如何根据输入打印出正确的响应？

Question

I'm new to Python and I have a very simple problem that I'd like explained to me. I'm making a text adventure game and I'd like to be able to call another function inside a function so my code is organized better.

This is what I have that's working:

def displayIntro():
    print("You come to a crossroads on your trip home.")
    print ("You can go LEFT towards the DEMON FOREST or RIGHT to SKULL MOUNTAIN")
    print()

def choosePath(): 
    path = ""
    while path != "LEFT" and path != "RIGHT":
        path = input("Which path will you choose? LEFT or RIGHT? --> ")
    if path == "RIGHT":
        print("You take the path to SKULL MOUNTAIN.")
    elif path == "LEFT":
        print ("You take the path to the DEMON FOREST.")




displayIntro()
choosePath()
choosenPath() 

But I'd like it to look something like this, but I don't know how to make it work:

def displayIntro():
    print("You come to a crossroads on your trip home.")
    print ("You can go LEFT towards the DEMON FOREST or RIGHT to SKULL MOUNTAIN")
    print()

def choosePath(): 
    path = ""
    while path != "LEFT" and path != "RIGHT":
        path = input("Which path will you choose? LEFT or RIGHT? --> ")
    return path


def checkPath(choosePath):

    if choosePath == "RIGHT":
        print("You take the path to SKULL MOUNTAIN.")
    elif choosePath == "LEFT": 
        print("You take the path to the DEMON FOREST.")



displayIntro()
choosePath()
checkPath() 

Answer 1

You're not passing the return value of choosePath() to checkPath() . It should be:

path = choosePath()
checkPath(path)

Or as a one-liner:

checkPath(choosePath())

Answer 2

You have to pass the return value of choosePath to checkPath .

checkPath(choosePath())

checkPath takes a parameter, so you need to supply a value. That value is the return value of choosePath , either "LEFT" or "RIGHT". Once you pass it in, the checkPath function will execute, checking the string and printing accordingly. Replace the choosePath and checkPath call with the above code.