Question

我有一些理解指针数组的问题，请参阅以下简化代码

#include <string>
#include <iostream>
using namespace std;
//holds two different datatypes
template <typename A, typename B>
class data {
public:
    A dataA;
    B dataB;
    data(){}
    data* add(A dataA, B dataB){
        this->dataA = dataA;
        this->dataB = dataB;
        return this;
    }
};
//stores an array of the data
template <typename  A, typename B>
class container {
    data<A, B> **dataArray;
    int ammount;
public:
    typedef data<A, B> * iterator;
    container(){
        //create dynamic array of pointers
        dataArray = new data<A, B>*[5];
        ammount = 0;
    }
    ~container(){
        for (int i = 0; i <= ammount; i++)
        {
            delete dataArray[i];
        }
    }

    void add(A dataA, B dataB){
        data<A, B> *newItem;
        if (ammount <= 5){
            newItem = new data<A, B>();
            dataArray[ammount] = newItem->add(dataA, dataB);
            ammount++;
        }
    }
    //basic iterators
    iterator begin(){
        return *dataArray;
    }
    iterator end(){
        return dataArray[5];
    }
};

void main(){
    string somestuff[] = { "one", "two", "three", "four", "five" };
    container<string, int> foo;
    for (int i = 0; i < 5; i++){
        foo.add(somestuff[i], i);
    }
    container<string, int>::iterator it = foo.begin();
    // why cant i just iterate through each item in the array
    for (it; it < foo.end(); it++)
    {
        cout << "Data A : " << it->dataA << " Data B : " << it->dataB << "\n";
    }

}

问题是主要的，我不能在显示时移动数组。我看这个问题都错了吗？

ps我知道STL容器在这里会更好但是我需要理解这个特定的场景，即指向对象的动态数组（不使用向量）

**编辑我添加了一些输出语句来显示我想要实现的目标

输出

Adding data
Added Data at address : 00779218
Added Data at address : 00779458
Added Data at address : 007794B8
Added Data at address : 00779560
Added Data at address : 00779730
Printing data
The address of 'it' is now: 00779218
Data A : one Data B : 0
incerementing from map.begin
The address of 'it' is now: 00779238
Data A :  Data B : 469784212
incerementing from map.begin
The address of 'it' is now: 00779258
Data A : X Data B : -33686019
incerementing from map.begin
The address of 'it' is now: 00779278

当我这样做时

for (it; it < foo.end(); it++){
        cout << "The address of 'it' is now: "<<it<<"\n";
        cout << "Data A : " << it->dataA << " Data B : " << it->dataB << "\n";
        cout << "incerementing from map.begin\n";

    }

我在dataArray的下一个项目中需要'it'变量顶点，它正在递增而不正确

Answer 1

试试这个程序：

#include <string>
#include <iostream>
using namespace std;
//holds two different datatypes
template <typename A, typename B>
class data {
  public:
    A dataA;
    B dataB;
    data* _next;
    data(){_next = NULL;}
    data* add(A dataA, B dataB){
      this->dataA = dataA;
      this->dataB = dataB;
      return this;
    }


    bool operator<(data& rhs)
    {
      if (this->dataB ==rhs.dataB);
      return true;
      return (this->dataB <rhs.dataB);
    }

    data* operator++()
    {
      if(this == NULL)
        return NULL;
      return _next;
    }
};
//stores an array of the data
template <typename  A, typename B>
class container {
  data<A, B> **dataArray;
  int ammount;
  public:
  typedef data<A, B> * iterator;
  container(){
    //create dynamic array of pointers
    dataArray = new data<A, B>*[5];
    ammount = 0;
  }
  ~container(){
    for (int i = 0; i < ammount; i++)
    {
      delete dataArray[i];
    }
  }

  void add(A dataA, B dataB){
    data<A, B> *newItem;
    if (ammount <= 5){
      newItem = new data<A, B>();
      dataArray[ammount] = newItem->add(dataA, dataB);
      if(ammount != 0)
      { 
        dataArray[ammount -1]->_next = dataArray[ammount];
      }
      ammount++;
    }
  }
  //basic iterators
  iterator begin(){
    return *dataArray;
  }
  iterator end(){
    return dataArray[4];
  }


};

int main(){
  string somestuff[] = { "one", "two", "three", "four", "five" };
  container<string, int> foo;
  for (int i = 0; i < 5; i++){
    foo.add(somestuff[i], i);
  }
  container<string, int>::iterator it = foo.begin();
  // why cant i just iterate through each item in the array
  for (it; *it < *foo.end(); it =++ *it)
  {
    if(it != NULL)
    cout << "Data A : " << it->dataA << " Data B : " << it->dataB << "\n";
    else
      return  0;
  }
  return 0;

}

迭代指针数组并显示

1 个答案: