Most elegant way of turning bits on and off

Question

I've got a register which can either be 0x02 or 0x01 (0010 or 0001). What is the most elegant way of setting the bits in a single operation? E.g. if the register is currently 0x02, I need to turn off bit 2 and turn on bit 1.

Many thanks

Simply assigning the value to the register isn't enough? xors and shifts may look cool and fast, but they make your code harder to understand. — Lucas Martins, Jun 26 '19 at 02:40
It's embedded and simply assigning a value doesn't seem to modify the register, only bitwise operators. — John Smith, Jun 26 '19 at 02:42
Consider posting an example code, to avoid misinterpretations. — Lucas Martins, Jun 26 '19 at 02:49
John, that appears to make no sense - if assigning (store) didn't work, neither would bitwise operations (load, modify, ***store***). — paxdiablo, Jun 26 '19 at 02:51

score 1 · Answer 1 · answered Jun 26 '19 at 02:35

1

x ^= 3 changes 1 to 2 and changes 2 to 1. ^ performs an exclusive or (XOR) operation.

answered Jun 26 '19 at 02:35

Eric Postpischil

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score 0 · Answer 2 · answered Jun 26 '19 at 04:58

The most elegant way is to define a structure that uses bit fields. Sadly this is rarely appropriate when dealing with hardware registers (e.g. order of bits in a bit-field isn't well defined, etc).

Without bitfields; use "and with compliment" (e.g. x = x & ~ 2; or x &= ~2;) to clear bits and "or" (e.g. x = x | 1; or x |= 1;) to set bits.

If you know the previous value you can optimize further. For example, if you know the previous value is 0x02 then you can just set a new value (e.g. x = 0x01; // It was 0x02 or maybe x = (0x02 & ~2) | 1; so compiler will work out the constant for you but the source code still reflects how the constant was determined).

Note that it's not possible to do it in a single operation unless you do know something about the previous value. For example, if you know that the previous value will have bit 0 clear and bit 1 set but don't know anything about the other bits (and need to preserve the other bits), then you can decrement (x--; // Clear bit 2, set bit 1) or use exclusive OR.

Most elegant way of turning bits on and off

2 Answers2