在运行时对向量执行C ++断言失败表达式：向量下标超出范围

Question

即时消息这真烦人的错误消息。 我知道我对此并不陌生，但似乎可以弄清楚。 谁能告诉我我哪里出问题了？

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行：932表达：矢量下标超出范围

和代码是

#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可以肯定地说，但是我怀疑失败的行是r >> numberVector[i] 。 我想你是说int j; r >> j; numberVector.push_back(j); int j; r >> j; numberVector.push_back(j);

问题恰恰是错误消息所说的：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，因此向量的大小始终为0。因此，对operator[]任何使用都将引用超出范围的元素。

Answer 2

您不能不先调用numberVector.resize()就使用numberVector[i] 。

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

从上面的评论看来，您似乎需要一个NumberOfInts大小的向量。 但是，请保留注释行-

//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[]的操作，应在声明时提及它的大小，并且大小应在有效范围内。 由于声明时未提及，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;
    // ....