python - Python3 multiple assignment and memory address

Question

After reading this and this, which are pretty similar to my question, I still cannot understand the following behaviour:

a = 257
b = 257
print(a is b) #False
a, b = 257, 257
print(a is b) #True

When printing id(a) and id(b) I can see that the variables, to which the values were assigned in separate lines, have different ids, whereas with multiple assignment both values have the same id:

a = 257
b = 257
print(id(a)) #139828809414512
print(id(b)) #139828809414224
a, b = 257, 257
print(id(a)) #139828809414416
print(id(b)) #139828809414416

But it's impossible to explain this behaviour by saying that multiple assignment of same values always creates pointers to the same id since:

a, b = -1000, -1000  
print(id(a)) #139828809414448
print(id(b)) #139828809414288

Is there a clear rule, which explains when the variables get the same id and when not?

edit

relevant info: The code in this question was run in interactive mode(ipython3)

Note: the behavior is inconsistent because it never matters. If there were any reason to care whether two 257s were the same object, there would be a simple, sensible pattern to it. — user2357112 supports Monica
peephole optimizations, irrelevant how it works as it is never something you would or should ever rely on. — Padraic Cunningham

user2357112 supports Monica user2357112 supports Monica · Accepted Answer · 2016-02-08T17:53:04

This is due to a constant folding optimization in the bytecode compiler. When the bytecode compiler compiles a batch of statements, it uses a dict to keep track of the constants it's seen. This dict automatically merges any equivalent constants.

Here's the routine responsible for recording and numbering constants (as well as a few related responsibilities):

static int
compiler_add_o(struct compiler *c, PyObject *dict, PyObject *o)
{
    PyObject *t, *v;
    Py_ssize_t arg;

    t = _PyCode_ConstantKey(o);
    if (t == NULL)
        return -1;

    v = PyDict_GetItem(dict, t);
    if (!v) {
        arg = PyDict_Size(dict);
        v = PyInt_FromLong(arg);
        if (!v) {
            Py_DECREF(t);
            return -1;
        }
        if (PyDict_SetItem(dict, t, v) < 0) {
            Py_DECREF(t);
            Py_DECREF(v);
            return -1;
        }
        Py_DECREF(v);
    }
    else
        arg = PyInt_AsLong(v);
    Py_DECREF(t);
    return arg;
}

You can see that it only adds a new entry and assigns a new number if it doesn't find an equivalent constant already present. (The _PyCode_ConstantKey bit makes sure things like 0.0, -0.0, and 0 are considered inequivalent.)

In interactive mode, a batch ends every time the interpreter has to actually run your command, so constant folding mostly doesn't happen across commands:

>>> a = 1000
>>> b = 1000
>>> a is b
False
>>> a = 1000; b = 1000 # 1 batch
>>> a is b
True

In a script, all top-level statements are one batch, so more constant folding happens:

a = 257
b = 257
print a is b

In a script, this prints True.

A function's code gets its constants tracked separately from code outside the function, which limits constant folding:

a = 257

def f():
    b = 257
    print a is b

f()

Even in a script, this prints False.

python - Python3 multiple assignment and memory address

3 Answers