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    <p><a href="http://docs.oracle.com/javase/7/docs/api/java/nio/ByteBuffer.html" rel="nofollow noreferrer"><code>ByteBuffer</code></a> works well for this: just put in 64 <code>long</code> values and get a <code>byte[]</code> out using the <code>array()</code> method. The <a href="http://docs.oracle.com/javase/7/docs/api/java/nio/ByteOrder.html" rel="nofollow noreferrer"><code>ByteOrder</code></a> class can handle endian issues effectively. For example, incorporating the approach suggested in a comment by <a href="https://stackoverflow.com/users/136264/wierob">wierob</a>:</p> <pre><code>private static byte[] xform(long[] la, ByteOrder order) { ByteBuffer bb = ByteBuffer.allocate(la.length * 8); bb.order(order); bb.asLongBuffer().put(la); return bb.array(); } </code></pre> <p>Addendum: The resulting <code>byte[]</code> components are signed, 8-bit values, but Java arrays require <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-10.html" rel="nofollow noreferrer">nonnegative integer index values</a>. Casting a <code>byte</code> to an <code>int</code> will result in sign extension, but masking the higher order bits will give the unsigned value of <code>byte b</code>:</p> <pre><code>int i = (int) b &amp; 0xFF; </code></pre> <p>This <a href="https://stackoverflow.com/questions/2137690/java-operator-precedence-guidelines">answer</a> elaborates on the applicable operator precedence rules. This related <a href="https://stackoverflow.com/questions/5201422">answer</a> demonstrates a similar approach for <code>double</code> values.</p>
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    1. POConverting long[64] to byte[512] in Java?
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    1. POConverting long[64] to byte[512] in Java?
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    1. CO+1 You don't even need the loop: bb.asLongBuffer().put(la);
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    2. COThank you! Your suggestion is easier to read and potentially more efficient. http://java.sun.com/javase/6/docs/api/java/nio/ByteBuffer.html#views
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