The data is uploaded by 18-bits ADC. One data is split into three bytes and the last 6 bits is useless. The reference voltage is 1 volt, that means 0x1FFFF represents 1 and 0x3FFFF represents -1. How to convert 18-bits twos-complement into float using java. I have written one and it works, but I think it is not efficient enough. My java is terrible.
float data;
int value = ((byte0 & 0xff) << 10) | ((byte1 & 0xff) << 2) | ((byte2 & 0xff) >> 6); // combine 3 bytes into int
int tmp = value & 0x2000; // judge positive or negative
if (tmp != 0) {
value = value - 262144 /* 2^18 */;
data = ((float)value) * 2 / 262143 /* 2^18-1 */;
} else {
data = ((float)value) * 2 / 262143;
}
You could try
double data = (value << 14) / (double) (0x1FFFF << 14);
This will use a shifted 32-bit 2s complement value.
NOTE: If it was 2s complement, 0x3ffff should be the smallest negative number, and 0x20000 is the largest negative number.
Put the sign bit of the ADC in the same place as the sign bit of a 32-bit integer and you can simplify by using the native sign.
public static final float ADC_RANGE = 1.0f; // From -1V to +1V
public static final int ADC_BITS = 18;
public static final int ADC_RANGE = 1 << (ADC_BITS - 1);
public static final int ADC_MASK = (ADC_RANGE - 1) << (32 - ADC_BITS);
int bits = ((byte0 & 0xFF) << 24) | ((byte1 & 0xFF) << 16) | ((byte2 & 0xC0) << 8); // Combine 3 bytes into int, left-aligned
float value = bits / (float)ADC_MASK * ADC_RANGE;
[edit] use constants for 'magic numbers'
What is the Big O of bit count? I'm not sure how the method works, but I would assume it is done in O(logn).
Specifically with this code (where x = 4, y = 1):
return Integer.bitCount(x^y);
Given its implementation, the method consists of six O(1) statements performed in sequence, so it's O(1).
public static int bitCount(int i) {
// HD, Figure 5-2
i = i - ((i >>> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
i = (i + (i >>> 4)) & 0x0f0f0f0f;
i = i + (i >>> 8);
i = i + (i >>> 16);
return i & 0x3f;
}
It is O(1), here is its implementation for JDK 1.5+:
public static int bitCount(int i) {
i = i - ((i >>> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
i = (i + (i >>> 4)) & 0x0f0f0f0f;
i = i + (i >>> 8);
i = i + (i >>> 16);
return i & 0x3f;
}
Any algorithm that work on input of limited size have complexity of O(1).
bitCount works with input of limited size.
Therefore bitCount have complexity of O(1).
This question is usually asked as a part of another question but it turns out that the answer is long. I've decided to answer it here so I can link to it elsewhere.
Although I'm not aware of a way that Java can produce audio samples for us at this time, if that changes in the future, this can be a place for it. I know that JavaFX has some stuff like this, for example AudioSpectrumListener, but still not a way to access samples directly.
I'm using javax.sound.sampled for playback and/or recording but I'd like to do something with the audio.
Perhaps I'd like to display it visually or process it in some way.
How do I access audio sample data to do that with Java Sound?
See also:
Java Sound Tutorials (Official)
Java Sound Resources (Unofficial)
Well, the simplest answer is that at the moment Java can't produce sample data for the programmer.
This quote is from the official tutorial:
There are two ways to apply signal processing:
You can use any processing supported by the mixer or its component lines, by querying for Control objects and then setting the controls as the user desires. Typical controls supported by mixers and lines include gain, pan, and reverberation controls.
If the kind of processing you need isn't provided by the mixer or its lines, your program can operate directly on the audio bytes, manipulating them as desired.
This page discusses the first technique in greater detail, because there is no special API for the second technique.
Playback with javax.sound.sampled largely acts as a bridge between the file and the audio device. The bytes are read in from the file and sent off.
Don't assume the bytes are meaningful audio samples! Unless you happen to have an 8-bit AIFF file, they aren't. (On the other hand, if the samples are definitely 8-bit signed, you can do arithmetic with them. Using 8-bit is one way to avoid the complexity described here, if you're just playing around.)
So instead, I'll enumerate the types of AudioFormat.Encoding and describe how to decode them yourself. This answer will not cover how to encode them, but it's included in the complete code example at the bottom. Encoding is mostly just the decoding process in reverse.
