Python控制台计算器输入1*5时输出不符合预期问题排查

问题复现 #

程序运行后输入算式1 * 5，预期输出：

You are ... lazy ... very lazy
5.0
Do you want to store the result? (y / n):

实际输出缺少... very lazy片段：

You are ... lazy
5.0
Do you want to store the result? (y / n):

附问题代码：

msg_0 = "Enter an equation"
msg_1 = "Do you even know what numbers are? Stay focused!"
msg_2 = "Yes ... an interesting math operation. You've slept through all classes, haven't you?"
msg_3 = "Yeah... division by zero. Smart move..."
msg_4 = "Do you want to store the result? (y / n):"
msg_5 = "Do you want to continue calculations? (y / n):"
msg_6 = " ... lazy"
msg_7 = " ... very lazy"
msg_8 = " ... very, very lazy"
msg_9 = "You are"
memory = 0


def is_one_digit(v):
    v = float(v)
    if -10 < v < 10 and v.is_integer():
        return True
    else:
        return False


def check(v1, v2, v3):
    msg = ""
    if is_one_digit(v1) and is_one_digit(v2):
        msg = msg + msg_6
    if (v1 == 1 or v2 == 1) and v3 == "*":
        msg = msg + msg_7
    if (v1 == 0 or v2 == 0) and (v3 == "*" or v3 == "+" or v3 == "-"):
        msg = msg + msg_8
    if msg != "":
        msg = msg_9 + msg
    print(msg)


while True:
    calc = input(msg_0)
    try:
        x = calc.split()[0]
        oper = calc.split()[1]
        y = calc.split()[2]

        if x == "M":
            x = memory
        if y == "M":
            y = memory

        float(x)
        float(y)

        if oper in ["+", "-", "*", "/"]:
            check(x, y, oper)
            if oper == "+":
                result = float(x) + float(y)
                print(result)
            elif oper == "-":
                result = float(x) - float(y)
                print(result)
            elif oper == "*":
                result = float(x) * float(y)
                print(result)
            elif oper == "/":
                if float(y) != 0:
                    result = float(x) / float(y)
                    print(result)
                else:
                    print(msg_3)
                    continue

            user_input = input(msg_4)
            if user_input == "y":
                memory = result
                user_i = input(msg_5)
                if user_i == "y":
                    continue
                elif user_i == "n":
                    break
                else:
                    user_i = input(msg_5)
            elif user_input == "n":
                user_i = input(msg_5)
                if user_i == "y":
                    continue
                elif user_i == "n":
                    break
                else:
                    user_i = input(msg_5)
            else:
                user_input = input(msg_5)

        else:
            print(msg_2)
    except ValueError:
        print(msg_1)
        continue

根因定位 #

问题出在变量类型不匹配：

• 从输入流拆分得到的x、y默认是字符串类型，代码中float(x)、float(y)仅做了类型合法性校验，没有将转换后的数值重新赋值给变量，传入check()函数的依然是字符串格式的"1"、"5"。
• is_one_digit()函数内部主动做了float(v)转换，所以判断两个操作数都是个位数的逻辑可以正常触发，成功拼接... lazy片段。
• 触发... very lazy的判断条件为(v1 == 1 or v2 == 1) and v3 == "*"，字符串"1"和整数1做等值比较结果为False，条件不触发，对应提示片段不会被拼接。

注意：判断操作数为0拼接... very, very lazy的逻辑也存在同样的类型问题，输入含0的简单算式时也会出现提示缺失。

修复方案 #

在类型校验环节将转换后的浮点数赋值回原变量即可，替换原代码中：

float(x)
float(y)

为：

x = float(x)
y = float(y)

修改后传入check()的x、y均为数值类型，所有等值判断可以正常命中，提示输出符合预期。

内容的提问来源于stack exchange，提问作者J. Moe