[英]template class operator overloading multiple typenames C++
我在surf.h中有followign代碼,其中聲明了具有兩種不同類型的模板類:
using namespace std;
template <typename T1, typename T2>
class surf;
template <typename T1, typename T2>
ostream & operator << (ostream & str, surf<T1,T2> & ov);
template <typename T1, typename T2>
class surf
{
public:
surf(T1 v1, T2 v2):
v1_(v1),
v2_(v2)
{}
friend ostream & operator << <T1, T2> (ostream & str, surf<T1,T2> & ov);
T1 v1_;
T2 v2_;
};
template <typename T1, typename T2>
ostream & operator << (ostream & str, surf<T1,T2> & ov)
{
str << "("<<ov.v1_<<","<<ov.v2_<<")";
return str;
}
typedef surf<int,double> intSurf;
然后定義一個新類,在其中創建類型為T的向量(在field.h中)
template<typename T>
class field;
template<typename T>
ostream & operator << (ostream & str, const field<T> & ov);
template<typename T>
class field
{
public:
field( int n, T val):
f_(n,val)
{}
friend ostream & operator << <T> (ostream & str, const field<T> & ov);
protected:
vector<T> f_;
};
template<typename T>
ostream & operator << (ostream & str, const field<T> & ov)
{
for(auto &fE: ov.f_)
{
str << fE << endl;
}
return str;
}
typedef field<intSurf> surfField;
在main.cpp中,我使用此字段。
#include "field.h"
int main()
{
surfField a(4, intSurf(2,5));
cout<< a << endl;
return true;
}
我使用g ++(5.4版)進行編譯,並得到以下錯誤:
在main.cpp:2:0:field.h中包含的文件中:'std :: ostream&運算符<<(std :: ostream&,const field&)的實例[T = surf; std :: ostream = std :: basic_ostream]':main.cpp:9:9:此處為必填字段。h:36:7:錯誤:“ operator <<”不匹配(操作數類型為“ std :: ostream” {aka std :: basic_ostream}'和'const surf')str << fE << endl;
我在做什么?
您因operator <<
重載而缺少const
operator <<
template <typename T1, typename T2>
ostream & operator << (ostream & str, const surf<T1,T2> & ov);
// ^^^^^
//...
friend ostream & operator << <T1, T2> (ostream & str, const surf<T1,T2> & ov);
// ^^^^^
//...
template <typename T1, typename T2>
ostream & operator << (ostream & str, const surf<T1,T2> & ov)
// ^^^^^
//...
需要此const,因為您嘗試顯示來自const field<T> & ov
元素
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