What is happening inside of my printFun function that is causing this behaviour-CodePudding

I am trying to figure out recursion and how it operates and I cant seem to figure out what is happening in this code.

def printFun(test):
 
    if (test < 1):
        return
    else:
 
        print(test, end="a ")        
        printFun(test-1)  # statement 2
        print(test, end="n ")
        return
 
# Driver Code
test = 3
printFun(test)

This outputs

3a 2a 1a 1n 2n 3n

I can make sense of the first 4 outputs. test = 3, which not less than 1, so print test(1a), then re-call the printFun function with test-1 being 2, which is not less than 1, so print test (2a), then (1a) then 0, which IS less than 1 so return. I assume this brings you back to the

print(test, end='n')

line? which now prints 1n.

This is where I am left perplexed... what is happening beyond this??? How does it start ascending and then stop again at 3? What is the flow and logic of this? Sorry if this is a ridiculous question and I am overlooking something blatantly obvious.

But I cannot wrap my mind around this... Anyone?

Thanks!

CodePudding user response：

Its because the stack unwinds depth first. In pseudocode, with each indentation being a new call to the function, you get

call printFun(3)
    print 3a
    call printFun(2)
        print 2a
        call printFun(1)
            print 1a
            call printFun(0)
                print nothing
                return
            (test still = 1 in this frame)
            print 1n
            return
        (test still = 2 in this frame)
        print 2n
        return
    (test still = 3 in this frame)
    print 3n
    return

When you return from the most recently called printFun, you get back to an older set of local variables holding the older value.

CodePudding user response：

You call printFun three times and each of it prints twice, so we should have 6 prints, isn't it?

Sometimes it's hard to unsolve recursion but it doesn't differ as calling another function:

def foo():
  print("before")
  other_foo()
  print("after")

Do you agree everything that is printed by other_foo will be between "before" and "after"? It's the same case, you could write it that way:

def printFun1():
    print("1a")
    print("1n")


def printFun2():
    print("2a")
    printFun1() # it prints "1a" and "1n" between "2a" and "2n"
    print("2n")


def printFun3():
    print("3a")
    printFun2()
    print("3n")

printFun3()