I need to convert a byte into an array of 4 booleans in Java. How might I go about this?

解決方案

Per Michael Petrotta's comment to your question, you need to decide which bits in the 8-bit byte should be tested for the resulting boolean array. For demonstration purposes, let's assume you want the four rightmost bits, then something like this should work:

public static boolean[] booleanArrayFromByte(byte x) {

boolean bs[] = new boolean[4];

bs[0] = ((x & 0x01) != 0);

bs[1] = ((x & 0x02) != 0);

bs[2] = ((x & 0x04) != 0);

bs[3] = ((x & 0x08) != 0);

return bs;

}

The hexadecimal values (0x01, 0x02, etc.) in this example are special bit masks that have only a single bit set at the desired location; so 0x01 has only the rightmost bit set, 0x08 has only the fourth-from-right bit set. By testing the given byte against these values with the bitwise AND operator (&) you will get that value back if the bit is set, or zero if not. If you want to check different bits, other than the rightmost four, then you'll have to create different bitmasks.

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