Bubble Sort Optimization C++

Question

I was just practising a little bit and tried to sort an array with the bubble sort algorithm. Compiler didn't give me any warnings nor errors and it worked well! First you type 10 times a number and then the program sorts them + prints them.

Code:

#include <iostream>
using namespace std;

void arr_sort(int* array, const int arr_size){

int temp = 0;   //Temporary integer to store (if necessary) the current element
int end = 0;    //Run time condition

while(end++ != arr_size){ // Will loop max. 10 times

    for(int i = 0; i < arr_size; i++){

        if(array[i] > array[i + 1]){    //If the current element 
            temp = array[i];    //is bigger than the next

            array[i] = array[i + 1];//Change the positions
            array[i + 1] = temp;    
        }
    }
}

}

int main(){

int arr_input[10];

for(int i = 0; i < 10;i++)      //The user has to type 10 numbers
    cin >> arr_input[i];        //which will be stored in this array

arr_sort(arr_input, 10);        //sorts the array

cout << endl << endl;

for(int i = 0; i < 10; i++)     //Print out the array!
    cout << arr_input[i] << ", ";
cout << endl;

return 0;
}

My only problem is the while-loop in the arr_sort function. I mean it sorts the array until end has the same value as arr_size. But often it doesn't need that long. My question now... How can I improve this function? How can I test if the array is completely sorted so the while-loop can stop without running another time and another time ...?

Answer 1

Before your for loop, assume it's sorted:

bool sorted = true;

In your if statement`, record that it's not sorted:

sorted = false;

After your for loop`, return if there was no evidence that it's not sorted:

if ( sorted ) return;