当我修改，修改和删除原始指针时，是否需要指针副本？

Question

Rao (2012, p. 180, listing 8.9)说：“您创建的10号线的副本的原因是，这样的循环修改指针正在通过增量运算符（++）。通过新的需要返回原来的指针用来存放完好相应delete[]在第26行需要使用新返回的地址来调用，而不是任何随机值“。

0: #include <iostream> 
1: using namespace std; 
2: 
3: int main() 
4: { 
5: cout << “How many integers you wish to enter? “; 
6: int InputNums = 0; 
7: cin >> InputNums; 
8: 
9: int* pNumbers = new int [InputNums]; // allocate requested integers 
10: int* pCopy = pNumbers; 
11: 
12: cout<<“Successfully allocated memory for “<<InputNums<< “ integers”<<endl; 
13: for(int Index = 0; Index < InputNums; ++Index) 
14: { 
15: cout << “Enter number “<< Index << “: “; 
16: cin >> *(pNumbers + Index); 
17: } 
18: 
19: cout << “Displaying all numbers input: “ << endl; 
20: for(int Index = 0, int* pCopy = pNumbers; Index < InputNums; ++Index) 
21: cout << *(pCopy++) << “ “; 
22: 
23: cout << endl; 
24: 
25: // done with using the pointer? release memory 
26: delete[] pNumbers; 
27: 
28: return 0; 
29: } 

是在pCopy真的有必要吗？我错过了什么？ 在下面的修改示例中，我不使用它并且delete[]似乎正常工作。

#include <iostream>  
using namespace std; 

int main() 
{ 

    cout << "How many integers do you want? " << endl; 
    int InputN = 0; 
    cin >> InputN; 

    int* pNumbers = new int [InputN]; 

    cout << "allocated memory for " << InputN << " integers" << endl; 

    for (int Idx = 0; Idx < InputN; Idx++) { 
     cout << "enter number for index " << Idx << endl; 
     cin >> *(pNumbers + Idx); 
    } 

    cout << "Display all input numbers: " << endl; 


    for (int Idx = 0; Idx < InputN + 2; ++Idx) { 
     cout << "integer with index: " << Idx << " has value " << *(pNumbers + Idx) << " and pointer: " << pNumbers + Idx << endl; 
     //cout << pNumbers - 50000 << " points to " << *(pNumbers - 50000) << endl; 
     } 

    delete[] pNumbers; 

    cout << "after the delete: " << endl; 
    for (int Idx = 0; Idx < InputN + 2; ++Idx) { 
     cout << "integer with index: " << Idx << " has value " << *(pNumbers + Idx) << " and pointer: " << pNumbers + Idx << endl; 
     //cout << pNumbers - 50000 << " points to " << *(pNumbers - 50000) << endl; 
     } 

    return 0; 
} 

Answer 1

在以下情况

{ 
    type* ptr = new type[size];  // line A 
    /* 
    some code, for example 
    ptr++;      // legal but dangerous/bad 
    */ 
    delete[] ptr;     // line B 
} 

可变ptr的值，即指向的地址，必须在线路A和B相同。否则，这是未定义的行为，并导致崩溃（如果你幸运的话）。在你的第二个代码清单中，你不使用指针变量来遍历元素，而是使用索引。所以，原始指针从未改变，并且您满足条件。

确保满足此要求有几种方法。

1. 您可以声明指针为const。

   { 
       type * const ptr = new type[size]; // not: type const*ptr 
       // ptr++;       // illegal: would give a compiler error 
       delete[] ptr; 
   } 
   

2. 您可以使用smart pointer

   { 
       std::unique_ptr<type[]> ptr = new type[size]; 
       /* some code */ 
   } // memory is automatically freed at end of scope of ptr 
   

3. 可以使用的容器，如std::vector

   { 
       std::vector<type> vec(size); 
       /* some code */ 
   } // memory is automatically freed at end of scope of ptr 
   

的最后一个选项是最方便的，因为它还允许您更改分配的内存量，或添加元素beyon d最初分配的内存（当矢量自动重新分配和扩展时）。