我目前正在尝试为python提交自己的“封送”代码,这样我就可以将编译好的python代码存储在上,以便动态地为脚本提供服务。正如大家都可以验证的那样,GAE不支持“封送”,而“泡菜”不能序列化代码对象。
我发现我可以用types.CodeType()
构造一个代码对象,但它需要12个参数。
尽管我已经尝试了很多,但是我找不到关于这个调用的任何文档,我真的需要构造代码对象,这样我就可以exec()
它了。我的问题是,有人知道这个types.CodeType()
“构造函数”的参数是什么吗?或者有什么方法可以自省吗?--我已经使用了定义了这里的info()
函数,但是它只显示了一般的信息!
快速常见问题:
更新
截至2011年7月7日,Google基础设施不允许对代码对象进行实例化,所以我的论点是毫无意义的。希望这将在未来的GAE上得到解决。
发布于 2011-07-07 15:18:36
我去取了代码,找到了这里,并删除了不推荐的“新”模块的依赖项。
import types, copy_reg
def code_ctor(*args):
# delegate to new.code the construction of a new code object
return types.CodeType(*args)
def reduce_code(co):
# a reductor function must return a tuple with two items: first, the
# constructor function to be called to rebuild the argument object
# at a future de-serialization time; then, the tuple of arguments
# that will need to be passed to the constructor function.
if co.co_freevars or co.co_cellvars:
raise ValueError, "Sorry, cannot pickle code objects from closures"
return code_ctor, (co.co_argcount, co.co_nlocals, co.co_stacksize,
co.co_flags, co.co_code, co.co_consts, co.co_names,
co.co_varnames, co.co_filename, co.co_name, co.co_firstlineno,
co.co_lnotab)
# register the reductor to be used for pickling objects of type 'CodeType'
copy_reg.pickle(types.CodeType, reduce_code)
if __name__ == '__main__':
# example usage of our new ability to pickle code objects
import cPickle
# a function (which, inside, has a code object, of course)
def f(x): print 'Hello,', x
# serialize the function's code object to a string of bytes
pickled_code = cPickle.dumps(f.func_code)
# recover an equal code object from the string of bytes
recovered_code = cPickle.loads(pickled_code)
# build a new function around the rebuilt code object
g = types.FunctionType(recovered_code, globals( ))
# check what happens when the new function gets called
g('world')
发布于 2012-10-13 22:13:38
问题是:
这个types.CodeType()“构造函数”的参数是什么?
来自python关于检查模块的文档
co_argcount: number of arguments (not including * or ** args)
co_code: string of raw compiled bytecode
co_consts: tuple of constants used in the bytecode
co_filename: name of file in which this code object was created
co_firstlineno: number of first line in Python source code
co_flags: bitmap: 1=optimized | 2=newlocals | 4=*arg | 8=**arg
co_lnotab: encoded mapping of line numbers to bytecode indices
co_name: name with which this code object was defined
co_names: tuple of names of local variables
co_nlocals: number of local variables
co_stacksize: virtual machine stack space required
co_varnames: tuple of names of arguments and local variables
这篇博文有更详细的解释:http://tech.blog.aknin.name/2010/07/03/pythons-innards-code-objects/
注意:博客文章谈到python 3,上面引用的python文档是python 2.7。
发布于 2011-07-07 15:35:30
code函数PyCode_New在这里(最少)有文档:http://docs.python.org/c-api/code.html --这个函数的C源代码(Python2.7)在这里:http://hg.python.org/cpython/file/b5ac5e25d506/Objects/codeobject.c#l43
PyCodeObject *
PyCode_New(int argcount, int nlocals, int stacksize, int flags,
PyObject *code, PyObject *consts, PyObject *names,
PyObject *varnames, PyObject *freevars, PyObject *cellvars,
PyObject *filename, PyObject *name, int firstlineno,
PyObject *lnotab)
但是,在Python构造函数中,最后六个参数似乎被交换了一些。这是提取Python:http://hg.python.org/cpython/file/b5ac5e25d506/Objects/codeobject.c#l247传递的参数的C代码。
if (!PyArg_ParseTuple(args, "iiiiSO!O!O!SSiS|O!O!:code",
&argcount, &nlocals, &stacksize, &flags,
&code,
&PyTuple_Type, &consts,
&PyTuple_Type, &names,
&PyTuple_Type, &varnames,
&filename, &name,
&firstlineno, &lnotab,
&PyTuple_Type, &freevars,
&PyTuple_Type, &cellvars))
return NULL;
化肥化:
def __init__(self, argcount, nlocals, stacksize, flags, code,
consts, names, varnames, filename, name,
firstlineno, lnotab, freevars=None, cellvars=None): # ...
https://stackoverflow.com/questions/6612449
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