Is this called recursion.

[bond5611]

I am also laughing at myself but really can this thing be called recursion?

int function3(int num){
            if(num==0)
            return 1;
            else {
            cout<<"*" <<" ";
            return function3(num-1);
            }
            }    
    
int function2(int num2)
{
    static int num=num2;
    static int acb=0;
    if(acb==num)
    return 1;
    else{
    acb=acb+1;
    function3(acb);     
    cout<<endl;
    return function2(acb);
    }  
}  
int function(int num){
     static int abc=num;
     static int acb=num;
     if(num==0 && abc==0)
     {
//     cout<<acb;
     
     return  function2(acb);
     }
     
     if(num==0){
                cout<<endl;
                abc=abc-1;
                
                num=abc;
                function(num);
                }
     else{
     cout<<"*"<<" ";
     return function(num-1);
     }
     }

calling function from main.

[firedraco]

Yeah, why? Seems like a really horrible way to implement something but, shrug.

[Stewbond]

Recursion is a function that calls itself. All of these functions do that.

You may get some problems with that specific code specifically I am seeing that if num==0 and abc!=0 then there will be no return value for int function(). but yes, it is recursive, even with the static variables in there.

[lukecian]

to use recursion correctly

you have to define the terminate condition, so that it wont

cause the stack overflow.

[andywestken]

Recursion does not require a function to call itself directly. You can also have indirect recursion, where one function calls another function which eventually calls the original function. e.g.

func1 -> func2 -> func3 -> func1 -> func2 -> func3 -> ...

But direct recursion (func1 -> func1 -> func1 -> ...) is much more common and a lot easier to get your head around.