Question

我的代码遇到问题。每当我从头开始尝试使用python 3时，我最终会遇到无限循环。我不确定我做错了什么。这是我想要开始工作的计算器......

loop = 1 # 1 means loop; anything else means don't loop.
choice = 0 # This variable holds the user's choice in the menu

while loop == 1:
    # Print what options you have
    print ("Welcome to StandardCalculator.py")

    print ("your options are:")
    print (" ")
    print ("1) Addition")
    print ("2) Subtraction")

    print ("3) Multiplication")

    print ("4) Division")
    print ("5) Raise to the Power")
    print ("6) Quit StandardCalculator.py")
    print (" ")

choice = input("Enter choice(1/2/3/4):")
if choice == 1:
        add1 = input("Add this: ")
        add2 = input("to this: ")
        print (add1, "+", add2, "=", add1 + add2)
elif choice == 2:
        sub2 = input("Subtract this: ")
        sub1 = input("from this: ")
        print (sub1, "-", sub2, "=", sub1 - sub2)
elif choice == 3:
        mul1 = input("Multiply this: ")
        mul2 = input("with this: ")
        print (mul1, "*", mul2, "=", mul1 * mul2)
elif choice == 4:
        div1 = input("Divide this: ")
        div2 = input("by this: ")
        print (div1, "/", div2, "=", div1 / div2)
elif choice == 5:
        pow1 = input ("Raise this: ")
        pow2 = input ("by: ")
        print (pow1, "**", pow2, "=", pow1 ** pow2)                     
elif choice == 6:
        loop = 0

print ("Thank-you for using StandardCalculator.py!")

Answer 1

这里的问题最终是缩进。目前，您的while循环只会继续打印您的提示，绝不允许您输入选择或将逻辑应用于该选项。

此外，当您获得choice输入时，它是String格式，您将它与int进行比较。让我们假设输入有效，您有两个选择：

int那个。
```
choice = int(choice)
```

与字符串比较

if choice == "1":
    ...
elif choice == "2":
    ...

修复缩进和类型问题后，请查看AndreL的答案。

loop = 1 # 1 means loop; anything else means don't loop.
choice = 0 # This variable holds the user's choice in the menu

while loop == 1:
    # Print what options you have
    print ("Welcome to StandardCalculator.py")

    print ("your options are:")
    print (" ")
    print ("1) Addition")
    print ("2) Subtraction")

    print ("3) Multiplication")

    print ("4) Division")
    print ("5) Raise to the Power")
    print ("6) Quit StandardCalculator.py")
    print (" ")

    choice = input("Enter choice(1/2/3/4):")
    choice = int(choice)
    if choice == 1:
            add1 = input("Add this: ")
            add2 = input("to this: ")
            print (add1, "+", add2, "=", add1 + add2)
    elif choice == 2:
            sub2 = input("Subtract this: ")
            sub1 = input("from this: ")
            print (sub1, "-", sub2, "=", sub1 - sub2)
    elif choice == 3:
            mul1 = input("Multiply this: ")
            mul2 = input("with this: ")
            print (mul1, "*", mul2, "=", mul1 * mul2)
    elif choice == 4:
            div1 = input("Divide this: ")
            div2 = input("by this: ")
            print (div1, "/", div2, "=", div1 / div2)
    elif choice == 5:
            pow1 = input ("Raise this: ")
            pow2 = input ("by: ")
            print (pow1, "**", pow2, "=", pow1 ** pow2)                     
    elif choice == 6:
            loop = 0

print ("Thank-you for using StandardCalculator.py!")

Answer 2

您尚未将选择部分包含在循环中，请在print (" ")之后标记所有内容，以便将其包含在while循环中。

同样将if语句从if choice == 1更改为if choice == "1"选项将是字符串而不是int。

最后，当你得到答案时，你需要将字符串输入更改为整数，然后更改

print (add1, "+", add2, "=", add1 + add2)

成像：

print (add1, "+", add2, "=", int(add1) + int(add2))

如果你懒得像我一样，如果用户输入了无法转换为字符串的内容，你的程序就会崩溃，所以你也必须在那里加入try。

Answer 3

编辑：误读了问题。

您不会更改＆＃39;循环＆＃39;的值，因此它总是等于1.我认为这就是您的目标：

while loop == 1:
    # Print what options you have
    print ("Welcome to StandardCalculator.py")

    print ("your options are:")
    print (" ")
    print ("1) Addition")
    print ("2) Subtraction")

    print ("3) Multiplication")

    print ("4) Division")
    print ("5) Raise to the Power")
    print ("6) Quit StandardCalculator.py")
    print (" ")

    choice = input("Enter choice(1/2/3/4):")

        if choice == 1:
            add1 = input("Add this: ")
            add2 = input("to this: ")
            print (add1, "+", add2, "=", add1 + add2)
        elif choice == 2:
            sub2 = input("Subtract this: ")
            sub1 = input("from this: ")
            print (sub1, "-", sub2, "=", sub1 - sub2)
        elif choice == 3:
            mul1 = input("Multiply this: ")
            mul2 = input("with this: ")
            print (mul1, "*", mul2, "=", mul1 * mul2)
        elif choice == 4:
            div1 = input("Divide this: ")
            div2 = input("by this: ")
            print (div1, "/", div2, "=", div1 / div2)
        elif choice == 5:
            pow1 = input ("Raise this: ")
            pow2 = input ("by: ")
            print (pow1, "**", pow2, "=", pow1 ** pow2)
        elif choice == 6:
            loop = 0
        else:
            loop = 1;


print ("Thank-you for using StandardCalculator.py!")

请注意，我添加了一个＆＃39; else＆＃39;将循环设置为1.这样，如果有人做出无效选择，它将不会中断，而是会再次继续循环。

小心缩进，你的代码被卡在打印六个选项上，因为循环最初只包含print语句。你的缩进并没有包含允许中断条件的其余代码。

while loop == 1:
    # Print what options you have
    print ("Welcome to StandardCalculator.py")

    print ("your options are:")
    print (" ")
    print ("1) Addition")
    print ("2) Subtraction")
    print ("3) Multiplication")
    print ("4) Division")
    print ("5) Raise to the Power")
    print ("6) Quit StandardCalculator.py")
    print (" ") # <- End of loop

choice = input("Enter choice(1/2/3/4):") # <- Outside loop, needed to be indented.

最后，在确定你的代码运行正常后，但我在你的else / if中删除了一些缩进，理想情况下Python应缩进4个空格。

理想：

    if choice == 1:
        add1 = input("Add this: ")
        add2 = input("to this: ")
        print (add1, "+", add2, "=", add1 + add2)
        loop = 0

不理想：

    if choice == 1:
            add1 = input("Add this: ")
            add2 = input("to this: ")
            print (add1, "+", add2, "=", add1 + add2)
            loop = 0

Python 3初学者 - 代码循环

3 个答案: