Reputation: 35
I'm asking for help with a problem implying an array sorting in the following manner: all even numbers must be in front of the odd ones. I've partially made the problem, but I did the sorting in the opposite manner and I can not manage to fix it.
#include "stdafx.h"
#include <iostream>
using namespace std;
int main()
{
int v[100], n, i, aux = 0, inv;
cout << "Number of elements: ";
cin >> n;
for (i = 0; i < n; i++)
{
cout << "v[" << i << "]=";
cin >> v[i];
}
do
{
inv = 0;
for (i = 0; i < n; i++)
{
if (v[i] % 2 == 1 && v[i + 1] % 2 == 0)
{
inv = 1;
aux = v[i];
v[i] = v[i + 1];
v[i + 1] = aux;
}
}
} while (inv != 0);
cout << endl;
for (i = 0; i < n; i++)
cout << v[i] << " ";
cout << endl;
system("pause");
return 0;
}
The output for this would be:
n = 8
1 3 2 4 7 8 4 2
Result: 2 4 8 4 2 -858993460 1 3
Upvotes: 2
Views: 53
Reputation: 26800
In the expression v[i + 1]
, you access v[n]
when i = n - 1
, this will result in an out-of-bounds memory access which results in undefined behaviour.
You should change the for
loop to this:
for (i = 0; i < n - 1; i++)
The output for the given input is:
a.exe
Number of elements: 8
v[0]=1
v[1]=3
v[2]=2
v[3]=4
v[4]=7
v[5]=8
v[6]=4
v[7]=2
2 4 8 4 2 1 3 7
Upvotes: 1