views:

152

answers:

3

Does Python have a pool of all strings and are they (strings) singletons there?

More precise, in the following code one or two strings were created in memory:

a = str(num)
b = str(num) 

?

+1  A: 

Strings are not interned in general. In your example two strings will be created (with the exception of values between 0 and 9). To test this we can use the is operator to see if the two strings are the same object:

>>> str(1056) is str(1056)
False
Mark Byers
What about this: In [1]: x = str(5) In [2]: y = str(5) In [3]: id(x) Out[3]: 3077925280L In [4]: id(y) Out[4]: 3077925280L?
gruszczy
gruszczy: That's a good question. This is a special case that only applies to the numbers 0 to 9. In general though, the statement is not true. I've clarified my answer.
Mark Byers
@Mark: 0 to 9 is a specific case on a specific compiler (though admittedly, it's the compiler most people use). Other compilers may choose a different number of pre-defined strings.
Brian
+14  A: 

Strings are immutable in Python, so the implementation can decide whether to intern (that's a term often associated with C#, meaning that some strings are stored in a pool) strings or not.

In your example, you're dynamically creating strings. CPython does not always look into the pool to detect whether the string is already there - it also doesn't make sense because you first have to reserve memory in order to create the string, and then compare it to the pool content (inefficient for long strings).

But for strings of length 1, CPython does look into the pool (cf. "stringobject.c"):

static PyStringObject *characters[UCHAR_MAX + 1];

...

PyObject *
PyString_FromStringAndSize(const char *str, Py_ssize_t size)
{

...

    if (size == 1 && str != NULL &&
    (op = characters[*str & UCHAR_MAX]) != NULL)
    {
        #ifdef COUNT_ALLOCS
            one_strings++;
        #endif

        Py_INCREF(op);
        return (PyObject *)op;
    }

...

So:

a = str(num)
b = str(num)
print a is b # <-- this will print False in most cases (but try str(1) is str(1))

But when using constant strings directly in your code, CPython uses the same string instance:

a = "text"
b = "text"
print a is b # <-- this will print True
AndiDog
@Andidog: If CPython does not look into the pool to check if the string is already there, then why does print a is b print true when num is equal to 5?
Brian
@Brian: Sorry, that was a bit inaccurate. Edited my answer to explain the way CPython implements that.
AndiDog
Good answer. The only detail I'd add is to note that Python does have `intern()`
keturn
@keturn: Thanks, didn't even know about `intern()` yet.
AndiDog
+3  A: 

In general, strings are not interned in Python, but they do sometimes seem to be:

>>> str(5) is str(5)
True
>>> str(50) is str(50)
False

This isn't uncommon in Python, where common objects might be optimized in ways that unusual ones are not:

>>> int(5+0) is int(5+0)
True
>>> int(50+0) is int(50+0)
True
>>> int(500+0) is int(500+0)
False

And keep in mind, all of these sorts of details will differ between implementations of Python, and even between versions of the same implementation.

Ned Batchelder