Question

我尝试使用sys.stdout.flush（）刷新stdout但它仍然无效。它工作正常，直到用户只输入破坏代码的C或F.后缀是第一个被调用的函数所以我确保如果用户只输入一个字符，则返回错误。但是一旦返回错误，用户就不能再键入“quit”或“q”。

#!/usr/local/bin/python
#Converts between Celsius and Fahrenheit
import random, sys

def check_version():
    """
    Make sure user is using Python 3k
    """
    if(sys.version_info[0] != 3):
        print("Stop peddling with your feet Fred!")
        print("Only Py3k supported")
        sys.exit()
    else:
        pass

def random_insult():
    """
    Returns a list of random insults with the sole purpose of insulting the user
    """
    insults = ["Kel", "stimpy", "knucklehead"]
    return insults[random.randrange(3)]

def suffix(temp):
    """
    Accepts the input temperature value which should be a string suffixed by C(c) or F(f)
    Returns the last element of the input string(C or F)
    """
    if(len(temp) >= 2):
        return temp[len(temp)-1]
    else:
        temperature("Input String TOO Small")

def temp_value(temp):
     """
     Accepts the input temperature value which should be a string suffixed by C(c) or F(f)
     Returns the actual temperature value
     """
     if(len(temp) >= 2):
         return temp[0:len(temp)-1]
     else:
         temperature("Input String TOO Small")

def cel_to_far(temp):
    """
    Accepts the input temperature value as Celsius and returns it in Fahrenheit
    """
    try:
        return ((temp * (9/5.0)) + 32)
    except TypeError:
        return "Has to be a number"

def far_to_cel(temp):
    """
    Accepts the input temperature value as Fahrenheit and returns it in Celsius
    """
    try:
        return ((temp - 32) * (5/9.0))
    except TypeError:
        return "Has to be a number"

def temperature(error=None):
    """
    Loops until the user enters quit or q. Allows the user to enter the temperature suffixed by either C(c) or F(f). 
    If suffixed with C then the temperature is taken as Celsius and converted to Fahrenheit.
    If suffixed with F then the temperature is taken as Fahrenheit and converted to Celsius.
    If the user enters anything else be sure to belittle him/her.
    """
    prompt1 = "Enter value suffixed by C or F *\n"
    prompt2 = "Type 'quit' or 'q' to quit     *\n"
    error1 = "What in the world are you doing "+ random_insult() + "?\n"
    error2 = "Did you forget to add C or F to the end of the value?\n"
    example = "Here's an example of input: 30F\n"
    stars = ("*" * 32) + "\n"
    temp = None
    if(error != None):
        print(error)
        print(example)
        temperature()        
    else:
        while(True):
            sys.stdout.flush()
            try:
                temp = input("\n"+ stars + prompt1 + prompt2 + stars + ">>") 
                if( (temp == 'quit') or (temp == 'q')):
                    return

                elif( (suffix(temp) == 'C') or (suffix(temp) == 'c') ):
                    print("Celsius:", temp_value(temp))
                    print("Fahrenheit: {0:.1f}".format(cel_to_far(float(temp_value(temp)))))

                elif( (suffix(temp) == 'F') or (suffix(temp) == 'f') ):
                    print("Fahrenheit:", temp_value(temp))
                    print("Celsius: {0:.1f}".format(far_to_cel(float(temp_value(temp)))))

                else:
                    print(error1 + error2 + example)

            except:
               print("Something went wrong and I don't care to fix it.\n")
               return

if(__name__  == '__main__'):
    check_version()
    temperature()

Answer 1

问题在于，当发生错误时，temperature会在不同的地方递归调用，并且要退出程序，用户必须输入q / quit的次数{ {1}}被召唤。

要解决此问题，我建议删除所有递归调用并以不同方式处理错误。例如，可能会检查temperature以确保它是来自用户的正确输入，如果不是这种情况，请打印错误消息并temp无限循环，直到continue被调用当输入为break或q时。

Answer 2

即使您没有做任何递归操作，您的程序也是递归的。

例如，在温度函数中，有这个位：

if(error != None):
    print(error)
    print(example)
    temperature()

因此温度函数调用温度函数。相反，你应该做一个

while temperature():
     pass

在main函数中，当某人退出时，温度函数返回false，并且永远不会从其他任何地方调用temperature（）函数。

当用户输入'q'或'quit'时，如何让我的小python程序退出？

2 个答案: