Optimize bitflag check

Question

How can I optimize the following code?

   ( (kbd_flags & KBD_FLAG_SHIFT) && !(kbd_flags & KBD_FLAG_CAPS))
|| (!(kbd_flags & KBD_FLAG_SHIFT) &&  (kbd_flags & KBD_FLAG_CAPS))

Basically, I want to check if either KBD_FLAG_SHIFT or KBD_FLAG_CAPS is set, but not both.

Answer 1

I would like to share my solution after doing some research. Zodoh's expression can be simplified to

kbd_flags & KBD_FLAG_SHIFT
    ? !(kbd_flags & KBD_FLAG_CAPS)
    :   kbd_flags & KBD_FLAG_CAPS

(Omitting unneeded parentheses and the true and false expressions)

Another interesting way I figured out is the following:

x = kbd_flags & (KBD_FLAG_SHIFT | KBD_FLAG_CAPS);
return x && (x & x - 1) == 0;

This works because of the way two's complement notation is designed. As an example, if kbd_flags is set to 001000, x - 1 will be 000111 and 001000 & 000111 is 000000. As a result, 000000 is equal to 0, thus returning true. The first x expression makes sure that the "no bit set" case is excluded.

It will also work with more than only two bit flags:

#define A 0x01
#define B 0x02
#define C 0x04
#define D 0x08
#define E 0x10
#define F 0x20

x = flags & (A | B | C | D | E | F);
return x && (x & x - 1) == 0;

Here, the expression x && (x & x - 1) == 0 will be true if and only if one of the flags A through F is set.

A quick test (f being the integer holding the flags to test for):

int f, x;
for (f = 0; f <= F + 1; f++) {
    x = f & (A | B | C | D | E | F);
    printf("0x%02x %d%d%d%d%d%d -> %d\n", f
        , (f & A) == A, (f & B) == B, (f & C) == C
        , (f & D) == D, (f & E) == E, (f & F) == F
        , x && (x & x - 1) == 0);
}

This code will output the follwing:

0x00 000000 -> 0
0x01 100000 -> 1
0x02 010000 -> 1
0x03 110000 -> 0
0x04 001000 -> 1
0x05 101000 -> 0
0x06 011000 -> 0
0x07 111000 -> 0
0x08 000100 -> 1
0x09 100100 -> 0
0x0a 010100 -> 0
0x0b 110100 -> 0
0x0c 001100 -> 0
0x0d 101100 -> 0
0x0e 011100 -> 0
0x0f 111100 -> 0
0x10 000010 -> 1
0x11 100010 -> 0
0x12 010010 -> 0
0x13 110010 -> 0
0x14 001010 -> 0
0x15 101010 -> 0
0x16 011010 -> 0
0x17 111010 -> 0
0x18 000110 -> 0
0x19 100110 -> 0
0x1a 010110 -> 0
0x1b 110110 -> 0
0x1c 001110 -> 0
0x1d 101110 -> 0
0x1e 011110 -> 0
0x1f 111110 -> 0
0x20 000001 -> 1
0x21 100001 -> 0

As you can see, x && (x & x - 1) == 0 is true iff one bit is set.