在Python中是否有goto或任何等价的东西能够跳转到特定的代码行?
当前回答
Python 2和3
pip3 install goto-statement
在Python 2.6到3.6和PyPy上测试。
链接:转向语句
foo.py
from goto import with_goto
@with_goto
def bar():
label .bar_begin
...
goto .bar_begin
其他回答
你可以使用用户定义异常来模拟goto
例子:
class goto1(Exception):
pass
class goto2(Exception):
pass
class goto3(Exception):
pass
def loop():
print 'start'
num = input()
try:
if num<=0:
raise goto1
elif num<=2:
raise goto2
elif num<=4:
raise goto3
elif num<=6:
raise goto1
else:
print 'end'
return 0
except goto1 as e:
print 'goto1'
loop()
except goto2 as e:
print 'goto2'
loop()
except goto3 as e:
print 'goto3'
loop()
不,有另一种方法来实现goto语句
class id:
def data1(self):
name=[]
age=[]
n=1
while n>0:
print("1. for enter data")
print("2. update list")
print("3. show data")
print("choose what you want to do ?")
ch=int(input("enter your choice"))
if ch==1:
n=int(input("how many elemet you want to enter="))
for i in range(n):
name.append(input("NAME "))
age.append(int(input("age ")))
elif ch==2:
name.append(input("NAME "))
age.append(int(input("age ")))
elif ch==3:
try:
if name==None:
print("empty list")
else:
print("name \t age")
for i in range(n):
print(name[i]," \t ",age[i])
break
except:
print("list is empty")
print("do want to continue y or n")
ch1=input()
if ch1=="y":
n=n+1
else:
print("name \t age")
for i in range(n):
print(name[i]," \t ",age[i])
n=-1
p1=id()
p1.data1()
使用评论中@bobince的建议来回答@ascobol的问题:
for i in range(5000):
for j in range(3000):
if should_terminate_the_loop:
break
else:
continue # no break encountered
break
else块的缩进是正确的。代码在循环Python语法后使用模糊的else。参见为什么python在for和while循环之后使用'else' ?
为了代替python的goto等效,我以以下方式使用break语句快速测试我的代码。这里假设您有结构化的代码库。测试变量在函数开始时初始化,我只是将“If test: break”块移动到我想测试的嵌套If -then块或循环的末尾,在代码末尾修改返回变量以反映我正在测试的块或循环变量。
def x:
test = True
If y:
# some code
If test:
break
return something
我想要同样的答案,我不想使用goto。所以我使用了下面的例子(来自learnpythonthehardway)
def sample():
print "This room is full of gold how much do you want?"
choice = raw_input("> ")
how_much = int(choice)
if "0" in choice or "1" in choice:
check(how_much)
else:
print "Enter a number with 0 or 1"
sample()
def check(n):
if n < 150:
print "You are not greedy, you win"
exit(0)
else:
print "You are nuts!"
exit(0)
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