Question

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;
}

Answer 1

在C ++中，如果至少有一个参数不是自定义类或者不是像enum这样的原始类，则不能重载运算符。您不能为+类型重载int，同样也不能为,类型重载double。

Answer 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this article）：

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

C ++覆盖全局运算符逗号给出错误

2 个答案: