Self-referencing lists

Question

Say you do the following:

a = [1]
a[0] = a

You end up with a as equal to [[...]]. What\'s going on here? How doe

Answer 1

You ain't seen nothing:

>>> a = []
>>> a[:] = [a] * 4
>>> a
[[...], [...], [...], [...]]

If you're interested in how this works in CPython, see list_repr in listobject.c and similar functions. Basically, any function that potentially prints a self-referential object calls Py_ReprEnter on the object before printing it and Py_ReprLeave when it is done. (See object.c for the definitions of these functions.) The former checks to see if the object is found in a thread-local stack of objects that are currently being printed (and if not, pushes it); the latter pops the object from the stack. So if Python is printing a list and discovers that the list is on the stack, that must mean that this is a self-referential list, and the list should be abbreviated, to avoid an infinite loop:

 i = Py_ReprEnter((PyObject*)v);
 if (i != 0) {
     return i > 0 ? PyString_FromString("[...]") : NULL;
 }

 // ...

 Py_ReprLeave((PyObject *)v);
 return result;