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
通过一些工作,在python中添加类似“goto”的语句在技术上是可行的。我们将使用"dis"和"new"模块,这两个模块对于扫描和修改python字节代码都非常有用。
实现背后的主要思想是首先将代码块标记为使用“goto”和“label”语句。一个特殊的“@goto”装饰器将用于标记“goto”函数。然后,我们扫描这两个语句的代码,并对底层字节代码应用必要的修改。这一切都发生在源代码编译时。
import dis, new
def goto(fn):
"""
A function decorator to add the goto command for a function.
Specify labels like so:
label .foo
Goto labels like so:
goto .foo
Note: you can write a goto statement before the correspnding label statement
"""
labels = {}
gotos = {}
globalName = None
index = 0
end = len(fn.func_code.co_code)
i = 0
# scan through the byte codes to find the labels and gotos
while i < end:
op = ord(fn.func_code.co_code[i])
i += 1
name = dis.opname[op]
if op > dis.HAVE_ARGUMENT:
b1 = ord(fn.func_code.co_code[i])
b2 = ord(fn.func_code.co_code[i+1])
num = b2 * 256 + b1
if name == 'LOAD_GLOBAL':
globalName = fn.func_code.co_names[num]
index = i - 1
i += 2
continue
if name == 'LOAD_ATTR':
if globalName == 'label':
labels[fn.func_code.co_names[num]] = index
elif globalName == 'goto':
gotos[fn.func_code.co_names[num]] = index
name = None
i += 2
# no-op the labels
ilist = list(fn.func_code.co_code)
for label,index in labels.items():
ilist[index:index+7] = [chr(dis.opmap['NOP'])]*7
# change gotos to jumps
for label,index in gotos.items():
if label not in labels:
raise Exception("Missing label: %s"%label)
target = labels[label] + 7 # skip NOPs
ilist[index] = chr(dis.opmap['JUMP_ABSOLUTE'])
ilist[index + 1] = chr(target & 255)
ilist[index + 2] = chr(target >> 8)
# create new function from existing function
c = fn.func_code
newcode = new.code(c.co_argcount,
c.co_nlocals,
c.co_stacksize,
c.co_flags,
''.join(ilist),
c.co_consts,
c.co_names,
c.co_varnames,
c.co_filename,
c.co_name,
c.co_firstlineno,
c.co_lnotab)
newfn = new.function(newcode,fn.func_globals)
return newfn
if __name__ == '__main__':
@goto
def test1():
print 'Hello'
goto .the_end
print 'world'
label .the_end
print 'the end'
test1()
希望这回答了问题。
其他回答
我在找一些类似的东西
for a in xrange(1,10):
A_LOOP
for b in xrange(1,5):
for c in xrange(1,5):
for d in xrange(1,5):
# do some stuff
if(condition(e)):
goto B_LOOP;
所以我的方法是使用一个布尔值来帮助打破嵌套的for循环:
for a in xrange(1,10):
get_out = False
for b in xrange(1,5):
if(get_out): break
for c in xrange(1,5):
if(get_out): break
for d in xrange(1,5):
# do some stuff
if(condition(e)):
get_out = True
break
不,Python不支持标签和goto。它是一种高度结构化的编程语言。
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
我认为while循环是“goto_Statement”的替代。因为3.6之后,goto循环不再工作了。我还写了一个while循环的例子。
str1 = "stop"
while str1 == "back":
var1 = int(input(" Enter Ist Number: "))
var2 = int(input(" Enter 2nd Number: "))
var3 = print(""" What is your next operation
For Addition Press And Enter : 'A'
For Muliplt Press And Enter : 'M'
For Division Press And Enter : 'D'
For Subtaction Press And Enter : 'S' """)
var4 = str(input("For operation press any number : "))
if(var1 == 45) and (var2 == 3):
print("555")
elif(var1 == 56) and (var2 == 9):
print("77")
elif(var1 == 56) and (var2 == 6):
print("4")
else:
if(var4 == "A" or "a"):
print(var1 + var2)
if(var4 == "M" or "m"):
print(var1 * var2)
if(var4 == "D" or "d"):
print(var1 / var2)
if(var4 == "S" or "s"):
print(var1 - var2)
print("if you want to continue then type 'stop'")
str1 = input()
print("Strt again")
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()
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