Thread: Numbers & Binary

If the max value is seven, that means there are 3 bits of interest.
Use a bit mask to test those three bits and set the corresponding slot in the boolean array true or false depending if the bit in that slot is on.

loop number of bits times
if bit on set true
if bit off set false
increment array index
shift bit by 1
end loop

Some examples (1 & 3 == 1) (4 & 3 == 0)

I assume you understand what the AND operator does: 1 & 1 == 1 // 1 & 0 = 0
Use the AND operator with what I call a "bit mask" (an int with one of its bits set to 1) to test if the corresponding bit in the value being tested is on or not.
EG 0x04 would test bit 5 with bits (0123 4567) in a byte
Shift the bit in the bit mask to the next position to test that position.

how it gives me my 3 booleans from the one int...

There are 32 bits in an int, with a loop all 32 bits could be tested (AND and shift) and a boolean value stored in a boolean array with 32 slots.

I think something just clicked in my brain
I was forgetting that the pattern I need is binary... so .. essentially I just create a int to binary (in the form of a boolean array, only 3 bits) converter...
Lightbulb moment.

EDIT:
I am .. close..

public static boolean[] sub(int a) {
		assert (a >= 0 && a <= 7);
		boolean[] arr = new boolean[3];
		int idx = 0;
		for (int i = 0; i <= 7; i++) {
			if ((a & (1 << i)) == 1) {
				arr[idx] = true;
				idx++;
			}
		}
 
		return arr;
	}

Doesn't quite work...

Doesn't quite work..

Please explain.
Use the Arrays class's toString method to show the array's contents:
System.out.println("an ID "+ java.util.Arrays.toString(theArrayName));

How many bits are to be tested?
How many times will the for statement loop?

That's giving me the same results.

Please explain.

This is what it's giving me now:

0 [false, false, false]
1 [true, false, false]
2 [false, false, false]
3 [true, false, false]
4 [false, false, false]
5 [true, false, false]
6 [false, false, false]
7 [true, false, false]

Wait, setting it to != 0 instead of == 1 makes it work? Heh, works for me!

Looks like only the odd numbers get a true for the low bit.

Do some debugging: print out the value of the bit mask that is used to see why none of the other positions are tested.

And print out the results of the AND to see what that is.

Great. Glad you got it working. I might have mislead in my explanation.
Not zero is a good compare for a single bit. But if more than one bit is being tested for, then the compare should be for equal to the "bit mask":
if( (someInt & bitMask) == bitMask)
For example. The number to test is on the left, bitmask = 3:
7 & 3 = 3 // here both bits are on
6 & 3 = 2 // here only one bit is on

how do you know which two bits are on

You know because you have built the "bit mask". The mask: 0x03 tests the 2 rightmost bits.