模糊运算符<<

Question

#include "stdafx.h"
#include "Record.h"

template<class T>//If I make instead of template regular fnc this compiles  
//otherwise I'm getting an error (listed on the very bottom) saying  
// that operator << is ambiguous, WHY?
ostream& operator<<(ostream& out, const T& obj)
{
    out << "Price: " 
        << (obj.getPrice()) << '\t'//this line causes this error
        << "Count: "
        << obj.getCount()
        << '\n';
    return out;
}

int _tmain(int argc, _TCHAR* argv[])
{
    vector<Record> v;
    v.reserve(10);
    for (int i = 0; i < 10; ++i)
    {
        v.push_back(Record(rand()%(10 - 0)));
    }
    copy(v.begin(),v.end(),ostream_iterator<Record>(cout, "\n"));
    return 0;
}

//Record class
class Record
{
    private:
        int myPrice_;
        int myCount_;
        static int TOTAL_;
    public:
        Record(){}
        Record(int price):myPrice_(price),myCount_(++TOTAL_)
        {/*Empty*/}
        int getPrice()const
        {
            return myPrice_;
        }

        int getCount()const
        {
            return myCount_;
        }
        bool operator<(const Record& right)
        {
            return (myPrice_ < right.myPrice_) && (myCount_ < right.myCount_);
        }
};

int Record::TOTAL_ = 0;

错误2错误C2593:'operator <<'不明确

Answer 1

背后的概念operator<<( ostream &, ... )是每个类都有自己的重载,以一种有意义的方式处理特定的类.

这意味着你可以operator<<( ostream &, const Record & )处理Record对象,operator<<( ostream &, const std::string & )处理标准字符串,以及operator<<( ostream &, const FooClass & )处理FooClass对象.这些函数中的每一个都知道如何处理它声明的对象类型,因为它们中的每一个都需要不同的处理.(例如getPrice()/ getCount()for Record,或getFoo()/ getBar()for FooClass.)

你的模板正在践踏整个概念.通过将其定义为模板函数(可以匹配任何类),您不仅要碰到operator<<()标准/代码库中已有的许多定义,而且还要碰撞所有可能的重载.

编译器如何决定是使用operator<<( ostream &, const std::string & )还是模板？它不能,所以它绝望地抛出手并放弃.这就是错误告诉你的.