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我一直在尝试解决这个问题一段时间，并决定是时候从外部角度看待问题了。我已经尽可能多地用谷歌搜索，从我读到的类似问题中，这（我认为）与我分配的内存少于所需的内存有关。我已经多次检查我的代码，但找不到我的错误在哪里。

我的教授编写了一个打印函数，并且仅在调用该函数时才会出现问题，因此我将搜索范围缩小到我编写的这个函数。

IntStore& IntStore::operator=(const IntStore& rhs)
{
   if(this != &rhs)
   {
      int* newData = new int[rhs.capacity];
      int* newFreq = new int[rhs.capacity];
      for(int i=0; i < rhs.used; i++)
      {
        newData[i] = rhs.data[i];
        newFreq[i] = rhs.freq[i];
      }
      delete [] data;
      delete [] freq;
      data = newData; 
      freq = newFreq;
      capacity = rhs.capacity;
      used = rhs.used;
   }
   return *this;      
}

他的打印功能如下：

void print_to_cout(IntStore src)
{
   // NOTE:
   // - first for-loop below looks silly but is purposely added
   // - don't try to remove/disable when doing the assignment
   // - Why is it added?
   for (int i = 1; i < 2; ++i)
   {
      src = src;
      IntStore copy1;
      copy1 = src;
   }

   int countDist = src.countDistinct();
   for (int i = 1; i <= countDist; ++i)
      cout << setw(5) << src.valAt(i);
   cout << endl;
   cout << "        (freq)";
   for (int i = 1; i <= countDist; ++i)
      cout << setw(5) << src.freqAt(i);
   cout << endl;
}

当我运行程序时，这是我得到的：

*** glibc detected *** ./a2: free(): invalid next size (fast): 0x0000000005714010 ***

任何帮助表示赞赏并提前感谢。

我包括构造函数以显示代码的更具体部分以帮助解决我的问题。至于打印功能。我不知道为什么它是这样写的。主 cpp 文件中的所有内容均由教授创建，不可编辑。我们所做的唯一编辑是对 Instore.cpp 文件。标题也是预制的。

#include "IntStore.h"
#include <iostream>
#include <cstdlib>
#include <cassert>
using namespace std;

void IntStore::resize(int new_capacity)
{
   if(new_capacity < used)
     new_capacity = used;
   if(new_capacity < 1)
     new_capacity = 1;
   capacity = new_capacity;
   int* newData = new int[capacity];
   int* newFreq = new int[capacity];
   for(int i=0; i < used; i++)
   {
       newData[i] = data[i];
       newFreq[i] = freq[i];
   }
   delete [] data;
   delete [] freq;
   data = newData;
   freq = newFreq;
}

IntStore::IntStore(int init_capacity) : capacity(init_capacity), used(0)
{
   if(capacity < 1)
     capacity = DEFAULT_CAPACITY;
   data = new int(capacity);
   freq = new int(capacity);
}

IntStore::IntStore(const IntStore& src) : capacity(src.capacity),
                                                  used(src.used)
{
   data = new int(capacity);
   freq = new int(capacity);
   for(int i = 0; i < used; i++)
   {
     data[i] = src.data[i];
     freq[i] = src.freq[i];
   }
}

IntStore::~IntStore()
{
   delete [] data;
   delete [] freq;
}

这是打印功能的情况：

  case 'p': case 'P':
     objectNum = get_object_num();
     switch (objectNum)
     {
     case 1:
        if ( is1.empty() )
           cout << "   is1: (empty)" << endl;
        else
        {
           cout << "   is1: (data)";
           print_to_cout(is1);
        }
        break;
     case 2:
        if ( is2.empty() )
           cout << "   is2: (empty)" << endl;
        else
        {
           cout << "   is2: (data)";
           print_to_cout(is2);
        }
        break;
     case 3:
        if ( is3.empty() )
           cout << "   is3: (empty)" << endl;
        else
        {
           cout << "   is3: (data)";
           print_to_cout(is3);
        }
     }
4

1 回答 1

1

在您的复制构造函数中new int(capacity)，您应该拥有new int[capacity]. 这些根本不是一回事。

您可能会考虑为您的赋值运算符使用复制和交换习语——您所拥有的大部分是正确的，但相当可怕，并且可能更短更简单。

于 2012-07-15T08:36:50.580 回答