C++运行时向量断言失败表达式：向量下标超出范围

Question

我收到了这个非常烦人的错误消息。我知道我对此很陌生，但这似乎是我能弄清楚的类型。谁能告诉我哪里出了问题吗？

运行时的消息是： 调试断言失败！ 程序： .... 文件：c:\program files\microsoft Visual Studio 10.0\vc\include\vector 线路：932 表达式：向量下标超出范围

，代码为

#include "VectorIntStorage.h"
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

void VectorIntStorage::Read(istream& r)
{
    char c[13];
    r >> c;
    r >> NumberOfInts; //gets number of ints for vector

    //numberVector = new std::vector<int> numberVector;

    for(int i = 0; i < NumberOfInts; i++)
    {
        r >> numberVector[i];
        cout << numberVector[i] << endl;

        if(_sortRead) //true
        {
            for(int k = 0; k < i; k++)
            {
                if(numberVector[i] < numberVector[k])
                {
                    int temp = numberVector[k];
                    numberVector[k] = numberVector[i];
                    numberVector[i] = temp;
                }
            }
        }
    }
}

void VectorIntStorage::Write(ostream& w)
{
    for(int i = 0; i < NumberOfInts; i++)
    {
        w << numberVector[i] << endl;
        cout << numberVector[i] << endl;
    }
}

void VectorIntStorage::sortStd()
{
    sort(numberVector.begin(), numberVector.end());
}

void VectorIntStorage::sortOwn()
{
    quickSort(0, NumberOfInts - 1);
}

void VectorIntStorage::setReadSort(bool sort)
{
    _sortRead = sort;
}

void VectorIntStorage::quickSort(int left, int right)
{
     int i = left, j = right;
      int tmp;
      int pivot = numberVector[(left + right) / 2];

      while (i <= j)
      {
            while (numberVector[i] < pivot)
                  i++;
            while (numberVector[j] > pivot)
                  j--;
            if (i <= j) 
            {
                  tmp = numberVector[i];
                  numberVector[i] = numberVector[j];
                  numberVector[j] = tmp;
                  i++;
                  j--;
            }
      }

      if (left < j)
      {
            quickSort(left, j);
      }
      if (i < right)
      {
            quickSort(i, right);
      }
}

VectorIntStorage::VectorIntStorage(const VectorIntStorage& copying)
{
    //int *duplicate = new int[(copying.NumberOfInts)];
    //vector<int> *duplicate = new vector<int>;

    //std::copy(numberVector.begin(), numberVector.end(), duplicate);
    //numberVector = duplicate;
    //NumberOfInts = copying.NumberOfInts;
}

VectorIntStorage::VectorIntStorage(void)
{
}


VectorIntStorage::~VectorIntStorage(void)
{
}

Answer 1

我们没有足够的信息来确定，但我怀疑失败的行是 r >>>; numberVector[i]。我想你的意思是 int j; r>> j; numberVector.push_back(j);

问题正是错误消息所说的：您的向量下标 (i) 超出范围。具体来说，您永远不会增加向量的大小，因此它的大小始终为 0。因此，operator[] 的任何使用都将引用超出范围的元素。

Answer 2

您不能只使用 numberVector[i] 而不先调用 numberVector.resize() 。

vector<int> vec;
vec[1] = 0; // fails - vec is empty so [1] is out of range
vec.resize(100);
vec[1] = 5; // ok, you can access vec[0] .. vec[99] now
vec.push_back(11); // Now the size is 101 elements, you can access vec[0] .. vec[100]

Answer 3

r >> NumberOfInts; //gets number of ints for vector

从上面的评论来看，您似乎需要一个大小为 NumberOfInts 的向量。但保留注释行 -

//numberVector = new std::vector<int> numberVector;

您将向量声明为 -

std::vector<int> numberVector; // The size of the vector is 0

要​​在 numberVector 上执行 [] 操作，应提及其大小并且应在有效范围内同时声明。由于在声明时没有提到，因此您需要执行 push_back 操作来动态增加向量的大小。

for(int i = 0; i < NumberOfInts; i++)
{
    r >> numberVector[i];    // Size isnot initially mentioned while declaration 
                             // of the vector to do an `[]` operation
    cout << numberVector[i] << endl;
    // ....