Assigning a given bit in a byte

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5.00/5 (1 vote)

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Setting or clearing a given bit i in a byte B is straightforward (and often documented), using shift and bitwise operations:

C++

B|= 1 << i;  // 1 shift, 1 or
B&= ~(1 << i); // 1 shift, 1 complement, 1 and

But copying the value of a one bit variable b (instead of the constant 0 or 1) is not so often mentioned. Here is a solution I can think of:

C++

B= (B & ~(1 << i)) | (b << i); // 2 shifts, 1 complement, 1 and, 1 or

Is it possible to do that in less operations (branches not allowed, of course) ?

Posted 5-Aug-13 23:08pm

YvesDaoust

Updated 5-Aug-13 23:10pm

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Jochen Arndt · Accepted Answer · 2013-08-06T00:00:00

Solution 1

A good source for such problems is Bit Twiddling Hacks[^]. See 'Conditionally set or clear bits without branching' for your task. Adding an instruction to generate the mask, this can be used:

C++

unsigned m = 1U << i;
B ^= (-b ^ B) & m;
// or
B = (B & ~m) | (-b & m);

Posted 6-Aug-13 0:00am

Jochen Arndt

Comments

YvesDaoust 6-Aug-13 6:22am

Nice try. The expressions are indeed correct. I'll remember the -b trick!

Unfortunately in my context I cannot precompute the mask m (and even less its complement), so that the operation count is not reduced.

Jochen Arndt 6-Aug-13 6:35am

Looking again over my solution I recognized that you can omit the mask generation line when using the first solution:
B ^= (-b ^ B) & (1U << i);
Still 5 operations. But I think that it can't be solved with less operations.

YvesDaoust 6-Aug-13 6:52am

Yep, seems difficult to lower to 4 operations. Given the small domain, it is thinkable to fully tabulate the function (2^8 * 8 * 2 entries = 4 kB), but even the table lookup has a cost...

nv3 6-Aug-13 8:26am

My 5 for the interesting link, Jochen. But it seems that there is no better way than the one OP has proposed in his question. I would go for the straight forward implementation suggested by OP, even if the -b variant might save a cycle or two. It is much easier to read and hence maintain.