Python if的行为不像预期的那样

Question

我在试着做一个蟒蛇计算器。但出于某种原因，它只会输出平方根选项。

我目前在我的iPhone上使用python3.4和3.3模拟器，但在这两种设备上都存在相同的问题。

print("This program is a calculator app \n Creator: AtPython")
print("...")
Print("please choose a       mathematical operation")
Opt = input("square-root. \n addition, \n subtraction, \n multiplication, \n division: ")

if opt == "square-root" or "squareroot":
    num = float(input('Enter a number: '))
    num_sqrt = num ** 0.5
    print('The square root of %0.3f is %0.3f'%(num ,num_sqrt))
    input("Press 'Enter' to exit")

If opt == "addition":
    add_1 = float(input("Please enter a number"))
    add_2 = float(input("Please enter another number")) 
    add_ans = add_1 + add_2
    print(add_ans)
    qus = input("Would you like to add   another number? [y/n]")
    
elif qus == "y":
    add_3 = float(input("Enter a third number"))
    input("Press 'Enter' to exit")
    
else:
    input("Press 'Enter' to exit")

Answer 1

if opt == "square-root" or "squareroot":

这不解析为if opt == "square-root" or opt == "squareroot":，此解析为if (opt == "square-root") or bool("squareroot")，这将返回True，因为调用字符串上的bool总是True，除非字符串是空的。(更多信息请参见链接 )

解决办法是：

if opt == "square-root" or opt == "squareroot":

或

if opt in ["square-root","squareroot"]: