如何让我的小Python程序在用户输入'q'或'quit'时退出？
我尝试用 sys.stdout.flush() 来刷新输出，但还是不行。代码在用户只输入 C 或 F 时就会出问题。Suffix 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，就会返回一个错误。但是一旦返回了错误，用户就无法再输入 'quit' 或 'q' 了。
#!/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()
2 个回答
4
你的程序是递归的，虽然你并没有做任何递归的事情。
举个例子，在温度函数里面，有这么一段：
if(error != None):
print(error)
print(example)
temperature()
所以温度函数会自己调用自己。你应该在
while temperature():
pass
的主函数里处理这个问题，让温度函数在有人退出时返回false，并且绝对不要从其他地方调用temperature()函数。
6
问题在于，当出现错误时，temperature这个函数在不同的地方被重复调用。这样一来，用户想要退出程序，就得输入q或quit的次数和temperature被调用的次数一样多。
为了解决这个问题，我建议去掉所有重复调用的部分，换一种方式来处理错误。比如，可以先检查temp这个输入，确保它是用户输入的正确值。如果不是，就打印出错误信息，然后继续无限循环，直到用户输入q或quit时，才会跳出这个循环。