How to remove odd numbers from array - c++

Question

It looks like every time there are consecutive odd numbers three lines in a
row in array. One in the middle that does not get removed.
I see why this is happening but I'm not too sure how I can fix it.
This is for C++.
#include <iostream>
using namespace std;
void removeodd(int arr[],int& number)//{1,2,3,6,7,9,5,8};
{int count=0;
for(int i=0; i<number; i++)
{
if(arr[i]%2!=0)
{
int j=i+1;
for(int k=i;k<number; k++)
{
arr[k]=arr[j];
j++;
}
count++;
}
}
number=number-count;
for(int i=0; i<number; i++){
cout<<arr[i]<<" ";
}
}
int main(){
int size=8;
int ary[8]={1,2,3,6,7,9,5,8};
removeodd(ary, size);
return 0;
}
output of the code above is 2 6 9 8

Answer 1

The problem is that you're always advancing your index, even if you remove an odd number. So if there are two odd numbers, you remove the first one and move the index past the second one. To fix this, decrease the index again when you remove and odd number. So after this:
count++;
Also put this:
i--;
And the output you get is:
2 6 8

Answer 2

in your for loop..
assume value when the value 1 is deleted.
and in array ary is { 2,3,6,7,9,5,8 }
but i value is 1, so you did't check value 2 is odd number.
so fix the loop check all values.

Answer 3

You are making things difficult on yourself using an array of fundamental types (e.g. int) instead of a C++ container which contains member functions that allow you to .erase() any element in your array. A choice of a vector<int> will allow you to remove odd values in a trivial manner.
For example,
#include <vector>
...
std::vector<int> arr {1,2,9,6,7,3,5,8};
will create your array as a vector<int>. As a vector container you can simply iterate over each element in your vector from arr.begin() to arr.end() using a simple iterator. The only caveat, is you must revalidate the iterator after each erasure as iterators are invalidated on container changes. To revalidate your iterator, you simply assign the result of the .erase() operation, e.g.
for (auto it = arr.begin(); it != arr.end();) { /* iterate vector */
if (*it & 1u) /* if value is odd */
it = arr.erase(it); /* erase it saving/revalidating iter */
else /* otherwise */
it++; /* simply advance iter */
}
Putting it together in a short example that outputs the original arr, removes odd numbers, and then outputs arr again with the odd numbers removed, you could do:
#include <iostream>
#include <vector>
int main (void) {
std::vector<int> arr {1,2,9,6,7,3,5,8};
for (auto& v: arr) /* output original vector contents */
std::cout << " " << v;
std::cout << '\n';
for (auto it = arr.begin(); it != arr.end();) { /* iterate vector */
if (*it & 1u) /* if value is odd */
it = arr.erase(it); /* erase it saving/revalidating iter */
else /* otherwise */
it++; /* simply advance iter */
}
/* output remaining even values */
for (auto& v: arr)
std::cout << " " << v;
std::cout << '\n';
}
Example Use/Output
$ ./bin/vect_remove_odd
1 2 9 6 7 3 5 8
2 6 8
Look things over and let me know if you have further questions.