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John Murphy
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Asked: 2022-05-18T16:33:34+00:00 In:

Resolved the problem of the c++ vector subscript out of range.

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I am working on cpp, but I found the following warning message:

Debug assertion failed. C++ vector subscript out of range

Is there any way to stabilize the issue “c++ vector subscript out of range”? I read a lot of topics about this, but all of them were trying to install anything. Is this the correct way, or any recommendation for me? Please find the beginning command below:

#include "stdafx.h"
 #include "iostream"
 #include "vector"
 using namespace std;
 
 int _tmain(int argc, _TCHAR * argv[])
 {
  vector<int> v;
 
  cout << "Hello India" << endl;
  cout << "Size of vector is: " << v.size() << endl;
  for (int i = 1; i <= 10; ++i)
  {
  v.push_back(i);
 
  }
  cout << "size of vector: " << v.size() << endl;
 
  for (int j = 10; j > 0; --j)
  {
  cout << v[j];
  }
 
  return 0;
 }
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2 Answers

  1. Best Answer
    Expert

    The cause:

    This error happens because arr[10] is out of range, it has not been assigned yet. No matter how you index pushbacks, your vector has 10 elements indexed within the range from 0 (0 or 1) to 9, and all other 0 elements. Therefore, your second loop contains v[j], from 1 to 9.

    Solution:

    To fix the error, you can do as below:

    for(int j = 9;j >= 0;--j)
{
 cout << v[j];
}

    It will be better if you think of indexes as 0-based. Therefore, we recommend you also replace your first loop by this:

    for(int i = 0;i < 10;++i)
{
 v.push_back(i);
}

    Moreover, iterators also can have access to the contents of a container. In this case, iterators can be used (in this example is a reverse iterator).

    for (vector<int>::reverse_iterator i = v.rbegin(); i != v.rend(); ++i)
{
 std::cout << *i << std::endl;
}
    • 5

  2. v contains 10 element. The index starts at 0 to 9. 

    for(int j=10;j>0;--j)
{
 cout<<v[j]; // v[10] out of range
}

    You should update the for loop 

    for(int j=9; j>=0; --j)
// ^^^^^^^^^^
{
 cout<<v[j]; // out of range
}

    You can also use reverse iterator for printing elements in reverse order 

    for (auto ri = v.rbegin(); ri != v.rend(); ++ri)
{
 std::cout << *ri << std::endl;
}
    • 15

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