Meaning of '\0\0' in Python?

Question

I'm looking at a 3rd party API and they have the following piece of code:

def array_u16 (n): return array('H', '\0\0'*n)

I understand that '\0' means NULL, does '\0\0' have any special meaning or does it just mean 2 NULLs?

Answer 1

The array class accepts a format character (called a typecode) followed by an initializer. H means an unsigned short, with a minimum size of 2 bytes so, '\0\0' satisfies that. The * n part is to initialize the entire array to NULL bytes.

Answer 2

It just assures that two bytes are provided n times so the size of the array will be equal to n. If '\0' was provided, the resulting array would have a size == n//2 (due to the type-code 'H' requiring 2 bytes); that is obviously counter intuitive:

>>> array('H', '\0' * 10)    # 5 elements
array('H', [0, 0, 0, 0, 0])
>>> array('H', '\0\0' * 10)  # 10 elements
array('H', [0, 0, 0, 0, 0, 0, 0, 0, 0, 0])

Note that, in Python 3, if you need the same snippet to work you must provide a bytes object as the initializer argument to array:

>>> array('H', b'\0\0' * 10)   
array('H', [0, 0, 0, 0, 0, 0, 0, 0, 0, 0])

As you also can't provide a u'' string in Python 2. Other than that, the behavior stays exactly the same.

So '\0\0' is for convenience reasons, nothing more. No semantics are attached to '\0\0'.

_{No semantics are really attached to '\0' either (as they do in, for example, C) '\0' is just another string in Python.}

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As a further example for this behavior, take the initialization of an array with a type-code of 'I' for unsigned ints with a minimum of 2 bytes but 4 on 64bit builds of Python.

In the spirit of the snippet you've provided, you'd initialize the array by doing something like this:

>>> array('I', b'\0\0\0\0' * 10)
array('I', [0, 0, 0, 0, 0, 0, 0, 0, 0, 0])

Yes, four times the b'\0' string to get 10 elements.

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As a final note -- the following timings are performed on Python 3 but 2 is the same -- you might be wondering why he used '\0\0\' * n instead of the more intuitive-looking [0] * n to initialize the array. Well, it's quite faster:

n = 10000
%timeit array('I', [0]*n)
1000 loops, best of 3: 212 µs per loop

%timeit array('I', b'\0\0\0\0'* n)
100000 loops, best of 3: 6.36 µs per loop

Of course, you can do better (for type-codes other than 'b') by feeding a bytearray to array. One way to initialize a bytearray is by providing an int as the number of items to initialize with null bytes:

%timeit array('I', bytearray(n))
1000000 loops, best of 3: 1.72 µs per loop

but, if I remember correctly, the bytearray(int) way of initializing a bytearray might get deprecated in 3.7+ :-).