[英]Overload operator<< for template class
I am having problem with overloading operator<< for a template class. 我有重载operator <<为模板类的问题。 I am using Visual Studio 2010, and here is my code.
我使用的是Visual Studio 2010,这是我的代码。
#ifndef _FINITEFIELD
#define _FINITEFIELD
#include<iostream>
namespace Polyff{
template <class T, T& n> class FiniteField;
template <class T, T& n> std::ostream& operator<< (std::ostream&, const FiniteField<T,n>&);
template <class T, T& n> class FiniteField {
public:
//some other functions
private:
friend std::ostream& operator<< <T,n>(std::ostream& out, const FiniteField<T,n>& obj);
T _val;
};
template <class T, T& n>
std::ostream& operator<< (std::ostream& out, const FiniteField<T,n>& f) {
return out<<f._val;
}
//some other definitions
}
#endif
In main I just have 在主要我只是
#include"FiniteField.h"
#include"Integer.h"
#include<iostream>
using std::cout;
using namespace Polyff;
Integer N(5);
int main () {
FiniteField<Integer, N> f1;
cout<< f1;
}
where Integer
is just a wrapper of int
with some special functionality I need. 其中
Integer
只是int
的包装器,具有我需要的一些特殊功能。
However, when I compile the above code, I got error C2679, which says binary '<<' : no operator found which takes a right-hand operand of type 'Polyff::FiniteField<T,n>' (or there is no acceptable conversion)
但是,当我编译上面的代码时,我得到了错误C2679,它表示
binary '<<' : no operator found which takes a right-hand operand of type 'Polyff::FiniteField<T,n>' (or there is no acceptable conversion)
I have also tried to remove the parameters in the friend declaration so the code becomes: 我还尝试删除友元声明中的参数,以便代码变为:
friend std::ostream& operator<< <> (std::ostream& out, const FiniteField<T,n>& obj);
But this produce another error: C2785: 'std::ostream &Polyff::operator <<(std::ostream &,const Polyff::FiniteField<T,n> &)' and '<Unknown>' have different return types
但这会产生另一个错误:C2785:
'std::ostream &Polyff::operator <<(std::ostream &,const Polyff::FiniteField<T,n> &)' and '<Unknown>' have different return types
so I am wondering how should I change the code so it compiles and why? 所以我想知道我应该如何更改代码以便编译以及为什么? Thanks!
谢谢!
------------------------- edited on 2012.12.31 --------------------------- -------------------------编辑于2012.12.31 -------------------- -------
The code compiles with g++ now. 代码现在用g ++编译。 Here is the github repository.
这是github存储库。
This seem to work as expected: 这似乎按预期工作:
namespace Polyff{
template <class T, T* n> class FiniteField;
template <class T, T* n> std::ostream& operator<< (std::ostream&, const FiniteField<T,n>&);
template <class T, T* n> class FiniteField {
public:
//some other functions
private:
friend std::ostream& operator<< <T,n>(std::ostream& out, const FiniteField<T,n>& obj);
T _val;
};
template <class T, T* n>
std::ostream& operator<< (std::ostream& out, const FiniteField<T,n>& f) {
return out << f._val.n; // I added the field of my Integer class
}
//some other definitions
}
struct Integer{
Integer() : n(0){}
Integer(int nn) : n(nn){}
int n;
};
using std::cout;
using namespace Polyff;
Integer N(5);
int main () {
FiniteField<Integer, &N> f1;
cout<< f1;
}
I just replaced the reference by the pointer of your object in the template arguments, since a pointer to a global object (static or not) is an information known at compile-time (or at least link time). 我刚刚在模板参数中用对象的指针替换了引用,因为指向全局对象(静态或非静态)的指针是在编译时(或至少链接时)已知的信息。 I am not aware of the language accepting references.
我不知道接受引用的语言。
Note that in this example, 0
will be printed because it corresponds to the default construction of _val
. 请注意,在此示例中,将打印
0
因为它对应于_val
的默认构造。
I tried to compile your code on my Visual C++ 2010. I got the same error as you did. 我试图在我的Visual C ++ 2010上编译你的代码。我得到了和你一样的错误。
There are actually two things that Visual did not like in your code: 在你的代码中,Visual实际上有两件事情不喜欢:
That made the job for me! 这使我的工作!
您应该使用普通变量( T n
而不是T& n
)替换引用。
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