Question

我正在尝试制作一个python计算器。但由于某种原因，它只会输出平方根选项。另外，如果您有任何有助于改进我的代码的提示，请提供。我目前在我的iPhone上使用python3.4和3.3模拟器，但它在两个设备上都有同样的问题，谢谢先进。所有代码都正确缩进，但堆栈交换让我编辑它，

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":这解析为if (opt == "square-root") or bool("squareroot")，返回True因为对字符串调用bool总是True，除非字符串是空。（有关详细信息，请参阅link）

修复将是：

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

或

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

Python代码输出错误的目标

1 个答案: