Convert Hex to Decimal Value

Question

If I have a hex value of 53, what is the best way in Java to see what the values of bit 6 and bit 7 are? Bit 1 will be considered as the Least Significant Bit (or the rightmost bit within a byte).

The goal is to take bit6 and bit7 and convert them to its combined decimal form.

You can mask each bit with the corresponding power of 2 number. For example, to know the value of the fourth bit you would use bitwise and with 8. x & 8 => x & (2^3) — Saleh Omar
– Saleh Omar, Commented Jan 8, 2013 at 21:21
The usual bit addressing is that the least significant bit is bit 0, not bit 1. — Ted Hopp
– Ted Hopp, Commented Jan 8, 2013 at 21:24
While not a direct answer to your question, you may also be interested in Integer.toBinaryString(int) — Greg Kopff
– Greg Kopff, Commented Jan 8, 2013 at 21:30

Ted Hopp · Accepted Answer · 2013-01-08 21:21:34Z

2

The easiest way, I think is:

int bits7And8 = (byteValue >>> 5) & 3;

That will give you bits 6 and 7 in the lowest two bit positions. If you need their value as a String:

String bits7And8String = Integer.toString(bits7And8);

edited Jan 8, 2013 at 21:21

answered Jan 8, 2013 at 21:19

Ted Hopp

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3 Comments

Ted Hopp Over a year ago

@c12 - Sorry - I was off by one because of your unconventional indexing of the least significant bit being 1 instead of 0.

c12 Over a year ago

byte byteValue = 53; is that what you intended for your example? assuming a hex of 53.

Ted Hopp Over a year ago

@c12 - If the hex value is 53 (binary 01010011), then bits 7 and 8 (6 and 7 in your indexing scheme) are binary 10, which is 2 in decimal. That's what I understood you to want. (The "7And8" in the variable names should be "6And7" to be consistent with your indexing scheme.)

Martijn Courteaux · Accepted Answer · 2013-01-08 21:21:54Z

That is easy:

byte b = 53; boolean bit6 = ((b >> 5) & 1) == 1; boolean bit7 = ((b >> 6) & 1) == 1;

Or:

byte b = 53; int bit6and7 = b & (0x3 << 5); // Will give: 0 bit7 bit6 0 0 0 0 0

Or:

byte b = 53; int bit6and7 = (b >> 5) & 0x3; // Will give: 0 0 0 0 0 0 bit7 bit6

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