重载运算符<<用于原始类型.那可能吗? [英] Overloading operator<< for primitive types. Is that possible?

问题描述

嘿.是否可能使运算符<<对于原始类型? Fx可以说我每次想写一个int时都要写一个std :: endl.我可以重载运算符<< for int，以便它自动将std :: endl放到输出中?我已经尝试过了，

Hey. Is it possible to overload operator<< for primitive types? Fx lets say that I want to write a std::endl each time want to write a int. Can I overload operator<< for int, so that it automatic puts a std::endl to the output? I have tried with this,

std::ostream& operator<<(std::ostream& strm, int & i)
{
   strm << i << std::endl;
   return strm;
}

但是它不起作用.我无法回忆起编译器错误消息，但我认为我正在让操作员以任何方式重载所有错误. 我尝试调用上述重载运算符<<这样，

but it doesn't work. I cant recall the compiler error message, but I think that I'm getting operator overloading all wrong any ways. I try to call the above overloaded operator<< in this way,

int main()
{
   int i = 2;
   std::out<<"Here is an int " << i;

   return 0;
}

但是它根本不起作用.也许我不能重载POD类型?

But it doesn't work at all. Maybe I can't overload POD types?

推荐答案

正如zabzonk所说，标准库提供了(ostream& int)重载，因此您无法定义另一个.

As zabzonk said, the standard library provides an (ostream&, int) overload so you can't define another.

模拟您在做什么(尽管以目前的形式它是毫无意义的):

To simulate what you were doing (though it is completely pointless in its present form :) :

class EndlinedInteger {
public:
    EndlinedInteger(int i) : i(i) { }
    friend ostream& operator<<(ostream&, EndlinedInteger const&);
private:
    int i;
};

ostream& operator<<(ostream& out, EndlinedInteger const& ei) {
    out << ei.i << endl;
    return out;
}

int main()
{
   EndlinedInteger i = 2;
   std::cout<<"Here is an int " << i;
}

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