python - 用禁果使int变得可迭代

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Python overload primitives

(1个答案)


已关闭6年。




我知道，这是错误的，但是可能吗？
我以为当.__iter__方法返回迭代器时，该对象被认为是可迭代的？那为什么不起作用呢？

>>> from forbiddenfruit import curse
>>> def __iter__(self):
...     for i in range(self):
...         yield i
>>> curse(int, "__iter__", __iter__)
>>> for x in 5:
...     print x
...
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'int' object is not iterable

int现在似乎确实具有__iter__方法:

>>> int(5).__iter__
<bound method int.__iter__ of 5>

最佳答案

for循环的反汇编为:

import dis

dis.dis("for _ in _: pass")
#>>>   1           0 SETUP_LOOP              14 (to 17)
#>>>               3 LOAD_NAME                0 (_)
#>>>               6 GET_ITER
#>>>         >>    7 FOR_ITER                 6 (to 16)
#>>>              10 STORE_NAME               0 (_)
#>>>              13 JUMP_ABSOLUTE            7
#>>>         >>   16 POP_BLOCK
#>>>         >>   17 LOAD_CONST               0 (None)
#>>>              20 RETURN_VALUE

因此，我们需要GET_ITER操作码。

TARGET(GET_ITER) {
    /* before: [obj]; after [getiter(obj)] */
    PyObject *iterable = TOP();
    PyObject *iter = PyObject_GetIter(iterable);
    Py_DECREF(iterable);
    SET_TOP(iter);
    if (iter == NULL)
        goto error;
    PREDICT(FOR_ITER);
    DISPATCH();
}

使用PyObject_GetIter:

PyObject *
PyObject_GetIter(PyObject *o)
{
    PyTypeObject *t = o->ob_type;
    getiterfunc f = NULL;
    f = t->tp_iter;
    if (f == NULL) {
        if (PySequence_Check(o))
            return PySeqIter_New(o);
        return type_error("'%.200s' object is not iterable", o);
    }
    else {
        PyObject *res = (*f)(o);
        if (res != NULL && !PyIter_Check(res)) {
            PyErr_Format(PyExc_TypeError,
                         "iter() returned non-iterator "
                         "of type '%.100s'",
                         res->ob_type->tp_name);
            Py_DECREF(res);
            res = NULL;
        }
        return res;
    }
}

这首先检查t->tp_iter是否为空。

现在，这就是使一切都点击的东西:

class X:
    pass

X.__iter__ = lambda x: iter(range(10))

list(X())
#>>> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

from forbiddenfruit import curse

class X:
    pass

curse(X, "__iter__", lambda x: iter(range(10)))

list(X())
#>>> Traceback (most recent call last):
#>>>   File "", line 16, in <module>
#>>> TypeError: 'X' object is not iterable

当您在类上正常设置属性时，它会调用PyType_Type->setattro:

static int
type_setattro(PyTypeObject *type, PyObject *name, PyObject *value)
{
    if (!(type->tp_flags & Py_TPFLAGS_HEAPTYPE)) {
        PyErr_Format(
            PyExc_TypeError,
            "can't set attributes of built-in/extension type '%s'",
            type->tp_name);
        return -1;
    }
    if (PyObject_GenericSetAttr((PyObject *)type, name, value) < 0)
        return -1;
    return update_slot(type, name);
}

看到update_slot吗？这样就更新了插槽，因此对GET_ITER的下一次调用将在tp->tp_iter上命中X。但是，forbiddenfruit绕过此过程，只是将字典注入(inject)类中。这意味着PyLong_Type保留其默认值:

PyTypeObject PyLong_Type = {
    ...
    0,                                          /* tp_iter */
    ...
};

所以

if (f == NULL)

被触发

if (PySequence_Check(o))

失败(因为它不是一个序列)，然后它只是

return type_error("'%.200s' object is not iterable", o);

关于python - 用禁果使int变得可迭代，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/29591349/