In C is there a branch-less technique to compute the absolute difference between two unsigned ints? For example given the variables a and b, I would like the value 2 for cases when a=3, b=5 or b=3, a=5. Ideally I would also like to be able to vectorize the computation using the SSE registers.
问题:
回答1:
There are several ways to do it, I'll just mention one:
SSE4
- Use
PMINUD
andPMAXUD
to separate the larger value in register #1, and the smaller value in register #2. - Subtract them.
MMX/SSE2
- Flip the sign bit of the two values because the next instruction only accepts signed integer comparison.
PCMPGTD
. Use this result as a mask.- Compute the results of both
(a-b)
and(b-a)
- Use
POR ( PAND ( mask, a-b ), PANDN ( mask, b-a ) )
to select the correct value for the absolute difference.
回答2:
From tommesani.com, one solution for this problem is to use saturating unsigned subtraction twice.
As the saturating subtraction never goes below 0, you compute: r1 = (a-b).saturating r2 = (b-a).saturating
If a is greater than b, r1 will contain the answer, and r2 will be 0, and vice-versa for b>a. ORing the two partial results together will yield the desired result.
According to the VTUNE users manual, PSUBUSB/PSUBUSW is available for 128-bit registers, so you should be able to get a ton of parallelization this way.
回答3:
max(i,j) - min(i,j)
(i>j)*(i-j) + (j>i)*(j-i)
you can certainly use SSE registers, but compiler may do this for you anyways
回答4:
SSE2:
Seems to be about the same speed as Phernost's second function. Sometimes GCC schedules it to be a full cycle faster, other times a little slower.
__m128i big = _mm_set_epi32( INT_MIN, INT_MIN, INT_MIN, INT_MIN );
a = _mm_add_epi32( a, big ); // re-center the variables: send 0 to INT_MIN,
b = _mm_add_epi32( b, big ); // INT_MAX to -1, etc.
__m128i diff = _mm_sub_epi32( a, b ); // get signed difference
__m128i mask = _mm_cmpgt_epi32( b, a ); // mask: need to negate difference?
mask = _mm_andnot_si128( big, mask ); // mask = 0x7ffff... if negating
diff = _mm_xor_si128( diff, mask ); // 1's complement except MSB
diff = _mm_sub_epi32( diff, mask ); // add 1 and restore MSB
SSSE3:
Ever so slightly faster than previous. There is a lot of variation depending on how things outside the loop are declared. (For example, making a
and b
volatile
makes things faster! It appears to be a random effect on scheduling.) But this is consistently fastest by a cycle or so.
__m128i big = _mm_set_epi32( INT_MIN, INT_MIN, INT_MIN, INT_MIN );
a = _mm_add_epi32( a, big ); // re-center the variables: send 0 to INT_MIN,
b = _mm_add_epi32( b, big ); // INT_MAX to -1, etc.
__m128i diff = _mm_sub_epi32( b, a ); // get reverse signed difference
__m128i mask = _mm_cmpgt_epi32( b, a ); // mask: need to negate difference?
mask = _mm_xor_si128( mask, big ); // mask cannot be 0 for PSIGND insn
diff = _mm_sign_epi32( diff, mask ); // negate diff if needed
SSE4 (thx rwong):
Can't test this.
__m128i diff = _mm_sub_epi32( _mm_max_epu32( a, b ), _mm_min_epu32( a, b ) );
回答5:
Try this (assumes 2nd complements, which is OK judgning by the fact that you're asking for SSE):
int d = a-b;
int ad = ((d >> 30) | 1) * d;
Explanation: sign-bit (bit 31) gets propagated down to 1st bit. the | 1 part ensures that the multiplier is either 1 or -1. Multiplications are fast on modern CPUs.
回答6:
compute the difference and return the absolute value
__m128i diff = _mm_sub_epi32(a, b);
__m128i mask = _mm_xor_si128(diff, a);
mask = _mm_xor_si128(mask, b);
mask = _mm_srai_epi32(mask, 31);
diff = _mm_xor_si128(diff, mask);
mask = _mm_srli_epi32(mask, 31);
diff = _mm_add_epi32(diff, mask);
This requires one less operation that using the signed compare op, and produces less register pressure.
Same amount of register pressure as before, 2 more ops, better branch and merging of dependency chains, instruction pairing for uops decoding, and separate unit utilization. Although this requires a load, which may be out of cache. I'm out of ideas after this one.
__m128i mask, diff;
diff = _mm_set1_epi32(-1<<31); // dependency branch after
a = _mm_add_epi32(a, diff); // arithmetic sign flip
b = _mm_xor_si128(b, diff); // bitwise sign flip parallel with 'add' unit
diff = _mm_xor_si128(a, b); // reduce uops, instruction already decoded
mask = _mm_cmpgt_epi32(b, a); // parallel with xor
mask = _mm_and_si128(mask, diff); // dependency merge, branch after
a = _mm_xor_si128(a, mask); // if 2 'bit' units in CPU, parallel with next
b = _mm_xor_si128(b, mask); // reduce uops, instruction already decoded
diff = _mm_sub_epi32(a, b); // result
After timing each version with 2 million iterations on a Core2Duo, differences are immeasurable. So pick whatever is easier to understand.
回答7:
One or more of the below will likely result in branchless code, depending on the machine and compiler, since the conditional expressions are all very simple.
I haven't been through all the sse answers but possibly some of the below are represented in the vector code; certainly all the below are vectorizable (if you have the unsigned compare to begin with, or fake it by toggling the msb first.). I thought it would be helpful to have some practical scalar answers to the question.
unsigned udiff( unsigned a, unsigned b )
{
unsigned result = a-b; // ok if a<b;
if(a <b ) result = -result;
return result;
}
unsigned udiff( unsigned a, unsigned b )
{
unsigned n =(a<b)? (unsigned)-1 : 0u;
unsigned result = a-b;
return (result^n)-n; // 'result' if n = 0; '-result' if n = 0xFFFFFFFF
}
unsigned udiff( unsigned a, unsigned b )
{
unsigned axb = a^b;
if( a < b ) axb = 0;
return (axb^b) - (axb^a); // a-b, or b-a
}
This will work on x86_64 (or anything where 64-bit temps are basically free)
unsigned udiff( unsigned a, unsigned b )
{
unsigned n= (unsigned)(
(long long)((unsigned long long)a - (unsigned long long)b)>>32
); // same n as 2nd example
unsigned result = a-b;
return (result^n)-n; // 'result' if n = 0; '-result' if n = 0xFFFFFFFF
}
回答8:
Erm ... its pretty easy ...
int diff = abs( a - b );
Easily vectorisable (Using SSE3 as):
__m128i sseDiff = _mm_abs_epi32( _mm_sub_epi32( a, b ) );
a and b are unsigned integers. Consider a=0 and b=0xffffffff. The correct absolute difference is 0xffffffff, but your solution will give 1.