c++ - 尝试重载 "+"运算符时收到错误消息说我的类没有命名类型

Question

当我为重载运算符 + 的友元函数编写定义时，给我错误消息的部分在实现文件中。它说统计学家没有命名类型。但它是一个友元函数，并且写在包含 header 的实现文件中，所以我不确定为什么它不识别这个。我还意识到我拼错了文件名的统计员，但不知道如何轻松地重命名代码块中的文件。

//header file   
#ifndef STATISTICIAN_H
#define STATISTICIAN_H
namespace GREGORY_STOCKER_STATICTICIAN{



class Statistician{

public:
    Statistician();
    void next_number(double);
    void erase_sequence();
    int get_length() const {return length_sequence;}
    double get_sum() const{return sum;}
    double get_mean() const;
    double get_largest() const;
    double get_smallest() const;
    double get_last() const;
    friend Statistician operator + (const Statistician &,const Statistician &);


private:
    int length_sequence;
    double sum;
    double smallest;
    double largest;
    double last;

};

#endif

}


//implementation file
using namespace std;
#include "Statictician.h"
#include <iostream>
#include <cassert>

namespace GREGORY_STOCKER_STATICTICIAN{

    Statistician :: Statistician()
    {
        length_sequence = 0;
        sum = 0;
        smallest = 0;
        largest = 0;
        last = 0;
    }

    void Statistician :: next_number(double num)
    {

        length_sequence += 1;
        sum += num;
        if(length_sequence == 1)
        {
            smallest = num;
            largest = num;
        }
        if (num < smallest)
            smallest = num;
        if (num > largest)
            largest = num;
        last = num;
    }

    void Statistician :: erase_sequence()
    {
        length_sequence = 0;
        sum = 0;
        smallest =0;
        largest = 0;
        last = 0;
    }

    double Statistician :: get_mean () const
    {
        assert(length_sequence > 0);
            return sum / 2;

    }

    double Statistician ::  get_largest() const
    {
        assert(length_sequence > 0);
        return largest;
    }
    double Statistician ::  get_smallest() const
    {
        assert(length_sequence > 0);
        return smallest;
    }

    double Statistician :: get_last() const
    {
        assert(length_sequence > 0);
        return last;
    }

}



//the part that is tripping the error message    
Statistician operator +(const Statistician &s1,const Statistician &s2)


{
    Statistician s3;
    s3.sum = (s1.sum + s2.sum);
    s3.sequence_length = (s1.sequence_length + s2.sequence_length;
    if(s1. largest > s2.largest)
        s3.largest = s1.largest;
    else
        s3.smallest = s2.smallest;
        if(s1. smallest < s2.smallest)
        s3.smallest = s1.smallest;
    else
        s3.smallest = s2.smallest;
    s3.last = s2.last;

    return s3;
}

Answer 1

类中的 friend 声明，在最小的封闭命名空间中声明函数。因此，您的 friend 声明实际上声明了和 friends GREGORY_STOCKER_STATICTICIAN::operator +。它既不声明也不友善 ::operator +。

但是您的代码试图在命名空间之外实现 ::operator +。这产生了一个完全不同的功能:它不会被任何试图将两个 Statisticians 加在一起的代码找到，因为该代码只能找到命名空间版本。此外，它甚至无法编译(因为您在 Arnav Borborah 的回答下发布了错误消息):因为 ::operator + 不是 friend，它无法访问私有(private)成员。

所以最简单的解决方案实际上是将 operator+ 定义放在命名空间中，以便它与声明相匹配:

namespace GREGORY_STOCKER_STATICTICIAN
{
    Statistician operator +(const Statistician &s1,const Statistician &s2)
    {
        // ...
    }
}

现在您也不需要资格 Statistician。