I want to check the size of int data type in python:
import sys sys.getsizeof(int)
It comes out to be "436", which doesn't make sense to me. Anyway, I want to know how many bytes (2,4,..?) int will take on my machine.
To be safe, Python allocates a fixed number of bytes of space in memory for each variable of a normal integer type, which is known as int in Python. Typically, an integer occupies four bytes, or 32 bits. Integers whose binary representations require fewer than 32 bits are padded to the left with 0s.
sys. getsizeof (object[, default]) Return the size of an object in bytes. The object can be any type of object. All built-in objects will return correct results, but this does not have to hold true for third-party extensions as it is implementation specific.
These represent numbers in the range -2147483648 through 2147483647. (The range may be larger on machines with a larger natural word size, but not smaller.)
28 bytes for the int and 24 bytes for the float. That means that the number 42 takes up 7 times as much memory in Python as in C. This is one of the reasons for why Python code is often slower than C code, and why it requires a lot more memory.
You're getting the size of the class, not of an instance of the class. Call int
to get the size of an instance:
>>> sys.getsizeof(int()) 24
If that size still seems a little bit large, remember that a Python int
is very different from an int
in (for example) c. In Python, an int
is a fully-fledged object. This means there's extra overhead.
Every Python object contains at least a refcount and a reference to the object's type in addition to other storage; on a 64-bit machine, that takes up 16 bytes! The int
internals (as determined by the standard CPython implementation) have also changed over time, so that the amount of additional storage taken depends on your version.
int
objects in Python 2 and 3Here's the situation in Python 2. (Some of this is adapted from a blog post by Laurent Luce). Integer objects are represented as blocks of memory with the following structure:
typedef struct { PyObject_HEAD long ob_ival; } PyIntObject;
PyObject_HEAD
is a macro defining the storage for the refcount and the object type. It's described in some detail by the documentation, and the code can be seen in this answer.
The memory is allocated in large blocks so that there's not an allocation bottleneck for every new integer. The structure for the block looks like this:
struct _intblock { struct _intblock *next; PyIntObject objects[N_INTOBJECTS]; }; typedef struct _intblock PyIntBlock;
These are all empty at first. Then, each time a new integer is created, Python uses the memory pointed at by next
and increments next
to point to the next free integer object in the block.
I'm not entirely sure how this changes once you exceed the storage capacity of an ordinary integer, but once you do so, the size of an int
gets larger. On my machine, in Python 2:
>>> sys.getsizeof(0) 24 >>> sys.getsizeof(1) 24 >>> sys.getsizeof(2 ** 62) 24 >>> sys.getsizeof(2 ** 63) 36
In Python 3, I think the general picture is the same, but the size of integers increases in a more piecemeal way:
>>> sys.getsizeof(0) 24 >>> sys.getsizeof(1) 28 >>> sys.getsizeof(2 ** 30 - 1) 28 >>> sys.getsizeof(2 ** 30) 32 >>> sys.getsizeof(2 ** 60 - 1) 32 >>> sys.getsizeof(2 ** 60) 36
These results are, of course, all hardware-dependent! YMMV.
The variability in integer size in Python 3 is a hint that they may behave more like variable-length types (like lists). And indeed, this turns out to be true. Here's the definition of the C struct
for int
objects in Python 3:
struct _longobject { PyObject_VAR_HEAD digit ob_digit[1]; };
The comments that accompany this definition summarize Python 3's representation of integers. Zero is represented not by a stored value, but by an object with size zero (which is why sys.getsizeof(0)
is 24
bytes while sys.getsizeof(1)
is 28
). Negative numbers are represented by objects with a negative size attribute! So weird.
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