It may seem obvious to some people, but to elaborate on the above answers:
If you are doing removal of std::vector elements using erase in a loop over the whole vector, you should process your vector in reverse order, that is to say using
for (int i = v.size() - 1; i >= 0; i--)
instead of (the classical)
for (int i = 0; i < v.size(); i++)
The reason is that indices are affected by erase so if you remove the 4-th element, then the former 5-th element is now the new 4-th element, and it won't be processed by your loop if you're doing i++.
Below is a simple example illustrating this where I want to remove all the odds element of an int vector;
#include <iostream>
#include <vector>
using namespace std;
void printVector(const vector<int> &v)
{
for (size_t i = 0; i < v.size(); i++)
{
cout << v[i] << " ";
}
cout << endl;
}
int main()
{
vector<int> v1, v2;
for (int i = 0; i < 10; i++)
{
v1.push_back(i);
v2.push_back(i);
}
// print v1
cout << "v1: " << endl;
printVector(v1);
cout << endl;
// print v2
cout << "v2: " << endl;
printVector(v2);
// Erase all odd elements
cout << "--- Erase odd elements ---" << endl;
// loop with decreasing indices
cout << "Process v2 with decreasing indices: " << endl;
for (int i = v2.size() - 1; i >= 0; i--)
{
if (v2[i] % 2 != 0)
{
cout << "# ";
v2.erase(v2.begin() + i);
}
else
{
cout << v2[i] << " ";
}
}
cout << endl;
cout << endl;
// loop with increasing indices
cout << "Process v1 with increasing indices: " << endl;
for (int i = 0; i < v1.size(); i++)
{
if (v1[i] % 2 != 0)
{
cout << "# ";
v1.erase(v1.begin() + i);
}
else
{
cout << v1[i] << " ";
}
}
return 0;
}
Output:
v1:
0 1 2 3 4 5 6 7 8 9
v2:
0 1 2 3 4 5 6 7 8 9
--- Erase odd elements ---
Process v2 with decreasing indices:
# 8 # 6 # 4 # 2 # 0
Process v1 with increasing indices:
0 # # # # #
Note that on the second version with increasing indices, even numbers are not displayed as they are skipped because of i++