Fastest and most efficient conversion of a byte array to a 29 bit integer in Java
Since 29 bit integers are popular in AMF, I would like to incorporate the fastest / best routine known. Two routines currently exist in our library and can be tested live on ideone http://ideone.com/KNmYT
Here is the source for quick reference
public static int readMediumInt(ByteBuffer in) {
ByteBuffer buf = ByteBuffer.allocate(4);
开发者_如何学JAVA buf.put((byte) 0x00);
buf.put(in.get());
buf.put(in.get());
buf.put(in.get());
buf.flip();
return buf.getInt();
}
public static int readMediumInt2(ByteBuffer in) {
byte[] bytes = new byte[3];
in.get(bytes);
int val = 0;
val += bytes[0] * 256 * 256;
val += bytes[1] * 256;
val += bytes[2];
if (val < 0) {
val += 256;
}
return val;
}
Usually I'd do this with bit operations. The second version might eventually get optimized to something close to this by the JVM, but one can't be sure. Now, this is only 24 bits, following your samples, but the question says "29 bit integer". I'm not sure which you really wanted.
public static int readMediumInt(ByteBuffer buf) {
return ((buf.get() & 0xFF) << 16)
| ((buf.get() & 0xFF) << 8)
| ((buf.get() & 0xFF);
}
If you do actually want to read AMF 29-bit integers, this should do the job (assuming i've understood the format correctly):
private static int readMediumInt(ByteBuffer buf) {
int b0, b1, b2;
if ((b0 = buf.get()) >= 0) return b0;
if ((b1 = buf.get()) >= 0) return ((b0 << 7) & ((~(-1 << 7)) << 7)) | b1;
if ((b2 = buf.get()) >= 0) return ((b0 << 14) & ((~(-1 << 7)) << 14)) | ((b1 << 7) & ((~(-1 << 7)) << 7)) | b2;
return ((b0 << 22) & ((~(-1 << 7)) << 22)) | ((b1 << 15) & ((~(-1 << 7)) << 15)) | ((b2 << 8) & ((~(-1 << 7)) << 8)) | (buf.get() & 0xff);
}
The most important change is to avoid allocating objects within the method. By the way your micro benchmark didn't reset "start", so the second result includes the time used for the first method. Also, you need to run micro benchmarks multiple times, otherwise the just in time compiler has no chance to run. I suggest to use a method similar to
public static int readMediumInt3(ByteBuffer buf) {
return ((buf.get() & 0xff) << 16) +
((buf.get() & 0xff) << 8) +
((buf.get() & 0xff));
}
The complete code is:
import java.nio.ByteBuffer;
public class Main {
public static int readMediumInt(ByteBuffer in) {
ByteBuffer buf = ByteBuffer.allocate(4);
buf.put((byte) 0x00);
buf.put(in.get());
buf.put(in.get());
buf.put(in.get());
buf.flip();
return buf.getInt();
}
public static int readMediumInt2(ByteBuffer in) {
byte[] bytes = new byte[3];
in.get(bytes);
int val = 0;
val += bytes[0] * 256 * 256;
val += bytes[1] * 256;
val += bytes[2];
if (val < 0) {
val += 256;
}
return val;
}
public static int readMediumInt3(ByteBuffer buf) {
return ((buf.get() & 0xff) << 16) +
((buf.get() & 0xff) << 8) +
((buf.get() & 0xff));
}
public static void main(String[] args) {
Main m = new Main();
for (int i = 0; i < 5; i++) {
// version 1
ByteBuffer buf = ByteBuffer.allocate(4);
buf.putInt(424242);
buf.flip();
long start;
start = System.nanoTime();
for (int j = 0; j < 10000000; j++) {
buf.position(0);
readMediumInt(buf);
}
start = System.nanoTime() - start;
System.out.printf("Ver 1: elapsed: %d ms\n", start / 1000000);
// version 2
ByteBuffer buf2 = ByteBuffer.allocate(4);
buf2.putInt(424242);
buf2.flip();
start = System.nanoTime();
for (int j = 0; j < 10000000; j++) {
buf2.position(0);
readMediumInt2(buf2);
}
start = System.nanoTime() - start;
System.out.printf("Ver 2: elapsed: %d ms\n", start / 1000000);
// version 3
ByteBuffer buf3 = ByteBuffer.allocate(4);
buf3.putInt(424242);
buf3.flip();
start = System.nanoTime();
for (int j = 0; j < 10000000; j++) {
buf3.position(0);
readMediumInt3(buf3);
}
start = System.nanoTime() - start;
System.out.printf("Ver 3: elapsed: %d ms\n", start / 1000000);
}
}
}
My results:
- Ver 1: elapsed: 556 ms
- Ver 2: elapsed: 187 ms
- Ver 3: elapsed: 3 ms
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