Integer to byte with given number of bits set

Question

I don't know what to call this, which makes googling harder.

I have an integer, say 3, and want to convert it to 11100000, that is, a byte with the value of the integers number of bits set, from the most significantly bit.

I guess it could be done with:

byte result = 0;
for(int i = 8; i > 8 - 3; i--)
    result += 2 ^ i;

but is there anything faster / more nice or, preferably, standard library included in .net?

Just a small point ^ is an Exclusive Or in C# not the Power operator. — Martin Brown
– Martin Brown, Commented Nov 27, 2011 at 17:51
@Martin Haha thanks, took me a few seconds just now to realize. Easy to forget sometimes, if you also happen to write equations for a paper while programming. — Max
– Max, Commented Nov 27, 2011 at 18:22
@Max, I only know this because I've made the same mistake 100s of times myself. — Martin Brown
– Martin Brown, Commented Dec 1, 2011 at 11:08

Henk Holterman · Accepted Answer · 2011-11-27 17:57:35Z

11

int n = 3; // 0..8 
int mask = 0xFF00;
byte result  = (byte) (mask >> n);

edited Nov 27, 2011 at 17:57

answered Nov 27, 2011 at 17:49

Henk Holterman

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2 Comments

Thomas Levesque Over a year ago

@Cameron, yes it is, since the result of mask >> n is of type int

Henk Holterman Over a year ago

@JimRhodes, I think the problem definition precludes that.

Matthew Strawbridge · Accepted Answer · 2011-11-27 18:11:15Z

5

Because there are only a few possibilities, you could just cache them:

// Each index adds another bit from the left, e.g. resultCache[3] == 11100000.
byte[] resultCache = { 0x00, 0x80, 0xC0, 0xE0, 0xF0, 0XF8, 0xFC, 0xFE, 0xFF };

You'd also get an exception instead of a silent error if you accidentally tried to get the value for n > 8.

answered Nov 27, 2011 at 18:11

Matthew Strawbridge

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2 Comments

Jason Williams Over a year ago

+1 consolation prize for pointing out a simple & effective approach to many of these types of problems (table lookup) that is easily overlooked, and which all programmers should have in their box-of-tricks. (In this case it's not as good a solution as Henk's, though)

Matthew Strawbridge Over a year ago

@JasonWilliams Thanks. I voted up Henk's elegant solution before posting this alternative :-).

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