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SIMD instructions for floating point equality comparison (with NaN == NaN)

Which instructions would be used for comparing two 128 bit vectors consisting of 4 * 32-bit floating point values?

Is there an instruction that considers a NaN value on both sides as equal? If not, how big would the performance impact of a workaround that provides reflexivity (i.e. NaN equals NaN) be?

I heard that ensuring reflexivity would have a significant performance impact compared with IEEE semantics, where NaN doesn't equal itself, and I'm wondering if big that impact would be.

I know that you typically want use epsilon comparisons instead of exact quality when dealing floating-point values. But this question is about exact equality comparisons, which you could for example use to eliminate duplicate values from a hash-set.

Requirements

  • +0 and -0 must compare as equal.
  • NaN must compare equal with itself.
  • Different representations of NaN should be equal, but that requirement might be sacrificed if the performance impact is too big.
  • The result should be a boolean, true if all four float elements are the same in both vectors and false if at least one element differs. Where true is represented by a scalar integer 1 and false by 0.

Test cases

(NaN, 0, 0, 0) == (NaN, 0, 0, 0) // for all representations of NaN
(-0,  0, 0, 0) == (+0,  0, 0, 0) // equal despite different bitwise representations
(1,   0, 0, 0) == (1,   0, 0, 0)
(0,   0, 0, 0) != (1,   0, 0, 0) // at least one different element => not equal 
(1,   0, 0, 0) != (0,   0, 0, 0)

My idea for implementing this

I think it might be possible to combine two NotLessThan comparisons (CMPNLTPS ?) using and to achieve the desired result. The assembler equivalent of AllTrue(!(x < y) and !(y < x)) or AllFalse((x < y) or (y > x).

Background

The background for this question is Microsoft's plan to add a Vector type to .NET. Where I'm arguing for a reflexive .Equals method and need a clearer picture of how big the performance impact of this reflexive equals over a IEEE equals would be. See Should Vector<float>.Equals be reflexive or should it follow IEEE 754 semantics? on programmers.se for the long story.

like image 871
CodesInChaos Avatar asked Jan 22 '16 16:01

CodesInChaos


2 Answers

Even AVX VCMPPS (with it's greatly enhanced choice of predicates) doesn't give us a single-instruction predicate for this. You have to do at least two compares and combine the results. It's not too bad, though.

  • different NaN encodings aren't equal: effectively 2 extra insns (adding 2 uops). Without AVX: One extra movaps beyond that.

  • different NaN encodings are equal: effectively 4 extra insns (adding 4 uops). Without AVX: Two extra movaps insn

An IEEE compare-and-branch is 3 uops: cmpeqps / movmskps / test-and-branch. Intel and AMD both macro-fuse the test-and-branch into a single uop/m-op.

With AVX512: bitwise-NaN is probably just one extra instruction, since normal vector compare and branch probably uses vcmpEQ_OQps / ktest same,same / jcc, so combining two different mask regs is free (just change the args to ktest). The only cost is the extra vpcmpeqd k2, xmm0,xmm1.

AVX512 any-NaN is just two extra instructions (2x VFPCLASSPS, with the 2nd one using the result of the first as a zeromask. See below). Again, then ktest with two different args to set flag.


My best idea so far: ieee_equal || bitwise_equal

If we give up on considering different NaN encodings equal to each other:

  • Bitwise equal catches two identical NaNs.
  • IEEE equal catches the +0 == -0 case.

There are no cases where either compare gives a false positive (since ieee_equal is false when either operand is NaN: we want just equal, not equal-or-unordered. AVX vcmpps provides both options, while SSE only provides a plain equal operation.)

We want to know when all elements are equal, so we should start with inverted comparisons. It's easier to check for at least one non-zero element than to check for all elements being non-zero. (i.e. horizontal AND is hard, horizontal OR is easy (pmovmskb / test, or ptest). Taking the opposite sense of a comparison is free (jnz instead of jz).) This is the same trick that Paul R used.