This is a long answer but I wanted to give a thorough overview.
A Little About Digital Audio
Generally when digital audio is explained, we're referring to Linear Pulse-Code Modulation (LPCM).
A continuous sound wave is sampled at regular intervals and the amplitudes are quantized to integers of some scale.
Shown here is a sine wave sampled and quantized to 4-bit:
(Notice that the most positive value in two's complement representation is 1 less than the most negative value. This is a minor detail to be aware of. For example if you're clipping audio and forget this, the positive clips will overflow.)
When we have audio on the computer, we have an array of these samples. A sample array is what we want to turn the byte array in to.
To decode PCM samples, we don't care much about the sample rate or number of channels, so I won't be saying much about them here. Channels are usually interleaved, so that if we had an array of them, they'd be stored like this:
Index 0: Sample 0 (Left Channel)
Index 1: Sample 0 (Right Channel)
Index 2: Sample 1 (Left Channel)
Index 3: Sample 1 (Right Channel)
Index 4: Sample 2 (Left Channel)
Index 5: Sample 2 (Right Channel)
...
In other words, for stereo, the samples in the array just alternate between left and right.
Some Assumptions
All of the code examples will assume the following declarations:
byte[] bytes; The byte array, read from the AudioInputStream.
float[] samples; The output sample array that we're going to fill.
float sample; The sample we're currently working on.
long temp; An interim value used for general manipulation.
int i; The position in the byte array where the current sample's data starts.
We'll normalize all of the samples in our float[] array to the range of -1f <= sample <= 1f. All of the floating-point audio I've seen comes this way and it's pretty convenient.
If our source audio doesn't already come like that (as is for e.g. integer samples), we can normalize them ourselves using the following:
sample = sample / fullScale(bitsPerSample);
Where fullScale is 2bitsPerSample - 1, i.e. Math.pow(2, bitsPerSample-1).
How do I coerce the byte array in to meaningful data?
The byte array contains the sample frames split up and all in a line. This is actually very straight-forward except for something called endianness, which is the ordering of the bytes in each sample packet.
Here's a diagram. This sample (packed in to a byte array) holds the decimal value 9999:
24-bit sample as big-endian:
bytes[i] bytes[i + 1] bytes[i + 2]
┌──────┐ ┌──────┐ ┌──────┐
00000000 00100111 00001111
24-bit sample as little-endian:
bytes[i] bytes[i + 1] bytes[i + 2]
┌──────┐ ┌──────┐ ┌──────┐
00001111 00100111 00000000
They hold the same binary values; however, the byte orders are reversed.
In big-endian, the more significant bytes come before the less significant bytes.
In little-endian, the less significant bytes come before the more significant bytes.
WAV files are stored in little-endian order and AIFF files are stored in big-endian order. Endianness can be obtained from AudioFormat.isBigEndian.
To concatenate the bytes and put them in to our long temp variable, we:
Bitwise AND each byte with the mask 0xFF (which is 0b1111_1111) to avoid sign-extension when the byte is automatically promoted. (char, byte and short are promoted to int when arithmetic is performed on them.) See also What does value & 0xff do in Java?
Bit shift each byte in to position.
Bitwise OR the bytes together.
Here's a 24-bit example:
long temp;
if (isBigEndian) {
temp = (
((bytes[i ] & 0xffL) << 16)
| ((bytes[i + 1] & 0xffL) << 8)
| (bytes[i + 2] & 0xffL)
);
} else {
temp = (
(bytes[i ] & 0xffL)
| ((bytes[i + 1] & 0xffL) << 8)
| ((bytes[i + 2] & 0xffL) << 16)
);
}
Notice that the shift order is reversed based on endianness.
This can also be generalized to a loop, which can be seen in the full code at the bottom of this answer. (See the unpackAnyBit and packAnyBit methods.)
Now that we have the bytes concatenated together, we can take a few more steps to turn them in to a sample. The next steps depend on the actual encoding.
How do I decode Encoding.PCM_SIGNED?
The two's complement sign must be extended. This means that if the most significant bit (MSB) is set to 1, we fill all the bits above it with 1s. The arithmetic right-shift (>>) will do the filling for us automatically if the sign bit is set, so I usually do it this way:
int bitsToExtend = Long.SIZE - bitsPerSample;
float sample = (temp << bitsToExtend) >> bitsToExtend.
(Where Long.SIZE is 64. If our temp variable wasn't a long, we'd use something else. If we used e.g. int temp instead, we'd use 32.)
To understand how this works, here's a diagram of sign-extending 8-bit to 16-bit:
11111111 is the byte value -1, but the upper bits of the short are 0.
Shift the byte's MSB in to the MSB position of the short.
0000 0000 1111 1111
<< 8
───────────────────
1111 1111 0000 0000
Shift it back and the right-shift fills all the upper bits with 1s.
We now have the short value of -1.
1111 1111 0000 0000
>> 8
───────────────────
1111 1111 1111 1111
Positive values (that had a 0 in the MSB) are left unchanged. This is a nice property of the arithmetic right-shift.
Then normalize the sample, as described in Some Assumptions.
You might not need to write explicit sign-extension if your code is simple
Java does sign-extension automatically when converting from one integral type to a larger type, for example byte to int. If you know that your input and output format are always signed, you can use the automatic sign-extension while concatenating bytes in the earlier step.
Recall from the section above (How do I coerce the byte array in to meaningful data?) that we used b & 0xFF to prevent sign-extension from occurring. If you just remove the & 0xFF from the highest byte, sign-extension will happen automatically.
For example, the following decodes signed, big-endian, 16-bit samples:
for (int i = 0; i < bytes.length; i++) {
int sample = (bytes[i] << 8) // high byte is sign-extended
| (bytes[i + 1] & 0xFF); // low byte is not
// ...
}
How do I decode Encoding.PCM_UNSIGNED?
We turn it in to a signed number. Unsigned samples are simply offset so that, for example:
An unsigned value of 0 corresponds to the most negative signed value.
An unsigned value of 2bitsPerSample - 1 corresponds to the signed value of 0.
An unsigned value of 2bitsPerSample corresponds to the most positive signed value.
So this turns out to be pretty simple. Just subtract the offset:
float sample = temp - fullScale(bitsPerSample);
Then normalize the sample, as described in Some Assumptions.
How do I decode Encoding.PCM_FLOAT?
This is new since Java 7.
In practice, floating-point PCM is typically either IEEE 32-bit or IEEE 64-bit and already normalized to the range of ±1.0. The samples can be obtained with the utility methods Float#intBitsToFloat and Double#longBitsToDouble.
// IEEE 32-bit
float sample = Float.intBitsToFloat((int) temp);
// IEEE 64-bit
double sampleAsDouble = Double.longBitsToDouble(temp);
float sample = (float) sampleAsDouble; // or just use double for arithmetic
How do I decode Encoding.ULAW and Encoding.ALAW?
These are companding compression codecs that are more common in telephones and such. They're supported by javax.sound.sampled I assume because they're used by Sun's Au format. (However, it's not limited to just this type of container. For example, WAV can contain these encodings.)
You can conceptualize A-law and μ-law like they're a floating-point format. These are PCM formats but the range of values is non-linear.
There are two ways to decode them. I'll show the way which uses the mathematical formula. You can also decode them by manipulating the binary directly which is described in this blog post but it's more esoteric-looking.
For both, the compressed data is 8-bit. Standardly A-law is 13-bit when decoded and μ-law is 14-bit when decoded; however, applying the formula yields a range of ±1.0.
Before you can apply the formula, there are three things to do:
Some of the bits are standardly inverted for storage due to reasons involving data integrity.
They're stored as sign and magnitude (rather than two's complement).
The formula also expects a range of ±1.0, so the 8-bit value has to be scaled.
For μ-law all the bits are inverted, so:
temp ^= 0xffL; // 0xff == 0b1111_1111
(Note that we can't use ~, because we don't want to invert the high bits of the long.)
For A-law, every other bit is inverted, so:
temp ^= 0x55L; // 0x55 == 0b0101_0101
(XOR can be used to do inversion. See How do you set, clear and toggle a bit?)
To convert from sign and magnitude to two's complement, we:
Check to see if the sign bit was set.
If so, clear the sign bit and negate the number.
// 0x80 == 0b1000_0000
if ((temp & 0x80L) != 0) {
temp ^= 0x80L;
temp = -temp;
}
Then scale the encoded numbers, the same way as described in Some Assumptions:
sample = temp / fullScale(8);
Now we can apply the expansion.
The μ-law formula translated to Java is then:
sample = (float) (
signum(sample)
*
(1.0 / 255.0)
*
(pow(256.0, abs(sample)) - 1.0)
);
The A-law formula translated to Java is then:
float signum = signum(sample);
sample = abs(sample);
if (sample < (1.0 / (1.0 + log(87.7)))) {
sample = (float) (
sample * ((1.0 + log(87.7)) / 87.7)
);
} else {
sample = (float) (
exp((sample * (1.0 + log(87.7))) - 1.0) / 87.7
);
}
sample = signum * sample;
Here's the full example code for the SimpleAudioConversion class.
package mcve.audio;
import javax.sound.sampled.AudioFormat;
import javax.sound.sampled.AudioFormat.Encoding;
import static java.lang.Math.*;
/**
* <p>Performs simple audio format conversion.</p>
*
* <p>Example usage:</p>
*
* <pre>{#code AudioInputStream ais = ... ;
* SourceDataLine line = ... ;
* AudioFormat fmt = ... ;
*
* // do setup
*
* for (int blen = 0; (blen = ais.read(bytes)) > -1;) {
* int slen;
* slen = SimpleAudioConversion.decode(bytes, samples, blen, fmt);
*
* // do something with samples
*
* blen = SimpleAudioConversion.encode(samples, bytes, slen, fmt);
* line.write(bytes, 0, blen);
* }}</pre>
*
* #author Radiodef
* #see Overview on Stack Overflow
*/
public final class SimpleAudioConversion {
private SimpleAudioConversion() {}
/**
* Converts from a byte array to an audio sample float array.
*
* #param bytes the byte array, filled by the AudioInputStream
* #param samples an array to fill up with audio samples
* #param blen the return value of AudioInputStream.read
* #param fmt the source AudioFormat
*
* #return the number of valid audio samples converted
*
* #throws NullPointerException if bytes, samples or fmt is null
* #throws ArrayIndexOutOfBoundsException
* if bytes.length is less than blen or
* if samples.length is less than blen / bytesPerSample(fmt.getSampleSizeInBits())
*/
public static int decode(byte[] bytes,
float[] samples,
int blen,
AudioFormat fmt) {
int bitsPerSample = fmt.getSampleSizeInBits();
int bytesPerSample = bytesPerSample(bitsPerSample);
boolean isBigEndian = fmt.isBigEndian();
Encoding encoding = fmt.getEncoding();
double fullScale = fullScale(bitsPerSample);
int i = 0;
int s = 0;
while (i < blen) {
long temp = unpackBits(bytes, i, isBigEndian, bytesPerSample);
float sample = 0f;
if (encoding == Encoding.PCM_SIGNED) {
temp = extendSign(temp, bitsPerSample);
sample = (float) (temp / fullScale);
} else if (encoding == Encoding.PCM_UNSIGNED) {
temp = unsignedToSigned(temp, bitsPerSample);
sample = (float) (temp / fullScale);
} else if (encoding == Encoding.PCM_FLOAT) {
if (bitsPerSample == 32) {
sample = Float.intBitsToFloat((int) temp);
} else if (bitsPerSample == 64) {
sample = (float) Double.longBitsToDouble(temp);
}
} else if (encoding == Encoding.ULAW) {
sample = bitsToMuLaw(temp);
} else if (encoding == Encoding.ALAW) {
sample = bitsToALaw(temp);
}
samples[s] = sample;
i += bytesPerSample;
s++;
}
return s;
}
/**
* Converts from an audio sample float array to a byte array.
*
* #param samples an array of audio samples to encode
* #param bytes an array to fill up with bytes
* #param slen the return value of the decode method
* #param fmt the destination AudioFormat
*
* #return the number of valid bytes converted
*
* #throws NullPointerException if samples, bytes or fmt is null
* #throws ArrayIndexOutOfBoundsException
* if samples.length is less than slen or
* if bytes.length is less than slen * bytesPerSample(fmt.getSampleSizeInBits())
*/
public static int encode(float[] samples,
byte[] bytes,
int slen,
AudioFormat fmt) {
int bitsPerSample = fmt.getSampleSizeInBits();
int bytesPerSample = bytesPerSample(bitsPerSample);
boolean isBigEndian = fmt.isBigEndian();
Encoding encoding = fmt.getEncoding();
double fullScale = fullScale(bitsPerSample);
int i = 0;
int s = 0;
while (s < slen) {
float sample = samples[s];
long temp = 0L;
if (encoding == Encoding.PCM_SIGNED) {
temp = (long) (sample * fullScale);
} else if (encoding == Encoding.PCM_UNSIGNED) {
temp = (long) (sample * fullScale);
temp = signedToUnsigned(temp, bitsPerSample);
} else if (encoding == Encoding.PCM_FLOAT) {
if (bitsPerSample == 32) {
temp = Float.floatToRawIntBits(sample);
} else if (bitsPerSample == 64) {
temp = Double.doubleToRawLongBits(sample);
}
} else if (encoding == Encoding.ULAW) {
temp = muLawToBits(sample);
} else if (encoding == Encoding.ALAW) {
temp = aLawToBits(sample);
}
packBits(bytes, i, temp, isBigEndian, bytesPerSample);
i += bytesPerSample;
s++;
}
return i;
}
/**
* Computes the block-aligned bytes per sample of the audio format,
* using Math.ceil(bitsPerSample / 8.0).
* <p>
* Round towards the ceiling because formats that allow bit depths
* in non-integral multiples of 8 typically pad up to the nearest
* integral multiple of 8. So for example, a 31-bit AIFF file will
* actually store 32-bit blocks.
*
* #param bitsPerSample the return value of AudioFormat.getSampleSizeInBits
* #return The block-aligned bytes per sample of the audio format.
*/
public static int bytesPerSample(int bitsPerSample) {
return (int) ceil(bitsPerSample / 8.0); // optimization: ((bitsPerSample + 7) >>> 3)
}
/**
* Computes the largest magnitude representable by the audio format,
* using Math.pow(2.0, bitsPerSample - 1). Note that for two's complement
* audio, the largest positive value is one less than the return value of
* this method.
* <p>
* The result is returned as a double because in the case that
* bitsPerSample is 64, a long would overflow.
*
* #param bitsPerSample the return value of AudioFormat.getBitsPerSample
* #return the largest magnitude representable by the audio format
*/
public static double fullScale(int bitsPerSample) {
return pow(2.0, bitsPerSample - 1); // optimization: (1L << (bitsPerSample - 1))
}
private static long unpackBits(byte[] bytes,
int i,
boolean isBigEndian,
int bytesPerSample) {
switch (bytesPerSample) {
case 1: return unpack8Bit(bytes, i);
case 2: return unpack16Bit(bytes, i, isBigEndian);
case 3: return unpack24Bit(bytes, i, isBigEndian);
default: return unpackAnyBit(bytes, i, isBigEndian, bytesPerSample);
}
}
private static long unpack8Bit(byte[] bytes, int i) {
return bytes[i] & 0xffL;
}
private static long unpack16Bit(byte[] bytes,
int i,
boolean isBigEndian) {
if (isBigEndian) {
return (
((bytes[i ] & 0xffL) << 8)
| (bytes[i + 1] & 0xffL)
);
} else {
return (
(bytes[i ] & 0xffL)
| ((bytes[i + 1] & 0xffL) << 8)
);
}
}
private static long unpack24Bit(byte[] bytes,
int i,
boolean isBigEndian) {
if (isBigEndian) {
return (
((bytes[i ] & 0xffL) << 16)
| ((bytes[i + 1] & 0xffL) << 8)
| (bytes[i + 2] & 0xffL)
);
} else {
return (
(bytes[i ] & 0xffL)
| ((bytes[i + 1] & 0xffL) << 8)
| ((bytes[i + 2] & 0xffL) << 16)
);
}
}
private static long unpackAnyBit(byte[] bytes,
int i,
boolean isBigEndian,
int bytesPerSample) {
long temp = 0;
if (isBigEndian) {
for (int b = 0; b < bytesPerSample; b++) {
temp |= (bytes[i + b] & 0xffL) << (
8 * (bytesPerSample - b - 1)
);
}
} else {
for (int b = 0; b < bytesPerSample; b++) {
temp |= (bytes[i + b] & 0xffL) << (8 * b);
}
}
return temp;
}
private static void packBits(byte[] bytes,
int i,
long temp,
boolean isBigEndian,
int bytesPerSample) {
switch (bytesPerSample) {
case 1: pack8Bit(bytes, i, temp);
break;
case 2: pack16Bit(bytes, i, temp, isBigEndian);
break;
case 3: pack24Bit(bytes, i, temp, isBigEndian);
break;
default: packAnyBit(bytes, i, temp, isBigEndian, bytesPerSample);
break;
}
}
private static void pack8Bit(byte[] bytes, int i, long temp) {
bytes[i] = (byte) (temp & 0xffL);
}
private static void pack16Bit(byte[] bytes,
int i,
long temp,
boolean isBigEndian) {
if (isBigEndian) {
bytes[i ] = (byte) ((temp >>> 8) & 0xffL);
bytes[i + 1] = (byte) ( temp & 0xffL);
} else {
bytes[i ] = (byte) ( temp & 0xffL);
bytes[i + 1] = (byte) ((temp >>> 8) & 0xffL);
}
}
private static void pack24Bit(byte[] bytes,
int i,
long temp,
boolean isBigEndian) {
if (isBigEndian) {
bytes[i ] = (byte) ((temp >>> 16) & 0xffL);
bytes[i + 1] = (byte) ((temp >>> 8) & 0xffL);
bytes[i + 2] = (byte) ( temp & 0xffL);
} else {
bytes[i ] = (byte) ( temp & 0xffL);
bytes[i + 1] = (byte) ((temp >>> 8) & 0xffL);
bytes[i + 2] = (byte) ((temp >>> 16) & 0xffL);
}
}
private static void packAnyBit(byte[] bytes,
int i,
long temp,
boolean isBigEndian,
int bytesPerSample) {
if (isBigEndian) {
for (int b = 0; b < bytesPerSample; b++) {
bytes[i + b] = (byte) (
(temp >>> (8 * (bytesPerSample - b - 1))) & 0xffL
);
}
} else {
for (int b = 0; b < bytesPerSample; b++) {
bytes[i + b] = (byte) ((temp >>> (8 * b)) & 0xffL);
}
}
}
private static long extendSign(long temp, int bitsPerSample) {
int bitsToExtend = Long.SIZE - bitsPerSample;
return (temp << bitsToExtend) >> bitsToExtend;
}
private static long unsignedToSigned(long temp, int bitsPerSample) {
return temp - (long) fullScale(bitsPerSample);
}
private static long signedToUnsigned(long temp, int bitsPerSample) {
return temp + (long) fullScale(bitsPerSample);
}
// mu-law constant
private static final double MU = 255.0;
// A-law constant
private static final double A = 87.7;
// natural logarithm of A
private static final double LN_A = log(A);
private static float bitsToMuLaw(long temp) {
temp ^= 0xffL;
if ((temp & 0x80L) != 0) {
temp = -(temp ^ 0x80L);
}
float sample = (float) (temp / fullScale(8));
return (float) (
signum(sample)
*
(1.0 / MU)
*
(pow(1.0 + MU, abs(sample)) - 1.0)
);
}
private static long muLawToBits(float sample) {
double sign = signum(sample);
sample = abs(sample);
sample = (float) (
sign * (log(1.0 + (MU * sample)) / log(1.0 + MU))
);
long temp = (long) (sample * fullScale(8));
if (temp < 0) {
temp = -temp ^ 0x80L;
}
return temp ^ 0xffL;
}
private static float bitsToALaw(long temp) {
temp ^= 0x55L;
if ((temp & 0x80L) != 0) {
temp = -(temp ^ 0x80L);
}
float sample = (float) (temp / fullScale(8));
float sign = signum(sample);
sample = abs(sample);
if (sample < (1.0 / (1.0 + LN_A))) {
sample = (float) (sample * ((1.0 + LN_A) / A));
} else {
sample = (float) (exp((sample * (1.0 + LN_A)) - 1.0) / A);
}
return sign * sample;
}
private static long aLawToBits(float sample) {
double sign = signum(sample);
sample = abs(sample);
if (sample < (1.0 / A)) {
sample = (float) ((A * sample) / (1.0 + LN_A));
} else {
sample = (float) ((1.0 + log(A * sample)) / (1.0 + LN_A));
}
sample *= sign;
long temp = (long) (sample * fullScale(8));
if (temp < 0) {
temp = -temp ^ 0x80L;
}
return temp ^ 0x55L;
}
}
This is how you get the actual sample data from the currently playing sound. The other excellent answer will tell you what the data means. Haven't tried it on another OS than my Windows 10 machine YMMV. For me it pulls the current system default recording device. On Windows set it to "Stereo Mix" instead of "Microphone" to get playing sound. You may have to toggle "Show Disabled Devices" to see "Stereo Mix".
import javax.sound.sampled.*;
public class SampleAudio {
private static long extendSign(long temp, int bitsPerSample) {
int extensionBits = 64 - bitsPerSample;
return (temp << extensionBits) >> extensionBits;
}
public static void main(String[] args) throws LineUnavailableException {
float sampleRate = 8000;
int sampleSizeBits = 16;
int numChannels = 1; // Mono
AudioFormat format = new AudioFormat(sampleRate, sampleSizeBits, numChannels, true, true);
TargetDataLine tdl = AudioSystem.getTargetDataLine(format);
tdl.open(format);
tdl.start();
if (!tdl.isOpen()) {
System.exit(1);
}
byte[] data = new byte[(int)sampleRate*10];
int read = tdl.read(data, 0, (int)sampleRate*10);
if (read > 0) {
for (int i = 0; i < read-1; i = i + 2) {
long val = ((data[i] & 0xffL) << 8L) | (data[i + 1] & 0xffL);
long valf = extendSign(val, 16);
System.out.println(i + "\t" + valf);
}
}
tdl.close();
}
}
Here is my problem:
I have a binary value
101001
and a mask
011100
I would like to compare them and get the result as an integer. In this case that would give:
1 **010** 01
0 **111** 00
= 010 => 2
My first idea consists of dealing with a character array. But I would like to know if there is a better way to achieve this aim in Java?
I would like to compare them and get the result as an integer
Assuming you meant 'mask' rather than 'compare':
int result = 0B011100 & 0B011100;
No char arrays required.
This is rather trivial.
I just improved the algorithm to be able to have mask where one bits can be splitted such as:
00111011011
Here is my function to get a value from a mask and a masked value
private static long getMaskedValue(long maskedValue, long mask){
long definitiveMaskedValue = 0;
int count=0;
maskedValue = mask & maskedValue;
while (mask != 0){
while ((mask & 1) == 0){
mask = mask >>> 1;
maskedValue = maskedValue >>> 1;
}
while ((mask & 1) == 1){
definitiveMaskedValue = definitiveMaskedValue + ((maskedValue & 1) << count);
count++;
mask = mask >>> 1;
maskedValue = maskedValue >>> 1;
}
}
return definitiveMaskedValue;
}
Here is my function to store a value in a variable thanks to a bitmask, it returns the old variable with the value stored inside.
I had to use BigInteger because of the shift operator that can not shift more than 32bits left in Java.
private static long setMaskedValue (long maskedValue, long mask, long valueToAdd) {
int nbZero=0;
int nbLeastSignificantBit=0;
long tmpMask=mask;
maskedValue = maskedValue & ~mask;
while (tmpMask != 0){
while ((tmpMask & 1) == 0){
tmpMask = tmpMask >>> 1;
nbLeastSignificantBit++;
nbZero ++;
}
while ((tmpMask & 1) == 1){
tmpMask = tmpMask >>> 1;
BigInteger bigValueToAdd = BigInteger.valueOf(valueToAdd).shiftLeft(nbZero);
long tmpValueToAdd = bigValueToAdd.longValue();
BigInteger bigMaskOneBit = BigInteger.valueOf(1).shiftLeft(nbLeastSignificantBit);
long maskOneBit = bigMaskOneBit.longValue();
long bitValueToSet = getMaskedValue(tmpValueToAdd, maskOneBit);
maskedValue = maskedValue | bitValueToSet << nbLeastSignificantBit;
nbLeastSignificantBit++;
}
}
return maskedValue;
}
Of course.
You need first AND your bits.
Shift right to avoid those zeros at right of mask.
You need value as integer already.
Then do the AND: int masked = value & mask;
Then shift right until the first 1 in the mask.
while (mask % 2 == 0) {
mask = mask >>> 1;
masked = masked >>> 1;
}
You can use while (mask & 1 == 0) { if you prefer :)
& is bitwise AND.
| is bitwise OR.
^ is bitwise XOR (if my memory doesn't fail :).
>>> is shifting right (unsigned integer)
I have a spec which reads the next two bytes are signed int.
To read that in java i have the following
When i read a signed int in java using the following code i get a value of 65449
Logic for calculation of unsigned
int a =(byte[1] & 0xff) <<8
int b =(byte[0] & 0xff) <<0
int c = a+b
I believe this is wrong because if i and with 0xff i get an unsigned equivalent
so i removed the & 0xff and the logic as given below
int a = byte[1] <<8
int b = byte[0] << 0
int c = a+b
which gives me the value -343
byte[1] =-1
byte[0]=-87
I tried to offset these values with the way the spec reads but this looks wrong.Since the size of the heap doesnt fall under this.
Which is the right way to do for signed int calculation in java?
Here is how the spec goes
somespec() { xtype 8 uint8 xStyle 16 int16 }
xStyle :A signed integer that represents an offset (in bytes) from the start of this Widget() structure to the start of an xStyle() structure that expresses inherited styles for defined by page widget as well as styles that apply specifically to this widget.
If you value is a signed 16-bit you want a short and int is 32-bit which can also hold the same values but not so naturally.
It appears you wants a signed little endian 16-bit value.
byte[] bytes =
short s = ByteBuffer.wrap(bytes).order(ByteOrder.LITTLE_ENDIAN).getShort();
or
short s = (short) ((bytes[0] & 0xff) | (bytes[1] << 8));
BTW: You can use an int but its not so simple.
// to get a sign extension.
int i = ((bytes[0] & 0xff) | (bytes[1] << 8)) << 16 >> 16;
or
int i = (bytes[0] & 0xff) | (short) (bytes[1] << 8));
Assuming that bytes[1] is the MSB, and bytes[0] is the LSB, and that you want the answer to be a 16 bit signed integer:
short res16 = ((bytes[1] << 8) | bytes[0]);
Then to get a 32 bit signed integer:
int res32 = res16; // sign extends.
By the way, the specification should say which of the two bytes is the MSB, and which is the LSB. If it doesn't and if there aren't any examples, you can't implement it!
Somewhere in the spec it will say how an "int16" is represented. Paste THAT part. Or paste a link to the spec so that we can read it ourselves.
Take a look on DataInputStream.readInt(). You can either steel code from there or just use DataInputStream: wrap your input stream with it and then read typed data easily.
For your convenience this is the code:
public final int readInt() throws IOException {
int ch1 = in.read();
int ch2 = in.read();
int ch3 = in.read();
int ch4 = in.read();
if ((ch1 | ch2 | ch3 | ch4) < 0)
throw new EOFException();
return ((ch1 << 24) + (ch2 << 16) + (ch3 << 8) + (ch4 << 0));
}
I can't compile it right now, but I would do (assuming byte1 and byte0 are realling of byte type).
int result = byte1;
result = result << 8;
result = result | byte0; //(binary OR)
if (result & 0x8000 == 0x8000) { //sign extension
result = result | 0xFFFF0000;
}
if byte1 and byte0 are ints, you will need to make the `&0xFF
UPDATE because Java forces the expression of an if to be a boolean
do you have a way of finding a correct output for a given input?
technically, an int size is 4 bytes, so with just 2 bytes you can't reach the sign bit.
I ran across this same problem reading a MIDI file. A MIDI file has signed 16 bit as well as signed 32 bit integers. In a MIDI file, the most significant bytes come first (big-endian).
Here's what I did. It might be crude, but it maintains the sign. If the least significant bytes come first (little-endian), reverse the order of the indexes.
pos is the position in the byte array where the number starts.
length is the length of the integer, either 2 or 4. Yes, a 2 byte integer is a short, but we all work with ints.
private int convertBytes(byte[] number, int pos, int length) {
int output = 0;
if (length == 2) {
output = ((int) number[pos]) << 24;
output |= convertByte(number[pos + 1]) << 16;
output >>= 16;
} else if (length == 4) {
output = ((int) number[pos]) << 24;
output |= convertByte(number[pos + 1]) << 16;
output |= convertByte(number[pos + 2]) << 8;
output |= convertByte(number[pos + 3]);
}
return output;
}
private int convertByte(byte number) {
return (int) number & 0xff;
}