How to compare two packed bitfields without having to unpack each field

This post has been republished via RSS; it originally appeared at: MSDN Blogs.

Suppose you are packing multiple bitfields into a single
integer.
Let's say you have a 16-bit integer that you have
packed three bitfields into:

Suppose you have two of these packed bitfields,
x and y,

and you want to know whether every field in x
is greater than or equal the corresponding field in y.
I.e., you want to determine whether
xr ≥ yr,
xg ≥ yg, and
xb ≥ yb.

One way would be to unpack the bitfields.

bool IsEveryComponentGreaterThanOrEqual(uint16_t x, uint16_t y)

{

 auto xr = x >> 11;

 auto yr = y >> 11;

 if (xr < yr) return false;

 auto xg = (x >> 5) & 0x3F;

 auto yg = (y >> 5) & 0x3F;

 if (xg < yg) return false;
 auto xb = x & 0x1F;

 auto yb = y & 0x1F;

 if (xb < yb) return false;
 return true;

}

There's an obvious optimization here,
which is to avoid the extra shifting.

bool IsEveryComponentGreaterThanOrEqual(uint16_t x, uint16_t y)

{

 auto xr = x & 0xF100;

 auto yr = y & 0xF100;

 if (xr < yr) return false;

 auto xg = x & 0x07E0;

 auto yg = y & 0x07E0;

 if (xg < yg) return false;
 auto xb = x & 0x001F;

 auto yb = y & 0x001F;

 if (xb < yb) return false;
 return true;

}

But maybe you can do even better.

Well, if you had planned ahead and inserted a zero padding bit
at the front of each field:

then you could subtract the two values and see if any padding bit
became set,
which indicates that an underflow occurred
somewhere to the right.

bool IsEveryComponentGreaterThanOrEqual(uint32_t x, uint32_t y)

{

 auto m = (x - y) & ((1 << 18) | (1 << 12) | (1 << 5));

 return m == 0;

}

However, this forces you to reserve padding bits,
and it seems silly to have padding bits all over your data
just for this purpose.
I mean, those are bits that could've been doing something useful!

In our example, those three extra bits forced us to use a larger
integral type,
which means our memory usage doubled.

Can you do it without inserting padding bits?

Indeed you can,
thanks to a trick from

emulator master
Darek Mihocka:
The carry-out vector.

You can read

the paper
or take the easier route and

read the presentation.

In this case, we want the subtraction carry-out vector
(which is really the borrow vector).
The formula is

right here in the Bochs emulator source code.

#define SUB_COUT_VEC(op1, op2, result)

  (((~(op1)) & (op2)) | ((~((op1) ^ (op2))) & (result)))

In the subtraction carry-out vector,
a bit is set if the subtraction resulted in a borrow
at that position.
We then check whether there was a borrow at the corresponding
high bits 4, 10, or 15.

Here we go:

bool IsEveryComponentGreaterThanOrEqual(uint16_t x, uint16_t y)

{

 auto c = ((~x & y) | (~(x ^ y) & (x - y));

 c &= 0x8410;

 return c == 0;

}

Slide 13 of the presentation linked above
shows how this technique can be used to implement
saturating bitfield arithmetic in general-purpose registers.
Who needs SIMD registers!

The carry-out vector is truly magical.

Bonus reading:

How Bochs Works Under the Hood.
The "Lazy flags handling" section has a useful diagram.