Question

假设我有一个Python函数f和fhelp。 fhelp旨在递归调用自身。 f不应该递归调用。有没有办法让f确定它是否已被递归调用？

Answer 1

使用traceback模块：

>>> import traceback
>>> def f(depth=0):
...     print depth, traceback.print_stack()
...     if depth < 2:
...         f(depth + 1)
...
>>> f()
0  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in f
 None
1  File "<stdin>", line 1, in <module>
  File "<stdin>", line 4, in f
  File "<stdin>", line 2, in f
 None
2  File "<stdin>", line 1, in <module>
  File "<stdin>", line 4, in f
  File "<stdin>", line 4, in f
  File "<stdin>", line 2, in f
 None

因此，如果堆栈中的任何条目指示代码是从f调用的，则调用是直接递归的。 traceback.extract_stack方法可让您轻松访问此数据。以下示例中的if len(l[2] ...语句只计算函数名称的完全匹配数。为了使它更漂亮（感谢agf的想法），你可以把它变成装饰者：

>>> def norecurse(f):
...     def func(*args, **kwargs):
...         if len([l[2] for l in traceback.extract_stack() if l[2] == f.func_name]) > 0:
...             raise Exception, 'Recursed'
...         return f(*args, **kwargs)
...     return func
...
>>> @norecurse
... def foo(depth=0):
...     print depth
...     foo(depth + 1)
...
>>> foo()
0
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 5, in func
  File "<stdin>", line 4, in foo
  File "<stdin>", line 5, in func
Exception: Recursed

Answer 2

您可以使用装饰器设置的标志：

def norecurse(func):
    func.called = False
    def f(*args, **kwargs):
        if func.called:
            print "Recursion!"
            # func.called = False # if you are going to continue execution
            raise Exception
        func.called = True
        result = func(*args, **kwargs)
        func.called = False
        return result
    return f

然后你可以做

@norecurse
def f(some, arg, s):
    do_stuff()

如果在f运行时再次调用called，则True将为{{1}}并且会引发一个例外。

Python方法可以检查它是否已从内部调用？

2 个答案: