Question

这是我用C ++编写的源文件。

#include<string>
#include<sstream>
#include "Lecture.hpp"
#include <iostream>

using namespace std;


Lecture::Lecture() {
  capacity=5;
  log = new int[capacity];
  used = 0;
}

/*
 * The following is copy constructor.
 */
Lecture::Lecture(Lecture& orig) {
   copy(&orig);

}

/*
 * This is an empty destructor
 */
Lecture::~Lecture() {
    // dereference dynamic memory
}

 Lecture & Lecture:: operator=(Lecture & other){
     this->copy(&other);
     return *this;
 }
 /*
  * Copy method.
  */
 void Lecture::copy(Lecture &other){    
     if(&other != this){
         capacity = other.capacity;
         log = new int[capacity];
         used = other.used;
         for(int x = 0; x < used; x++){
            log[x]= other.log[x]; 
         }
     }
 }

string Lecture::getLogs() {
  ostringstream ans;
  ans << "[";
  for (int i=0; i<used-1; ++i) {
    ans << log[i] <<", ";
  }
  if (used>0) 
    ans << log[used-1] << "]";
  else
    ans << "empty log]";
  return ans.str();
}

void Lecture::addLogEntry(int b) {
   if (used==capacity) {
    capacity *= 2;
    int* temp= new int[capacity];
    for (int i=0; i<used; ++i) {
      temp[i]=log[i];
    }
    delete[] log;
    log=temp;
  }
  log[used]=b;
  used++;
}

从copy构造函数和overloaded = operator函数我试图调用copy（）函数。它给了我以下错误：

Lecture.cpp: In copy constructor `Lecture::Lecture(Lecture&)':
Lecture.cpp:26: error: no matching function for call to `Lecture::copy(Lecture*)'
Lecture.hpp:21: note: candidates are: void Lecture::copy(Lecture&)
Lecture.cpp: In member function `Lecture& Lecture::operator=(Lecture&)':
Lecture.cpp:38: error: no matching function for call to `Lecture::copy(Lecture*)'
Lecture.hpp:21: note: candidates are: void Lecture::copy(Lecture&)
make[2]: Leaving directory `/cygdrive/g/Aristotelis/C++/Assessment_2'
make[1]: Leaving directory `/cygdrive/g/Aristotelis/C++/Assessment_2'
make[2]: *** [build/Debug/Cygwin-Windows/Lecture.o] Error 1
make[1]: *** [.build-conf] Error 2
make: *** [.build-impl] Error 2

由于复制方法的某些原因，它需要一个指针。这是为什么？这是我的头文件：

#ifndef LECTURE_HPP
#define LECTURE_HPP

#include<string>
using namespace std;

class Lecture{ 

 public : 
  Lecture();                 // default constructor
  Lecture(Lecture &other);  // copy constructor
  string getLogs();
  void addLogEntry(int);
  void copy(Lecture &other); // copy method
  ~Lecture();                // destructor
  Lecture& operator=(Lecture& other); // overloading of '='

 private: 
  int* log;
  int used;
  int capacity;



};

#endif  /* LECTURE_HPP */

这是主要方法：

#include<iostream>
#include<string>
//#include "University.hpp"
#include <cmath>
#include <cstdlib>



using namespace std;

/*
 * 
 */
int main(int argc, char** argv) {


  Lecture c1;

  cout << "Information of c1: " <<c1.getLogs() << endl;

  c1.addLogEntry(20);

  cout << "Information of c1: " <<c1.getLogs() << endl;

  Lecture c2=c1;

  cout << "Information of c2: " <<c2.getLogs() << endl;

  Lecture c3;
  c3=c1;

  cout << "Information of c3: " <<c3.getLogs() << endl;

  c1.addLogEntry(-4);
  c2.addLogEntry(10);

  cout << "-----------------------------------------------"<< endl;

  cout << "Information of c1: " <<c1.getLogs() << endl;
  cout << "Information of c2: " <<c2.getLogs() << endl;
  cout << "Information of c3: " <<c3.getLogs() << endl;


    return 0;
}

可能是什么问题？

Answer 1

因为你传递了一个指针：

Lecture::Lecture(Lecture& orig) {
   copy(&orig);  // The & here is taking the address of orig (so remove it!)
}

[旁注1：除非你有充分的理由，否则你的复制构造函数等应该优先考虑非const参考的const参考。

[旁注2：请参阅this question了解在C ++中实现复制构造函数和复制赋值运算符的惯用方法。]

Answer 2

编译器说：

Lecture.cpp: In copy constructor `Lecture::Lecture(Lecture&)':
Lecture.cpp:26: error: no matching function for call to `Lecture::copy(Lecture*)'
Lecture.hpp:21: note: candidates are: void Lecture::copy(Lecture&)

你从错误中得出结论“复制方法......需要一个指针”，但你错了。复制方法需要引用。我们知道因为给出了“候选人”的名单。你试图用指针调用它，编译器说没有匹配带有指针的copy函数。所以不要传递指针。传递你已经拥有的参考资料。

调用复制方法

2 个答案: