读取Java中的字节

Question

I am trying to understand how the following line of code works:

for (int i = 0; i < numSamples; i++) {
        short ampValue = 0;
        for (int byteNo = 0; byteNo < 2; byteNo++) {
            ampValue |= (short) ((data[pointer++] & 0xFF) << (byteNo * 8));
        }
        amplitudes[i] = ampValue;
    }

As far as I understand, this is reading 2 bytes (as 2 bytes per sample) in a inclusive manner, ie the ampValue is composed of two byte reads. The data is the actual data sample (file) and the pointer is increasing to read it upto the last sample. But I don't understand this part:

"data[pointer++] & 0xFF) << (byteNo * 8)); "

Also, I am wondering whether it makes any difference if I want to read this as a double instead of short?

Answer 1

In Java, all bytes are signed. The expression (data[pointer++] & 0xFF) converts the signed byte value to an int with the value of the byte if it were unsigned. Then the expression << (byteNo * 8) left-shifts the resulting value by zero or eight bits depending on the value of byteNo . The value of the whole expression is assigned with bitwise or to ampValue .

There appears to be a bug in this code. The value of ampValue is not reset to zero between iterations. And amplitude is not used. Are those identifiers supposed to be the same?

Answer 2

Looks like data[] is the array of bytes.

data[pointer++] gives you a byte value in the range [-128..127].

0xFF is an int contstant, so...

data[pointer++] & 0xFF promotes the byte value to an int value in the range [-128..127]. Then the & operator zeroes out all of the bits that are not set in 0xFF (ie, it zeroes out the 24 upper bits, leaving only the low 8 bits.

The value of that expression now will be in the range [0..255].

The << operator shifts the result to the left by (byteNo * 8) bits. That's the same as saying, it multiplies the value by 2 raised to the power of (byteNo * 8). When byteNo==0, it will multiply by 2 to the power 0 (ie, it will multiply by 1). When byteNo==1, it will multiply by 2 to the power 8 (ie, it will multiply by 256).

This loop is creating an int in the range [0..65535] (16 bits) from each pair of bytes in the array, taking the first member of each pair as the low-order byte and the second member as the high-order byte.

It won't work to declare ampValue as double, because the |= operator will not work on a double, but you can declare the amplitudes[] array to be an array of double, and the assignment amplitudes[i] = ampValue will implicitly promote the value to a double value in the range [0.0..65535.0].

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Additional info: Don't overlook @KevinKrumwiede's comment about a bug in the example.

Answer 3

Let's break down the statement:

• |= is the bitwise or and assignment operator. a |= b is equivalent to a = a | b a = a | b .
• (short) casts the int element from the data array to a short .
• pointer++ is a post-increment operation. The value of pointer will be returned and used and then immediately incremented every single time it's accessed in this fashion - this is beneficial in this case because the outer-loop is cycling through 2-byte samples (via the inner loop) from the contiguous data buffer, so this keeps incrementing.
• & is the bitwise AND operator and 0xFF is the hexadecimal value for the byte 0b11111111 (255 in decimal); the expression data[pointer++] & 0xFF is basically saying, for each bit in the byte retrieved from the data array, AND it with 1 . In this context, it forces Java, which by default stores signed byte objects (ie values from -128 to 127 in decimal), to return the value as an unsigned byte (ie values from 0 to 255 decimal).
• Since your samples are 2 bytes long, you need to shift the second lot of 8 bits left, as the most significant bits, using the left bit-shift operator << . The byteNo * 8 ensures that you're only shifting bits when it's the second of the two bytes.

After the two bytes have been read, ampValue will now contain the value of the sample as a short .