问题描述
Int这个类,其中operator<<在尝试使用gcc 4.6.1编译它时定义(见代码)我得到以下错误:没有匹配'操作符<<'in'std :: cout< a'。这是怎么回事?
Int this class where operator<< is defined (see code) while trying to compile it with gcc 4.6.1 I'm getting following error: no match for 'operator<<' in 'std::cout << a'. What's going on?
template<class Int_T = int, typename Best_Fit<Int_T>::type Min_Range = std::numeric_limits<Int_T>::min(),
typename Best_Fit<Int_T>::type Max_Range = std::numeric_limits<Int_T>::max()>
class Int
{
Int_T data_;
Int_T get_data()const
{
return data_;
}
};
//Here is this operator defined
template<class Int_T>
std::ostream& operator<<(std::ostream& out, const Int<Int_T, Best_Fit<Int_T>::type, Best_Fit<Int_T>::type>& obj)
{
out << obj.get_data();
return out;
}
其中Best_Fit如下所示:
where Best_Fit looks like:
#ifndef BEST_FIT_H_INCLUDED
#define BEST_FIT_H_INCLUDED
struct Signed_Type
{
typedef long long type;
};
struct Unsigned_Type
{
typedef unsigned long long type;
};
template<bool Cond, class First, class Second>
struct if_
{
typedef typename First::type type;
};
template<class First, class Second>
struct if_<false,First,Second>
{
typedef typename Second::type type;
};
template<class Int_T>
struct Best_Fit
{//evaluate it lazily ;)
typedef typename if_<std::is_signed<Int_T>::value,Signed_Type,Unsigned_Type>::type type;
};
#endif // BEST_FIT_H_INCLUDED
编辑:
#include <iostream>
int main(int argc, char* argv[])
{
Int<signed char,1,20> a(30);
cout << a;
}
推荐答案
一个类型和两个常见的已知最佳拟合类型,但您的模板运算符< 三种类型。
Your template has three arguments, a type, and two constants of a known best fit type, but your templated
operator<<
takes an instantiation of the template with three types.
template<class Int_T = int, typename Best_Fit<Int_T>::type Min_Range
= std::numeric_limits<Int_T>::min(), // constant!
typename Best_Fit<Int_T>::type Max_Range
= std::numeric_limits<Int_T>::max() // constant!
>
class Int
//...
template<class Int_T>
std::ostream& operator<<(std::ostream& out,
const Int<Int_T,
Best_Fit<Int_T>::type, // type!
Best_Fit<Int_T>::type // type!
>& obj)
我通常建议类模板在类定义中定义(使用 friend
在该上下文中定义一个自由函数)因为这个特殊的原因,在类模板中获取类型是很容易的,容易在其外失效。还有一些其他的差异(像事实,如果运算符被定义在类中,那么它只能通过ADL访问 - 无论你也决定在外部声明)
I usually recommend that operator overloads of class templates are defined inside the class definition (use
friend
to define a free function in that context) for this particular reason, it is trivial to get the types right inside the class template, and easy to fail outside of it. There are a couple other differences (like the fact that if the operator is defined inside the class then it will only be accessible through ADL --unless you also decide to declare it outside)
template<class Int_T = int, typename Best_Fit<Int_T>::type Min_Range
= std::numeric_limits<Int_T>::min(), // constant!
typename Best_Fit<Int_T>::type Max_Range
= std::numeric_limits<Int_T>::max() // constant!
>
class Int {
friend // allows you to define a free function inside the class
std::ostream& operator<<( std::ostream& out,
Int const & obj ) { // Can use plain Int to refer to this
// intantiation. No need to redeclare
// all template arguments
return out << obj.get_data();
}
};
这篇关于缺少运算符<<但它在那里的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!