Using return vs no return in recursion?-CodePudding

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?

CodePudding user response：

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();
}

CodePudding user response：

#include <iostream>
using namespace::std;

void rev();

void rev(){
    return 1;
}
int main(){
    int val=rev();
    cout<<rev<<endl;
    return 0;
}

g   test.cpp -o test
test.cpp: In function ‘void rev()’:
test.cpp:7:9: error: return-statement with a value, in function returning ‘void’ [-fpermissive]
    7 |  return 1;
      |         ^
test.cpp: In function ‘int main()’:
test.cpp:10:13: error: void value not ignored as it ought to be
   10 |  int val=rev();

Compiler is generating error that if return type is void we cannot return any value and there is no need to return any value because the array which is passed as an argument is being modified.

Why your compiler is not generating error because both of your methods are returning noting.

void foo(){
    return;//returns nothing
}

#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);//doesn't returns any value
}

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);
}