在surf.h中有我的后续代码,其中声明了具有两种不同类型的模板类:
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;
然后定义一个新类,在其中创建类型为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)进行编译,并收到以下错误:
在main.cpp包含的文件中:2:0: field.h:在‘std :: ostream&运算符<<(std :: ostream&,const field&)的实例中[T = surf; std :: ostream = std :: basic_ostream]’: main.cpp:9:9:从这里开始 field.h:36:7:错误:“ operator <<”不匹配(操作数类型为“ std :: ostream {aka std :: basic_ostream}”和“ const surf”) str << fE << endl;
我在做什么?
答案 0 :(得分:2)
您因const
超载而错过了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)
// ^^^^^
//...
此const是必需的,因为您尝试显示来自const field<T> & ov