如何将字节数组转换为其数值(Java) ？

我有一个8字节的数组，我想把它转换成相应的数值。

例如:。

byte[] by = new byte[8];  // the byte array is stored in 'by'


// CONVERSION OPERATION
// return the numeric value

我想要一个方法，将执行上述转换操作。

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One could use the Buffers that are provided as part of the java.nio package to perform the conversion.

Here, the source byte[] array has a of length 8, which is the size that corresponds with a long value.

First, the byte[] array is wrapped in a ByteBuffer, and then the ByteBuffer.getLong method is called to obtain the long value:

ByteBuffer bb = ByteBuffer.wrap(new byte[] {0, 0, 0, 0, 0, 0, 0, 4});
long l = bb.getLong();


System.out.println(l);

Result

4

I'd like to thank dfa for pointing out the ByteBuffer.getLong method in the comments.

Although it may not be applicable in this situation, the beauty of the Buffers come with looking at an array with multiple values.

For example, if we had a 8 byte array, and we wanted to view it as two int values, we could wrap the byte[] array in an ByteBuffer, which is viewed as a IntBuffer and obtain the values by IntBuffer.get:

ByteBuffer bb = ByteBuffer.wrap(new byte[] {0, 0, 0, 1, 0, 0, 0, 4});
IntBuffer ib = bb.asIntBuffer();
int i0 = ib.get(0);
int i1 = ib.get(1);


System.out.println(i0);
System.out.println(i1);

Result:

1
4
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最佳答案

Assuming the first byte is the least significant byte:

long value = 0;
for (int i = 0; i < by.length; i++)
{
value += ((long) by[i] & 0xffL) << (8 * i);
}

Is the first byte the most significant, then it is a little bit different:

long value = 0;
for (int i = 0; i < by.length; i++)
{
value = (value << 8) + (by[i] & 0xff);
}

Replace long with BigInteger, if you have more than 8 bytes.

Thanks to Aaron Digulla for the correction of my errors.

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If this is an 8-bytes numeric value, you can try:

BigInteger n = new BigInteger(byteArray);

If this is an UTF-8 character buffer, then you can try:

BigInteger n = new BigInteger(new String(byteArray, "UTF-8"));
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Complete java converter code for all primitive types to/from arrays http://www.daniweb.com/code/snippet216874.html

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Simply, you could use or refer to guava lib provided by google, which offers utiliy methods for conversion between long and byte array. My client code:

    long content = 212000607777l;
byte[] numberByte = Longs.toByteArray(content);
logger.info(Longs.fromByteArray(numberByte));
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You can also use BigInteger for variable length bytes. You can convert it to Long, Integer or Short, whichever suits your needs.

new BigInteger(bytes).intValue();

or to denote polarity:

new BigInteger(1, bytes).intValue();
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Each cell in the array is treated as unsigned int:

private int unsignedIntFromByteArray(byte[] bytes) {
int res = 0;
if (bytes == null)
return res;




for (int i=0;i<bytes.length;i++){
res = res | ((bytes[i] & 0xff) << i*8);
}
return res;
}
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public static long byteArrayToLong(byte[] bytes) {
return ((long) (bytes[0]) << 56)
+ (((long) bytes[1] & 0xFF) << 48)
+ ((long) (bytes[2] & 0xFF) << 40)
+ ((long) (bytes[3] & 0xFF) << 32)
+ ((long) (bytes[4] & 0xFF) << 24)
+ ((bytes[5] & 0xFF) << 16)
+ ((bytes[6] & 0xFF) << 8)
+ (bytes[7] & 0xFF);
}

convert bytes array (long is 8 bytes) to long

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You can try use the code from this answer: https://stackoverflow.com/a/68393576/7918717

It parses bytes as a signed number of arbitrary length. A few examples:

bytesToSignedNumber(false, 0xF1, 0x01, 0x04) returns 15794436 (3 bytes as int)

bytesToSignedNumber(false, 0xF1, 0x01, 0x01, 0x04) returns -251592444 (4 bytes as int)

bytesToSignedNumber(false, 0xF1, 0x01, 0x01, 0x01, 0x01, 0x01, 0x01, 0x01, 0x04) returns -1080581331768770303 (8 of 9 bytes as long)


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