-declaration 和该函数的定义中):

template < class U, unsigned int M >
std::ostream& operator<< ( std::ostream &out, Vector< U, M > const& cVector )
//                                                           ^^^^^

您的程序无法编译的原因是您的重载operator <<接受对 non- 的左值引用const作为其第二个参数,并且对 non- 的左值引用const 不能绑定到右值。

由于operator +两个实例之间的结果Vector是一个临时的,而一个临时的是一个右值,编译器无法调用您的operator <<,因此无法解析调用。

第二个问题:

解决上述问题后,您将不得不解决第二个问题:您的Vector类模板不提供 的const版本operator [],因此您的 rewritten operator <<(现在接受对const向量的引用)将无法访问向量的元素.

template < class T, unsigned int N >
class Vector
{
    // ...

    T& operator[] ( const unsigned int& );

    T const& operator[] ( const unsigned int& ) const; // <== ADD THIS!

    // ...
};

当然还有相应的定义:

template < class T, unsigned int N >
T const& Vector< T, N >::operator[] ( const unsigned int &index ) const
{
    return _values[ index ];
}
于 2013-03-25T21:29:15.217 回答
3

改变这个:

std::ostream& operator<< ( std::ostream&, Vector< U, M >& );

对此:

std::ostream& operator<< ( std::ostream&, const Vector< U, M >& );
//                                        ^^^^^

编译器告诉您,C++ 不允许您将临时变量绑定Vectoru + vnon -const Vector&

而且你不修改它Vector,所以它应该const开始的。

于 2013-03-25T21:27:47.000 回答