Using return vs no return in recursion?

Here is code to reverse an array using recursion

Using return rev(arr,++start,–end);

#include <iostream>
using namespace std;

void rev(int arr[],int start,int end)
{
    if(start >= end)
    {
        return;
    }
    int temp = arr[start];
    arr[start] = arr[end];
    arr[end] = temp;
    return rev(arr,++start,--end);
}

void reverse(int arr[],int size)
{
    rev(arr,0,size-1);
}

Using rev(arr,++start,–end);

void rev(int arr[],int start,int end)
{
    if(start >= end)
    {
        return;
    }
    int temp = arr[start];
    arr[start] = arr[end];
    arr[end] = temp;
    rev(arr,++start,--end);
}

void reverse(int arr[],int size)
{
    rev(arr,0,size-1);
}

They both give same output 7 6 5 4 3 2 1

What is the difference between using return and not using return with rev here?

>Solution :

There is no difference.

From 9.6.3 [stmt.return]:

A return statement with no operand shall be used only in a function
whose return type is cv void, a constructor (15.1), or a destructor
(15.4). A return statement with an operand of type void shall be used
only in a function whose return type is cv void.

[…]

Flowing off
the end of a constructor, a destructor, or a function with a cv void
return type is equivalent to a return with no operand.

Because the type of the function call expression is the cv-qualified return type of the function as defined in its signature, and because your function is defined to return void, then the three that follow are equivalent:

void f() {
    //stuff
    f();
    return;
}

void g() {
    //same stuff
    return g();
}

void h() {
    //same stuff
    h();
}