為什么這會給我無限循環？

Question

我試圖編寫一個簡單的程序：

import random

x = raw_input("How many rounds:")
rounds = 0

while rounds < x:
    # rock=0, paper=1, scissors=2
    computer1 = random.randint(0,2)
    computer2 = random.randint(0,2)

    if computer1 == computer2:
        print "draw"
    elif computer1 == 0 and computer2 == 1:
        print "lose"
    elif computer1 == 1 and computer2 == 2:
        print "lose"
    elif computer1 == 2 and computer2 == 0:
        print "lose"
    else:
        print "win"
    rounds = rounds + 1

為什么這會給我一個無限循環？ 當我取出輸入線並將x替換為某個值（例如10）時，輸出確實給了我10個結果。 但是為什么我不能用raw_input呢？

Answer 1

raw_input返回一個字符串，您需要將其raw_input轉換為int

x = int(raw_input("How many rounds:"))

請注意，在python中：

>>> 159 < "8"
True
>>> 159 < 8
False
>>> 

為了更好地理解int string比較，可以在此處查看 。

Answer 2

將輸入行更改為：

x = int(raw_input("How many rounds:"))

而且你應該很好走。 正如評論中指出的那樣， 默認的raw_input是string ，將整數與字符串進行比較不會得到所需的內容。

Answer 3

由於raw_input()始終返回一個str因此為了進行均勻比較，您需要將x的類型更改為int

import random

x = int(raw_input("How many rounds:"))

rounds = 0

Answer 4

為了確保用戶輸入所需的值（例如“ qwerty”或2.6），您可以添加異常處理：

import random
try:
    x = input("How many rounds: ")
    rounds = 0
    while rounds < int(x):
        # rock=0, paper=1, scissors=2
        computer1 = random.randint(0,2)
        computer2 = random.randint(0,2)

        if computer1 == computer2:
            print ("draw")
        elif computer1 == 0 and computer2 == 1:
            print ("lose")
        elif computer1 == 1 and computer2 == 2:
            print ("lose")
        elif computer1 == 2 and computer2 == 0:
            print ("lose")
        else:
            print ("win")
        rounds = rounds + 1
except (ValueError):
    print('Wrong input')

這段代碼是Python 3。