Thread: Need help understanding function

I am reviewing material for finals when i cam across this question to find out what prints from the following function

when print(4)

Code:

void print(int i)
{
     if(i>=1)
     {
             printf("%d",i);
             print(i-1);
             printf("%d",i);
     }
}

I understand that it enters with 4, the function is called within itself for i-1 so it reenters with 3, prints 3, than does it until i is 1 enters for the last time which then i which is 1 and 1-1 = 0, so it cannot enter the function no more however when i put compile the function i get the following


43211234, can anyone explain this output

i understand the 4321, thereafter is where i am lost

thanks in advance

Originally Posted by anduril462

Patience child. Don't bump your thread, it's against the forum guidelines. Read this: Announcements - C Programming.

Now, think about what happens after the call to print(i-1). Where do you return to when the recursive function is done? What happens after that point?

my bad, to be honest we didnt cover recursion in class, so that i guess why it doesnt make sense to me, anyways upon first look i thought that it enters at 4, print out the int then it would get to that function call and restart with 3 until it got to 0 and then wouldnt enter and it would always skip the second printf because of the function call