What is the ideal growth rate for a dynamically allocated array?

Question

C++ has std::vector and Java has ArrayList, and many other languages have their own form of dynamically allocated array. When a dynamic array runs out of space, it gets reallocated into a larger area and the old values are copied into the new array. A question central to the performance of such an array is how fast the array grows in size. If you always only grow large enough to fit the current push, you'll end up reallocating every time. So it makes sense to double the array size, or multiply it by say 1.5x.

Is there an ideal growth factor? 2x? 1.5x? By ideal I mean mathematically justified, best balancing performance and wasted memory. I realize that theoretically, given that your application could have any potential distribution of pushes that this is somewhat application dependent. But I'm curious to know if there's a value that's "usually" best, or is considered best within some rigorous constraint.

I've heard there's a paper on this somewhere, but I've been unable to find it.

Answer 1

It will entirely depend on the use case. Do you care more about the time wasted copying data around (and reallocating arrays) or the extra memory? How long is the array going to last? If it's not going to be around for long, using a bigger buffer may well be a good idea - the penalty is short-lived. If it's going to hang around (e.g. in Java, going into older and older generations) that's obviously more of a penalty.

There's no such thing as an "ideal growth factor." It's not just theoretically application dependent, it's definitely application dependent.

2 is a pretty common growth factor - I'm pretty sure that's what ArrayList and List<T> in .NET uses. ArrayList<T> in Java uses 1.5.

EDIT: As Erich points out, Dictionary<,> in .NET uses "double the size then increase to the next prime number" so that hash values can be distributed reasonably between buckets. (I'm sure I've recently seen documentation suggesting that primes aren't actually that great for distributing hash buckets, but that's an argument for another answer.)

Answer 2

I remember reading many years ago why 1.5 is preferred over two, at least as applied to C++ (this probably doesn't apply to managed languages, where the runtime system can relocate objects at will).

The reasoning is this:

1. Say you start with a 16-byte allocation.
2. When you need more, you allocate 32 bytes, then free up 16 bytes. This leaves a 16-byte hole in memory.
3. When you need more, you allocate 64 bytes, freeing up the 32 bytes. This leaves a 48-byte hole (if the 16 and 32 were adjacent).
4. When you need more, you allocate 128 bytes, freeing up the 64 bytes. This leaves a 112-byte hole (assuming all previous allocations are adjacent).
5. And so and and so forth.

The idea is that, with a 2x expansion, there is no point in time that the resulting hole is ever going to be large enough to reuse for the next allocation. Using a 1.5x allocation, we have this instead:

1. Start with 16 bytes.
2. When you need more, allocate 24 bytes, then free up the 16, leaving a 16-byte hole.
3. When you need more, allocate 36 bytes, then free up the 24, leaving a 40-byte hole.
4. When you need more, allocate 54 bytes, then free up the 36, leaving a 76-byte hole.
5. When you need more, allocate 81 bytes, then free up the 54, leaving a 130-byte hole.
6. When you need more, use 122 bytes (rounding up) from the 130-byte hole.

Answer 3

Ideally (in the limit as n → ∞), it's the golden ratio: ϕ = 1.618...

In practice, you want something close, like 1.5.

The reason is that you want to be able to reuse older memory blocks, to take advantage of caching and avoid constantly making the OS give you more memory pages. The equation you'd solve to ensure this reduces to x^{n − 1} − 1 = x^{n + 1} − xⁿ, whose solution approaches x = ϕ for large n.

Answer 4

One approach when answering questions like this is to just "cheat" and look at what popular libraries do, under the assumption that a widely used library is, at the very least, not doing something horrible.

So just checking very quickly, Ruby (1.9.1-p129) appears to use 1.5x when appending to an array, and Python (2.6.2) uses 1.125x plus a constant (in Objects/listobject.c):

/* This over-allocates proportional to the list size, making room
 * for additional growth.  The over-allocation is mild, but is
 * enough to give linear-time amortized behavior over a long
 * sequence of appends() in the presence of a poorly-performing
 * system realloc().
 * The growth pattern is:  0, 4, 8, 16, 25, 35, 46, 58, 72, 88, ...
 */
new_allocated = (newsize >> 3) + (newsize < 9 ? 3 : 6);

/* check for integer overflow */
if (new_allocated > PY_SIZE_MAX - newsize) {
    PyErr_NoMemory();
    return -1;
} else {
    new_allocated += newsize;
}

newsize above is the number of elements in the array. Note well that newsize is added to new_allocated, so the expression with the bitshifts and ternary operator is really just calculating the over-allocation.

Answer 5

Let's say you grow the array size by x. So assume you start with size T. The next time you grow the array its size will be T*x. Then it will be T*x^2 and so on.

If your goal is to be able to reuse the memory that has been created before, then you want to make sure the new memory you allocate is less than the sum of previous memory you deallocated. Therefore, we have this inequality:

T*x^n <= T + T*x + T*x^2 + ... + T*x^(n-2)

We can remove T from both sides. So we get this:

x^n <= 1 + x + x^2 + ... + x^(n-2)

Informally, what we say is that at nth allocation, we want our all previously deallocated memory to be greater than or equal to the memory need at the nth allocation so that we can reuse the previously deallocated memory.

For instance, if we want to be able to do this at the 3rd step (i.e., n=3), then we have

x^3 <= 1 + x 

This equation is true for all x such that 0 < x <= 1.3 (roughly)

See what x we get for different n's below:

n  maximum-x (roughly)

3  1.3

4  1.4

5  1.53

6  1.57

7  1.59

22 1.61

Note that the growing factor has to be less than 2 since x^n > x^(n-2) + ... + x^2 + x + 1 for all x>=2.

