调用自身内部的函数以使用if语句检查if输入，如果if语句永不中断且无限循环

Question

I'm writing a coin toss function that only accepts "h" and "t" as input.

To do this I use an if statement to check if the input is "h", and one nested in it to check if it is "t".

If both conditions aren't met, it calls the function again.

The problem is that even if the first input is "t" or "h", it still calls the function no matter what I do?

def coin_flip():
    player_coin = raw_input("Choose heads (h) or tails (t)!: ").lower
    if player_coin != "h":
        if player_coin != "t":
            coin_flip()

    coin = randint(1, 2)
    if coin == 1:
        coin = "h"
    else:
        coin = "t"

    if player_coin == coin:
        print "You won the coin toss! You get to go first!"
        player_turn()

    else:
        print "You guessed the wrong answer! The Computer goes first!"
        comp_turn()

Answer 1

Instead of calling the function again, just stick that part in a while loop with the following condition:

player_coin = None
while player_coin not in ('h', 't'):
    player_coin = raw_input("Choose heads (h) or tails (t)!: ").lower()

I used x in tuple instead of multiple checks, and made sure you're invoking lower() since it isn't in your example.