; inputs in xmm0, xmm1
movaps    xmm2, xmm0    ; unneeded with 3-operand AVX instructions

cmpneqps  xmm2, xmm1    ; 0:A and B are ordered and equal.  -1:not ieee_equal.  predicate=NEQ_UQ in VEX encoding expanded notation
pcmpeqd   xmm0, xmm1    ; -1:bitwise equal  0:otherwise

; xmm0   xmm2
;   0      0   -> equal   (ieee_equal only)
;   0     -1   -> unequal (neither)
;  -1      0   -> equal   (bitwise equal and ieee_equal)
;  -1     -1   -> equal   (bitwise equal only: only happens when both are NaN)

andnps    xmm0, xmm2    ; NOT(xmm0) AND xmm2
; xmm0 elements are -1 where  (not bitwise equal) AND (not IEEE equal).
; xmm0 all-zero iff every element was bitwise or IEEE equal, or both
movmskps  eax, xmm0
test      eax, eax      ; it's too bad movmsk doesn't set EFLAGS according to the result
jz no_differences

For double-precision, ...PS and pcmpeqQ will work the same.

If the not-equal code goes on to find out which element isn't equal, a bit-scan on the movmskps result will give you the position of the first difference.

With SSE4.1 PTEST you can replace andnps/movmskps/test-and-branch with:

ptest    xmm0, xmm2   ; CF =  0 == (NOT(xmm0) AND xmm2).
jc no_differences

I expect this is the first time most people have ever seen the CF result of PTEST be useful for anything. :)

It's still three uops on Intel and AMD CPUs ( (2ptest + 1jcc) vs (pandn + movmsk + fused-test&branch)), but fewer instructions. It is more efficient if you're going to setcc or cmovcc instead of jcc, since those can't macro-fuse with test.

That makes a total of 6 uops (5 with AVX) for a reflexive compare-and-branch, vs. 3 uops for an IEEE compare-and-branch. (cmpeqps / movmskps / test-and-branch.)

PTEST has a very high latency on AMD Bulldozer-family CPUs (14c on Steamroller). They have one cluster of vector execution units shared by two integer cores. (This is their alternative to hyperthreading.) This increases the time until a branch mispredict can be detected, or the latency of a data-dependency chain (cmovcc / setcc).

PTEST sets ZF when 0==(xmm0 AND xmm2): set if no elements were both bitwise_equal AND IEEE (neq or unordered). i.e. ZF is unset if any element was bitwise_equal while also being !ieee_equal. This can only happen when a pair of elements contain bitwise-equal NaNs (but can happen when other elements are unequal).

    movaps    xmm2, xmm0
    cmpneqps  xmm2, xmm1    ; 0:A and B are ordered and equal.
    pcmpeqd   xmm0, xmm1    ; -1:bitwise equal

    ptest    xmm0, xmm2
    jc   equal_reflexive   ; other cases

...

equal_reflexive:
    setnz  dl               ; set if at least one both-nan element

There's no condition that tests CF=1 AND anything about ZF. ja tests CF=0 and ZF=1. It's unlikely that you'd only want to test that anyway, so putting a jnz in the jc branch target works fine. (And if you did only want to test equal_reflexive AND at_least_one_nan, a different setup could probably set flags appropriately).


Considering all NaNs equal, even when not bitwise equal:

This is the same idea as Paul R's answer, but with a bugfix (combine NaN check with IEEE check using AND rather than OR.)

; inputs in xmm0, xmm1
movaps      xmm2, xmm0
cmpordps    xmm2, xmm2      ; find NaNs in A.  (0: NaN.  -1: anything else).  Same as cmpeqps since src and dest are the same.
movaps      xmm3, xmm1
cmpordps    xmm3, xmm3      ; find NaNs in B
orps        xmm2, xmm3      ; 0:A and B are both NaN.  -1:anything else

cmpneqps    xmm0, xmm1      ; 0:IEEE equal (and ordered).  -1:unequal or unordered
; xmm0 AND xmm2  is zero where elements are IEEE equal, or both NaN
; xmm0   xmm2 
;   0      0     -> equal   (ieee_equal and both NaN (impossible))
;   0     -1     -> equal   (ieee_equal)
;  -1      0     -> equal   (both NaN)
;  -1     -1     -> unequal (neither equality condition)

ptest    xmm0, xmm2        ; ZF=  0 == (xmm0 AND xmm2).  Set if no differences in any element
jz   equal_reflexive
; else at least one element was unequal

;     alternative to PTEST:  andps  xmm0, xmm2 / movmskps / test / jz

So in this case we don't need PTEST's CF result after all. We do when using PCMPEQD, because it doesn't have an inverse (the way cmpunordps has cmpordps).

9 fused-domain uops for Intel SnB-family CPUs. (7 with AVX: use non-destructive 3-operand instructions to avoid the movaps.) However, pre-Skylake SnB-family CPUs can only run cmpps on p1, so this bottlenecks on the FP-add unit if throughput is a concern. Skylake runs cmpps on p0/p1.

andps has a shorter encoding than pand, and Intel CPUs from Nehalem to Broadwell can only run it on port5. That may be desirable to prevent it from stealing a p0 or p1 cycle from surrounding FP code. Otherwise pandn is probably a better choice. On AMD BD-family, andnps runs in the ivec domain anyway, so you don't avoid the bypass delay between int and FP vectors (which you might otherwise expect to manage if you use movmskps instead of ptest, in this version that only uses cmpps, not pcmpeqd). Also note that instruction ordering is chosen for human readability here. Putting the FP compare(A,B) earlier, before the ANDPS, might help the CPU get started on that a cycle sooner.

If one operand is reused, it should be possible to reuse its self-NaN-finding result. The new operand still needs its self-NaN check, and a compare against the reused operand, so we only save one movaps/cmpps.

If the vectors are in memory, at least one of them needs to be loaded with a separate load insn. The other one can just be referenced twice from memory. This sucks if it's unaligned or the addressing mode can't micro-fuse, but could be useful. If one of the operands to vcmpps is a vector known to not have any NaNs (e.g. a zeroed register), vcmpunord_qps xmm2, xmm15, [rsi] will find NaNs in [rsi].

If we don't want to use PTEST, we can get the same result by using the opposite comparisons, but combining them with the opposite logical operator (AND vs. OR).

; inputs in xmm0, xmm1
movaps      xmm2, xmm0
cmpunordps  xmm2, xmm2      ; find NaNs in A (-1:NaN  0:anything else)
movaps      xmm3, xmm1
cmpunordps  xmm3, xmm3      ; find NaNs in B
andps       xmm2, xmm3      ; xmm2 = (-1:both NaN  0:anything else)
; now in the same boat as before: xmm2 is set for elements we want to consider equal, even though they're not IEEE equal

cmpeqps     xmm0, xmm1      ; -1:ieee_equal  0:unordered or unequal
; xmm0   xmm2 
;  -1      0     -> equal   (ieee_equal)
;  -1     -1     -> equal   (ieee_equal and both NaN (impossible))
;   0      0     -> unequal (neither)
;   0     -1     -> equal   (both NaN)

orps        xmm0, xmm2      ; 0: unequal.  -1:reflexive_equal
movmskps    eax, xmm0
test        eax, eax
jnz  equal_reflexive

Other ideas: unfinished, non-viable, broken, or worse-than-the-above

The all-ones result of a true comparison is an encoding of NaN. (Try it out. Perhaps we can avoid using POR or PAND to combine results from cmpps on each operand separately?

; inputs in A:xmm0 B:xmm1
movaps      xmm2, xmm0
cmpordps    xmm2, xmm2      ; find NaNs in A.  (0: NaN.  -1: anything else).  Same as cmpeqps since src and dest are the same.
; cmpunordps wouldn't be useful: NaN stays NaN, while other values are zeroed.  (This could be useful if ORPS didn't exist)

; integer -1 (all-ones) is a NaN encoding, but all-zeros is 0.0
cmpunordps  xmm2, xmm1
; A:NaN B:0   ->  0   unord 0   -> false
; A:0   B:NaN ->  NaN unord NaN -> true

; A:0   B:0   ->  NaN unord 0   -> true
; A:NaN B:NaN ->  0   unord NaN -> true

; Desired:   0 where A and B are both NaN.

cmpordps xmm2, xmm1 just flips the final result for each case, with the "odd-man-out" still on the 1st row.

We can only get the result we want (true iff A and B are both NaN) if both inputs are inverted (NaN -> non-NaN and vice versa). This means we could use this idea for cmpordps as a replacement for pand after doing cmpordps self,self on both A and B. This isn't useful: even if we have AVX but not AVX2, we can use vandps and vandnps (and vmovmskps since vptest is AVX2 only). Bitwise booleans are only single-cycle latency, and don't tie up the vector-FP-add execution port(s) which is already a bottleneck for this code.


VFIXUPIMMPS

I spent a while with the manual grokking its operation.

It can modify a destination element if a source element is NaN, but that can't be conditional on anything about the dest element.

I was hoping I could think of a way to vcmpneqps and then fixup that result, once with each source operand (to elide the boolean instructions that combine the results of 3 vcmpps instructions). I'm now fairly sure that's impossible, because knowing that one operand is NaN isn't enough by itself make a change to the IEEE_equal(A,B) result.

