2

第二个函数给出错误 C2803 http://msdn.microsoft.com/en-us/library/zy7kx46x%28VS.80%29.aspx : 'operator ,' 必须至少有一个类类型的形式参数。任何线索?

template<class T,class A = std::allocator<T>> 
class Sequence : public std::vector<T,A> {
 public:
    Sequence<T,A>& operator,(const T& a) {
        this->push_back(a);
        return *this;
    }
    Sequence<T,A>& operator,(const Sequence<T,A>& a) {
        for(Sequence<T,A>::size_type i=0 ; i<a.size() ; i++) {
            this->push_back(a.at(i));
        }
        return *this;
    }
};

//this works!
template<typename T> 
Sequence<T> operator,(const T& a, const T&b) {
    Sequence<T> seq;
    seq.push_back(a);
    seq.push_back(b);
    return seq;
}

//this gives error C2803!
Sequence<double> operator,(const double& a, const double& b) {
    Sequence<double> seq;
    seq.push_back(a);
    seq.push_back(b);
    return seq;
}
4

2 回答 2

4

在 C++ 中,如果运算符的至少一个参数不是自定义类或不是像枚举这样的原始类型,则不能重载运算符。你不能重载类型+int同样你不能重载类型,double

于 2010-03-09T01:57:35.040 回答
2

将其更改为:

Sequence<double> operator,(const Sequence<double>& a, const double& b)
{
    Sequence<double> seq(a);
    seq.push_back(b);
    return seq;
}

或(基于本文):

Sequence<double> operator,(Sequence<double> seq, const double& b)
{
    seq.push_back(b);
    return seq;
}
于 2010-03-09T01:57:00.203 回答