Tuple slicing not returning a new object as opposed to list slicing

user_10704952 0
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In Python (2 and 3). Whenever we use list slicing it returns a new object, e.g.:

l1 = [1,2,3,4]
print(id(l1))
l2 = l1[:]
print(id(l2))

Output

>>> 140344378384464
>>> 140344378387272

If the same thing is repeated with tuple, the same object is returned, e.g.:

t1 = (1,2,3,4)
t2 = t1[:]
print(id(t1))
print(id(t2))

Output

>>> 140344379214896
>>> 140344379214896

It would be great if someone can shed some light on why this is happening, throughout my Python experience I was under the impression empty slice returns a new object.

My understanding is that it's returning the same object as tuples are immutable and there's no point of creating a new copy of it. But again, it's not mentioned in the documents anywhere.

cpython slice tuples list python
4 Answers

Implementations are free to return identical instances for immutable types (in CPython, you may sometimes see similar optimizations for strings and integers). Since the object can not be changed, there is nothing in user code that needs to care whether it holds a unique instance or just another reference to an existing instance.

You can find the short-circuit in the C code here.

static PyObject*
tuplesubscript(PyTupleObject* self, PyObject* item)
{
    ... /* note: irrelevant parts snipped out */
    if (start == 0 && step == 1 &&
                 slicelength == PyTuple_GET_SIZE(self) &&
                 PyTuple_CheckExact(self)) {
            Py_INCREF(self);          /* <--- increase reference count */
            return (PyObject *)self;  /* <--- return another pointer to same */
        }
    ...

This is an implementation detail, note that pypy does not do the same.

user_6740390

It's an implementation detail. Because lists are mutable, l1[:] must create a copy, because you wouldn't expect changes to l2 to affect l1.

Since a tuple is immutable, though, there's nothing you can do to t2 that would affect t1 in any visible way, so the compiler is free (but not required) to use the same object for t1 and t1[:].

user_11268410

In Python 3.* my_list[:] is syntactic sugar for type(my_list).__getitem__(mylist, slice_object) where: slice_object is a slice object built from my_list's attributes (length) and the expression [:]. Objects that behave this way are called subscriptable in the Python data model see here. For lists and tuples __getitem__ is a built-in method.

In CPython, and for lists and tuples, __getitem__ is interpreted by the bytecode operation BINARY_SUBSCR which is implemented for tuples in here and for lists in here.

In case of tuples, walking through the code you will see that in this code block, static PyObject* tuplesubscript(PyTupleObject* self, PyObject* item) will return a reference to the same PyTupleObject that it got as input argument, if item is of type PySlice and the slice evaluates to the whole tuple. 

    static PyObject*
    tuplesubscript(PyTupleObject* self, PyObject* item)
    {
        /* checks if item is an index */ 
        if (PyIndex_Check(item)) { 
            ...
        }
        /* else it is a slice */ 
        else if (PySlice_Check(item)) { 
            ...
        /* unpacks the slice into start, stop and step */ 
        if (PySlice_Unpack(item, &start, &stop, &step) < 0) { 
            return NULL;
        }
       ...
        }
        /* if we start at 0, step by 1 and end by the end of the tuple then !! look down */
        else if (start == 0 && step == 1 &&
                 slicelength == PyTuple_GET_SIZE(self) && 
                 PyTuple_CheckExact(self)) {
            Py_INCREF(self); /* increase the reference count for the tuple */
            return (PyObject *)self; /* and return a reference to the same tuple. */
        ...
}

Now you examine the code for static PyObject * list_subscript(PyListObject* self, PyObject* item) and see for yourself that whatever the slice, a new list object is always returned.

user_72330220

Not sure about this but it seems that Python provides you with a new pointer to the same object to avoid copying since the tuples are identical (and since the object is a tuple, it's immutable).

user_78970420
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