我是python的新手,我想出了这个问题。我已经为计算器做了一个简单的程序。在添加功能中,我使用了try- except。当遇到这一行时(如果决定=='不'或决定=='n':),它会显示打印行“return(”你已退出“)”但它也会抛出异常。我不明白为什么。
import sys
def menu():
print "calculator using functions"
print "Choose your option:"
print " "
print "1) Addition"
print "2) Subtraction"
print "3) Multiplication"
print "4) Division"
print "5) Quit calculator.py"
print " "
return input ("Choose your option: ")
def add(a,b):
try:
print a, "+", b, "=", a + b
print " Do you want to continue: "
decide=raw_input("yes or no: ")
if decide== 'no' or decide== 'n':
return(" You have exited ")
sys.exit(0)
elif decide=='yes' or decide== 'y':
menu()
untrusted.execute()
except:
print "wrong choice!!!"
e = sys.exc_info()[0]
print "Error: %s" % e
sys.exit(0)
答案 0 :(得分:0)
这对我有用....我认为return(" You have exited ")
在这里是问题,而不是sys.exit(0)
尝试return 0
我不知道它为什么会出现问题但它有效。我尝试了既可以也可以没有条件。
import sys
def menu():
print "calculator using functions"
print "Choose your option:"
print " "
print "1) Addition"
print "2) Subtraction"
print "3) Multiplication"
print "4) Division"
print "5) Quit calculator.py"
print " "
return input ("Choose your option: ")
def add(a,b):
try:
print a, "+", b, "=", a + b
print " Do you want to continue: "
decide=raw_input("yes or no: ")
if decide== 'no' or decide== 'n':
return "n"
#sys.exit(0)
elif decide=='yes' or decide== 'y':
return "y"
untrusted.execute()
except:
print "wrong choice!!!"
e = sys.exc_info()[0]
print "Error: %s" % e
sys.exit(0)
while(True):
selected_option=menu()
if(selected_option == 1):
print "Enter first number:"
no1 = raw_input()
print "Enter second number:"
no2 = raw_input()
option=add(no1,no2)
if(option == 'y'):
print "yes selected"
continue
if(option=='n'):
print "no selected"
sys.exit(0)
输出: 你想继续吗?
是或否:重新
错误的选择!!!<type 'exceptions.NameError'>
你想继续吗?
是或否:y
是选择
使用函数的计算器
选择您的选项:
1)增加 2)减法 3)乘法 4)分部 5)退出calculator.py
你想继续吗?
是或否:n
没有选择
root @ yogesh-System-model:〜#