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C++覆盖全局运算符逗号给出错误

comma-operator operators c++

1

uray · 技术社区 · 16 年前

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参。有什么线索吗?

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;
}

2 回复 | 直到 16 年前

1

2

R Samuel Klatchko 16 年前

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;
}

2

3

vava 16 年前

在C++中,如果它的至少一个参数不是自定义类或非原始的枚举类,则不能重载运算符。你不能超载 + 对于 int 类型,类似地,您不能重载 , 对于 double 类型。