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I am a software developer from Poland, currently working in banking industry. For the past few years I have been writing software in Java, however I actively seek for a close alternative. Certified in SCJP, SCJD, SCWCD and SCBCD, used to be active on StackOverflow. I feel comfortable at the back-end, however recently rediscovered front-end development. In spare time I love cycling. Tomasz is a DZone MVB and is not an employee of DZone and has posted 84 posts at DZone. You can read more from them at their website. View Full User Profile

Building Extremely Large In-Memory InputStream for Testing Purposes

07.23.2014
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For some reason I needed extremely large, possibly even infinite InputStream that would simply return the same byte[]over and over. This way I could produce insanely big stream of data by repeating small sample. Sort of similar functionality can be found in Guava: Iterable<T> Iterables.cycle(Iterable<T>) and Iterator<T> Iterators.cycle(Iterator<T>). For example if you need an infinite source of 0 and 1, simply sayIterables.cycle(0, 1) and get 0, 1, 0, 1, 0, 1... infinitely. Unfortunately I haven't found such utility forInputStream, so I jumped into writing my own. This article documents many mistakes I made during that process, mostly due to overcomplicating and overengineering straightforward solution.

We don't really need an infinite InputStream, being able to create very large one (say, 32 GiB) is enough. So we are after the following method:

public static InputStream repeat(byte[] sample, int times)

It basically takes sample array of bytes and returns an InputStream returning these bytes. However when sample runs out, it rolls over, returning the same bytes again - this process is repeated given number of times, until InputStreamsignals end. One solution that I haven't really tried but which seems most obvious:

public static InputStream repeat(byte[] sample, int times) {
    final byte[] allBytes = new byte[sample.length * times];
    for (int i = 0; i < times; i++) {
        System.arraycopy(sample, 0, allBytes, i * sample.length, sample.length);
    }
    return new ByteArrayInputStream(allBytes);
}

I see you laughing there! If sample is 100 bytes and we need 32 GiB of input repeating these 100 bytes, generatedInputStream shouldn't really allocate 32 GiB of memory, we must be more clever here. As a matter of fact repeat()above has another subtle bug. Arrays in Java are limited to 231-1 entries (int), 32 GiB is way above that. The reason this program compiles is a silent integer overflow here: sample.length * times. This multiplication doesn't fit in int.

OK, let's try something that at least theoretically can work. My first idea was as follows: what if I create manyByteArrayInputStreams sharing the same byte[] sample (they don't do an eager copy) and somehow join them together? Thus I needed some InputStream adapter that could take arbitrary number of underlying InputStreams and chain them together - when first stream is exhausted, switch to next one. This awkward moment when you look for something in Apache Commons or Guava and apparently it was in the JDK forever... java.io.SequenceInputStream is almost ideal. However it can only chain precisely two underlying InputStreams. Of course since SequenceInputStreamis an InputStream itself, we can use it recursively as an argument to outer SequenceInputStream. Repeating this process we can chain arbitrary number of ByteArrayInputStreams together:

public static InputStream repeat(byte[] sample, int times) {
    if (times <= 1) {
        return new ByteArrayInputStream(sample);
    } else {
        return new SequenceInputStream(
                new ByteArrayInputStream(sample),
                repeat(sample, times - 1)
        );
    }
}

If times is 1, just wrap sample in ByteArrayInputStream. Otherwise use SequenceInputStream recursively. I think you can immediately spot what's wrong with this code: too deep recursion. Nesting level is the same as times argument, which will reach millions or even billions. There must be a better way. Luckily minor improvement changes recursion depth from O(n) to O(logn):

public static InputStream repeat(byte[] sample, int times) {
    if (times <= 1) {
        return new ByteArrayInputStream(sample);
    } else {
        return new SequenceInputStream(
                repeat(sample, times / 2),
                repeat(sample, times - times / 2)
        );
    }
}

Honestly this was the first implementation I tried. It's a simple application of divide and conquer principle, where we produce result by evenly splitting it into two smaller sub-problems. Looks clever, but there is one issue: it's easy to prove we create t (t = timesByteArrayInputStreams and O(t) SequenceInputStreams. While sample byte array is shared, millions of various InputStream instances are wasting memory. This leads us to alternative implementation, creating just one InputStream, regardless value of times:

import com.google.common.collect.Iterators;
import org.apache.commons.lang3.ArrayUtils;

public static InputStream repeat(byte[] sample, int times) {
    final Byte[] objArray = ArrayUtils.toObject(sample);
    final Iterator<Byte> infinite = Iterators.cycle(objArray);
    final Iterator<Byte> limited = Iterators.limit(infinite, sample.length * times);
    return new InputStream() {
        @Override
        public int read() throws IOException {
            return limited.hasNext() ?
                    limited.next() & 0xFF :
                    -1;
        }
    };
}

We will use Iterators.cycle() after all. But before we have to translate byte[] into Byte[] since iterators can only work with objets, not primitives. There is no idiomatic way to turn array of primitives to array of boxed types, so I useArrayUtils.toObject(byte[]) from Apache Commons Lang. Having an array of objects we can create an infiniteiterator that cycles through values of sample. Since we don't want an infinite stream, we cut off infinite iterator usingIterators.limit(Iterator<T>, int), again from Guava. Now we just have to bridge from Iterator<Byte> toInputStream - after all semantically they represent the same thing.

This solution suffers two problems. First of all it produces tons of garbage due to unboxing. Garbage collection is not that much concerned about dead, short-living objects, but still seems wasteful. Second issue we already faced previously:sample.length * times multiplication can cause integer overflow. It can't be fixed because Iterators.limit() takesint, not long - for no good reason. BTW we avoided third problem by doing bitwise and with 0xFF - otherwise byte with value -1 would signal end of stream, which is not the case. x & 0xFF is correctly translated to unsigned 255 (int). 

So even though implementation above is short and sweet, declarative rather than imperative, it's too slow and limited. If you have a C background, I can imagine how uncomfortable you were seeing me struggle. After all the most straightforward, painfully simple and low-level implementation was the one I came up with last:

public static InputStream repeat(byte[] sample, int times) {
    return new InputStream() {
        private long pos = 0;
        private final long total = (long)sample.length * times;

        public int read() throws IOException {
            return pos < total ?
                    sample[(int)(pos++ % sample.length)] :
                    -1;
        }
    };
}

GC free, pure JDK, fast and simple to understand. Let this be a lesson for you: start with the simplest solution that jumps to your mind, don't overengineer and don't be too smart. My previous solutions, declarative, functional, immutable, etc. - maybe they looked clever, but they were neither fast nor easy to understand.

The utility we just developed was not just a toy project, it will be used later in subsequent article. 

Published at DZone with permission of Tomasz Nurkiewicz, author and DZone MVB. (source)

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