Question

在泛型编程的家庭作业中，程序没有编译。我应该通过仅将代码插入指定位置并更改其他内容来解决此问题。我试图通过重载运算符来使编译器理解它们对该自定义类型的含义，但我仍然得到相同的错误。我现在不知所措。

#include <iostream>

using namespace std;

template <typename T>
struct TripleData {
    T m_first;
    T m_second;
    T m_third;  
    T getMean() {
        return (m_first + m_second + m_third)/3;
    }

    // INSERT CODE HERE
    void operator = (const T & value) {
        m_first = m_second = m_third = value;  
    }

    void operator += (const TripleData & tridata) {
        m_first += tridata.m_first;
        m_second += tridata.m_second;
        m_third += tridata.m_third;
    }

    T & operator / (const T & divisor) {
        return m_first/divisor;
    }
    //END OF INSERTED CODE
};

template <typename T, int N>
class GenericStaticMatrix {
    T m_data[N];
public:
    T & operator()(int i) {
        return m_data[i];
    }
    int getSize() const {
        return N;
    }
    T getMean() const {
        T sum = 0;                            //Error: conversion from 'int' to non-scalar type 'TripleData<double>' requested
        for (int i=0;i<N;i++) sum+=m_data[i]; 
        T ret = sum/(double)N;                //Error: conversion from 'double' to non-scalar type 'TripleData<double>' requested
        return ret;
    }
};


int main() {

    const int size = 10;
    int i;     
    GenericStaticMatrix<TripleData<double>,size> gsm_tdd;

    for (i=0;i<size;i++) {
        gsm_tdd(i).m_first = 1.1 + i;
        gsm_tdd(i).m_second = 2.2 + i;
        gsm_tdd(i).m_third = 3.3 + i;
    }

    cout << gsm_tdd.getMean().m_first <<endl;
    cout << gsm_tdd.getMean().m_second <<endl;
    cout << gsm_tdd.getMean().m_third <<endl;

    return 0;
}

提前致谢！

Answer 1

需要为TripleData<double>一般运行的代码是：

T sum = 0;                // (A)
sum += m_data[i];         // (B)
T ret = sum / (double)N;  // (C)

只需从此代码暗示的隐式接口向后工作。

// (A) implies non-explicit construction from an int
TripleData(int ); 

// (B) implies operator+=, you have this one almost right
TripleData& operator+=(const TripleData& rhs);

// (C) implies division by a **scalar** - this should NOT modify the object
TripleData operator/(double ) const;

您完全错过了(A) - 您编写了一个赋值运算符，但T x = y;并非分配，而是copy-initialization。 (B)基本上没问题，(C)您采用了错误类型的操作数。我们除以double，而不是TripleData<double>。

转换为请求的非标量类型

1 个答案: