why isn't my vector pushbacking even and odd numbers - c++

Question

i tried to separate even and odd numbers using vectors from a array ==>
so i made a function that returns true is number is even and false for if number is odd
then i used an if else statement where if the function returns true then it pushbacks the value in a vector and if the function returns false then it pushbacks the value in another vector , finally i printed all the elements in the vector but the output does not show any element except it shows one in the odd vector.
#include <iostream>
#include <vector>
using namespace std;
bool sort(int arr[] , int i){
if(arr[i] %2 == 0){
return true;
}
return false;
}
int main(){
int n;
cin >> n;
int *arr = new int[n];
for(int i=1 ; i<=n ; i++){
arr[i-1] = i;
}
vector <int> even , odd;
int i=0 ;
if(sort(arr , i)){
even.push_back(arr[i]);
sort(arr , i+1);
}else{
odd.push_back(arr[i]);
sort(arr,i+1);
}
cout << "the even numbers are : " << endl;
for(auto element:even){
cout << element << " ";
}
cout << endl;
cout << "the odd numbers are : " << endl;
for(auto element:odd){
cout << element << " ";
}
}

Answer 1

As #TonyDelroy said, you have to make for loop around call to sort(arr, i). Also first loop should go up to i <= n instead of i < n.
Your fixed working code below (see also std::partition_copy variant afterwards):
Try it online!
#include <iostream>
#include <vector>
using namespace std;
bool sort(int arr[] , int i){
if(arr[i] %2 == 0){
return true;
}
return false;
}
int main(){
int n;
cin >> n;
int *arr = new int[n];
for(int i=1 ; i<=n ; i++){
arr[i-1] = i;
}
vector <int> even , odd;
for (int i = 0; i < n; ++i)
if (sort(arr, i))
even.push_back(arr[i]);
else
odd.push_back(arr[i]);
cout << "the even numbers are : " << endl;
for(auto element:even){
cout << element << " ";
}
cout << endl;
cout << "the odd numbers are : " << endl;
for(auto element:odd){
cout << element << " ";
}
}
Input:
10
Output:
the even numbers are :
2 4 6 8 10
the odd numbers are :
1 3 5 7 9
As #chris said you can also use std::partition_copy to implement your algorithm:
Try it online!
#include <algorithm>
#include <iostream>
#include <iterator>
#include <vector>
int main() {
int n = 0;
std::cin >> n;
std::vector<int> arr(n), odd, even;
for (int i = 1; i <= n; ++i)
arr[i - 1] = i;
std::partition_copy(arr.cbegin(), arr.cend(),
std::back_insert_iterator(odd), std::back_insert_iterator(even),
[](auto const & x){ return (x & 1) == 1; });
std::cout << "the even numbers are : " << std::endl;
for (auto element: even)
std::cout << element << " ";
std::cout << std::endl << "the odd numbers are : " << std::endl;
for (auto element: odd)
std::cout << element << " ";
}
Input:
10
Output:
the even numbers are :
2 4 6 8 10
the odd numbers are :
1 3 5 7 9

Answer 2

You only push one element - the first.
Your partitioning code is equivalent to
if(sort(arr , 0)){
even.push_back(arr[0]);
sort(arr , 1);
}else{
odd.push_back(arr[0]);
sort(arr,1);
}
You need to loop over all the input numbers.
You can also simplify matters with a more generally useful evenness function that doesn't depend on an array:
bool is_even(int x) { return x % 2 == 0; }
and then there is no need to store all the inputs before processing them:
int main(){
vector <int> even , odd;
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
int x;
cin >> x;
if (is_even(x)) {
even.push_back(x);
}
else {
odd.push_back(x);
}
}
cout << "the even numbers are : " << endl;
for (auto element:even){
cout << element << " ";
}
cout << endl;
cout << "the odd numbers are : " << endl;
for (auto element:odd){
cout << element << " ";
}
}