我怎样才能将一个字节数组转换成双精度和返回？

为了将一个字节数组转换成一个double，我发现这个：

//convert 8 byte array to double int start=0;//??? int i = 0; int len = 8; int cnt = 0; byte[] tmp = new byte[len]; for (i = start; i < (start + len); i++) { tmp[cnt] = arr[i]; //System.out.println(java.lang.Byte.toString(arr[i]) + " " + i); cnt++; } long accum = 0; i = 0; for ( int shiftBy = 0; shiftBy < 64; shiftBy += 8 ) { accum |= ( (long)( tmp[i] & 0xff ) ) << shiftBy; i++; } return Double.longBitsToDouble(accum); 

但是我找不到任何将double转换为字节数组的东西。

Answer 1

甚至更简单，

 import java.nio.ByteBuffer; public static byte[] toByteArray(double value) { byte[] bytes = new byte[8]; ByteBuffer.wrap(bytes).putDouble(value); return bytes; } public static double toDouble(byte[] bytes) { return ByteBuffer.wrap(bytes).getDouble(); } 

Answer 2

 long bits = Double.doubleToLongBits(myDouble); 

Answer 3

 public static byte[] toByteArray(double d) { long l = Double.doubleToRawLongBits(d); return new byte[] { (byte)((l >> 56) & 0xff), (byte)((l >> 48) & 0xff), (byte)((l >> 40) & 0xff), (byte)((l >> 32) & 0xff), (byte)((l >> 24) & 0xff), (byte)((l >> 16) & 0xff), (byte)((l >> 8) & 0xff), (byte)((l >> 0) & 0xff), }; } 

Answer 4

该function已经在API中实现了。 将字节数组包装在ByteBuffer并使用ByteBuffer.putLong和ByteBuffer.getLong ：

 import java.nio.*; import java.util.Arrays; public class Test { public static void main(String... args) throws Exception { long[] longArray = { 1234, 2345, 3456 }; // Longs to bytes byte[] bytes = new byte[longArray.length * 8]; ByteBuffer buf = ByteBuffer.wrap(bytes); for (long l : longArray) buf.putLong(l); System.out.println(Arrays.toString(bytes)); // Bytes to longs ByteBuffer buf2 = ByteBuffer.wrap(bytes); long[] longs = new long[bytes.length / 8]; for (int i = 0; i < longs.length; i++) longs[i] = buf2.getLong(i*8); System.out.println(Arrays.toString(longs)); } } 

输出：

 [0, 0, 0, 0, 0, 0, 4, -46, 0, 0, 0, 0, 0, 0, 9, 41, 0, 0, 0, 0, 0, 0, 13, -128] [1234, 2345, 3456] 

Answer 5

 public static final short byteArrayToShort(byte[] bytes) { return ByteBuffer.wrap(bytes).getShort(); } public static final int byteArrayToInt(byte[] bytes) { return ByteBuffer.wrap(bytes).getInt(); } public static final float byteArrayToFloat(byte[] bytes) { return ByteBuffer.wrap(bytes).getFloat(); } public static double byteArrayToDouble(byte[] bytes) { return ByteBuffer.wrap(bytes).getDouble(); } public static final long byteArrayToLong(byte[] bytes) { return ByteBuffer.wrap(bytes).getLong(); } 

去享受。

Answer 6

我实际上遇到了上下两个问题，这似乎是我所看到的唯一的代码纠正了这一点。 我希望能帮助其他人在这方面寻找答案。 如果您确实需要其他代码，请确保您testing了所有的值，您应该编写一个可以转换为所有值的循环，并声明它们是肯定的。

 // byte2Double method - extracts doubles from byte array // source: http://www.java2s.com/Code/Java/Data-Type/bytetoDouble.htm public static final double[] byte2Double(byte[] inData, boolean byteSwap) { int j = 0, upper, lower; int length = inData.length / 8; double[] outData = new double[length]; if (!byteSwap) for (int i = 0; i < length; i++) { j = i * 8; upper = (((inData[j] & 0xff) << 24) + ((inData[j + 1] & 0xff) << 16) + ((inData[j + 2] & 0xff) << 8) + ((inData[j + 3] & 0xff) << 0)); lower = (((inData[j + 4] & 0xff) << 24) + ((inData[j + 5] & 0xff) << 16) + ((inData[j + 6] & 0xff) << 8) + ((inData[j + 7] & 0xff) << 0)); outData[i] = Double.longBitsToDouble((((long) upper) << 32) + (lower & 0xffffffffl)); } else for (int i = 0; i < length; i++) { j = i * 8; upper = (((inData[j + 7] & 0xff) << 24) + ((inData[j + 6] & 0xff) << 16) + ((inData[j + 5] & 0xff) << 8) + ((inData[j + 4] & 0xff) << 0)); lower = (((inData[j + 3] & 0xff) << 24) + ((inData[j + 2] & 0xff) << 16) + ((inData[j + 1] & 0xff) << 8) + ((inData[j] & 0xff) << 0)); outData[i] = Double.longBitsToDouble((((long) upper) << 32) + (lower & 0xffffffffl)); } return outData; }