is open(...).read() a resource leak?

[Benjamin Rutt]

If I did the following in an infinite loop, would the host system/user
account soon run out of file descriptors?  (I'm thinking no, since I'd
imagine that a file object has a __del__-like method that will call
close() automatically since it goes out of scope):

    open('/home/rutt/.bashrc,'r').read()

Can anyone confirm that I'm right in seeing (1) the file object's
actual destructor is below and seems to call a close method if not
already done (at the <-- indicated lines) and (2) the method table
that makes this method the destructor?  Both are in
Objects/fileobject.c in python sources.

(1)
static void
file_dealloc(PyFileObject *f)
{
	if (f->f_fp != NULL && f->f_close != NULL) { <----
		Py_BEGIN_ALLOW_THREADS               
		(*f->f_close)(f->f_fp);              <----
		Py_END_ALLOW_THREADS
	}
	Py_XDECREF(f->f_name);
	Py_XDECREF(f->f_mode);
	Py_XDECREF(f->f_encoding);
	drop_readahead(f);
	f->ob_type->tp_free((PyObject *)f);
}


(2)
PyTypeObject PyFile_Type = {
	PyObject_HEAD_INIT(&PyType_Type)
	0,
	"file",
	sizeof(PyFileObject),
	0,
	(destructor)file_dealloc,		/* tp_dealloc */  <----
	0,					/* tp_print */
	0,			 		/* tp_getattr */
	0,			 		/* tp_setattr */
	0,					/* tp_compare */
	(reprfunc)file_repr, 			/* tp_repr */
	0,					/* tp_as_number */
[...]

Thanks,
-- 
Benjamin Rutt