Answer 6

It really depends. Some people analyze common usage cases to find the optimal number.

I've seen 1.5x 2.0x phi x, and power of 2 used before.

Answer 7

If you have a distribution over array lengths, and you have a utility function that says how much you like wasting space vs. wasting time, then you can definitely choose an optimal resizing (and initial sizing) strategy.

The reason the simple constant multiple is used, is obviously so that each append has amortized constant time. But that doesn't mean you can't use a different (larger) ratio for small sizes.

In Scala, you can override loadFactor for the standard library hash tables with a function that looks at the current size. Oddly, the resizable arrays just double, which is what most people do in practice.

I don't know of any doubling (or 1.5*ing) arrays that actually catch out of memory errors and grow less in that case. It seems that if you had a huge single array, you'd want to do that.

I'd further add that if you're keeping the resizable arrays around long enough, and you favor space over time, it might make sense to dramatically overallocate (for most cases) initially and then reallocate to exactly the right size when you're done.

Answer 8

I agree with Jon Skeet, even my theorycrafter friend insists that this can be proven to be O(1) when setting the factor to 2x.

The ratio between cpu time and memory is different on each machine, and so the factor will vary just as much. If you have a machine with gigabytes of ram, and a slow CPU, copying the elements to a new array is a lot more expensive than on a fast machine, which might in turn have less memory. It's a question that can be answered in theory, for a uniform computer, which in real scenarios doesnt help you at all.

Answer 9

I know it is an old question, but there are several things that everyone seems to be missing.

First, this is multiplication by 2: size << 1. This is multiplication by anything between 1 and 2: int(float(size) * x), where x is the number, the * is floating point math, and the processor has to run additional instructions for casting between float and int. In other words, at the machine level, doubling takes a single, very fast instruction to find the new size. Multiplying by something between 1 and 2 requires at least one instruction to cast size to a float, one instruction to multiply (which is float multiplication, so it probably takes at least twice as many cycles, if not 4 or even 8 times as many), and one instruction to cast back to int, and that assumes that your platform can perform float math on the general purpose registers, instead of requiring the use of special registers. In short, you should expect the math for each allocation to take at least 10 times as long as a simple left shift. If you are copying a lot of data during the reallocation though, this might not make much of a difference.

Second, and probably the big kicker: Everyone seems to assume that the memory that is being freed is both contiguous with itself, as well as contiguous with the newly allocated memory. Unless you are pre-allocating all of the memory yourself and then using it as a pool, this is almost certainly not the case. The OS might occasionally end up doing this, but most of the time, there is going to be enough free space fragmentation that any half decent memory management system will be able to find a small hole where your memory will just fit. Once you get to really bit chunks, you are more likely to end up with contiguous pieces, but by then, your allocations are big enough that you are not doing them frequently enough for it to matter anymore. In short, it is fun to imagine that using some ideal number will allow the most efficient use of free memory space, but in reality, it is not going to happen unless your program is running on bare metal (as in, there is no OS underneath it making all of the decisions).

My answer to the question? Nope, there is no ideal number. It is so application specific that no one really even tries. If your goal is ideal memory usage, you are pretty much out of luck. For performance, less frequent allocations are better, but if we went just with that, we could multiply by 4 or even 8! Of course, when Firefox jumps from using 1GB to 8GB in one shot, people are going to complain, so that does not even make sense. Here are some rules of thumb I would go by though:

If you cannot optimize memory usage, at least don't waste processor cycles. Multiplying by 2 is at least an order of magnitude faster than doing floating point math. It might not make a huge difference, but it will make some difference at least (especially early on, during the more frequent and smaller allocations).

Don't overthink it. If you just spent 4 hours trying to figure out how to do something that has already been done, you just wasted your time. Totally honestly, if there was a better option than *2, it would have been done in the C++ vector class (and many other places) decades ago.

Lastly, if you really want to optimize, don't sweat the small stuff. Now days, no one cares about 4KB of memory being wasted, unless they are working on embedded systems. When you get to 1GB of objects that are between 1MB and 10MB each, doubling is probably way too much (I mean, that is between 100 and 1,000 objects). If you can estimate expected expansion rate, you can level it out to a linear growth rate at a certain point. If you expect around 10 objects per minute, then growing at 5 to 10 object sizes per step (once every 30 seconds to a minute) is probably fine.

What it all comes down to is, don't over think it, optimize what you can, and customize to your application (and platform) if you must.

Answer 10

Another two cents

• Most computers have virtual memory! In the physical memory you can have random pages everywhere which are displayed as a single contiguous space in your program's virtual memory. The resolving of the indirection is done by the hardware. Virtual memory exhaustion was a problem on 32 bit systems, but it is really not a problem anymore. So filling the hole is not a concern anymore (except special environments). Since Windows 7 even Microsoft supports 64 bit without extra effort. @ 2011
• O(1) is reached with any r > 1 factor. Same mathematical proof works not only for 2 as parameter.
• r = 1.5 can be calculated with old*3/2 so there is no need for floating point operations. (I say /2 because compilers will replace it with bit shifting in the generated assembly code if they see fit.)
• MSVC went for r = 1.5, so there is at least one major compiler that does not use 2 as ratio.

As mentioned by someone 2 feels better than 8. And also 2 feels better than 1.1.

My feeling is that 1.5 is a good default. Other than that it depends on the specific case.