I think the only way we could use vfixupimmps is for detecting NaNs in each source operand separately, like vcmpunord_qps but worse. Or as a really stupid replacement for andps, detecting either 0 or all-ones(NaN) in the mask results of previous compares.


AVX512 mask registers

Using AVX512 mask registers could help combine the results of compares. Most AVX512 compare instructions put the result into a mask register instead of a mask vector in a vector reg, so we actually have to do things this way if we want to operate in 512b chunks.

VFPCLASSPS k2 {k1}, xmm2, imm8 writes to a mask register, optionally masked by a different mask register. By setting only the QNaN and SNaN bits of the imm8, we can get a mask of where there are NaNs in a vector. By setting all the other bits, we can get the inverse.

By using the mask from A as a zero-mask for the vfpclassps on B, we can find the both-NaN positions with only 2 instructions, instead of the usual cmp/cmp/combine. So we save an or or andn instruction. Incidentally, I wonder why there's no OR-NOT operation. Probably it comes up even less often than AND-NOT, or they just didn't want porn in the instruction set.

Neither yasm nor nasm can assemble this, so I'm not even sure if I have the syntax correct!

; I think this works

;  0x81 = CLASS_QNAN|CLASS_SNAN (first and last bits of the imm8)
VFPCLASSPS    k1,     zmm0, 0x81 ; k1 = 1:NaN in A.   0:non-NaN
VFPCLASSPS    k2{k1}, zmm1, 0x81 ; k2 = 1:NaNs in BOTH
;; where A doesn't have a NaN, k2 will be zero because of the zeromask
;; where B doesn't have a NaN, k2 will be zero because that's the FPCLASS result
;; so k2 is like the bitwise-equal result from pcmpeqd: it's an override for ieee_equal

vcmpNEQ_UQps  k3, zmm0, zmm1
;; k3= 0 only where IEEE equal (because of cmpneqps normal operation)

;  k2   k3   ; same logic table as the pcmpeqd bitwise-NaN version
;  0    0    ->  equal   (ieee equal)
;  0    1    ->  unequal (neither)
;  1    0    ->  equal   (ieee equal and both-NaN (impossible))
;  1    1    ->  equal   (both NaN)

;  not(k2) AND k3 is true only when the element is unequal (bitwise and ieee)

KTESTW        k2, k3    ; same as PTEST: set CF from 0 == (NOT(k2) AND k2)
jc .reflexive_equal

We could reuse the same mask register as both zeromask and destination for the 2nd vfpclassps insn, but I used different registers in case I wanted to distinguish between them in a comment. This code needs a minimum of two mask registers, but no extra vector registers. We could also use k0 instead of k3 as the destination for vcmpps, since we don't need to use it as a predicate, only as a dest and src. (k0 is the register that can't be used as a predicate, because that encoding means instead means "no masking".)

I'm not sure we could create a single mask with the reflexive_equal result for each element, without a k... instruction to combine two masks at some point (e.g. kandnw instead of ktestw). Masks only work as zero-masks, not one-masks that can force a result to one, so combining the vfpclassps results only works as an AND. So I think we're stuck with 1-means-both-NaN, which is the wrong sense for using it as a zeromask with vcmpps. Doing vcmpps first, and then using the mask register as destination and predicate for vfpclassps, doesn't help either. Merge-masking instead of zero-masking would do the trick, but isn't available when writing to a mask register.

;;; Demonstrate that it's hard (probably impossible) to avoid using any k... instructions
vcmpneq_uqps  k1,    zmm0, zmm1   ; 0:ieee equal   1:unequal or unordered

vfpclassps    k2{k1}, zmm0, 0x81   ; 0:ieee equal or A is NaN.  1:unequal
vfpclassps    k2{k2}, zmm1, 0x81   ; 0:ieee equal | A is NaN | B is NaN.  1:unequal
;; This is just a slow way to do vcmpneq_Oqps: ordered and unequal.

vfpclassps    k3{k1}, zmm0, ~0x81  ; 0:ieee equal or A is not NaN.  1:unequal and A is NaN
vfpclassps    k3{k3}, zmm1, ~0x81  ; 0:ieee equal | A is not NaN | B is not NaN.  1:unequal & A is NaN & B is NaN
;; nope, mixes the conditions the wrong way.
;; The bits that remain set don't have any information from vcmpneqps left: both-NaN is always ieee-unequal.

If ktest ends up being 2 uops like ptest, and can't macro-fuse, then kmov eax, k2 / test-and-branch will probably be cheaper than ktest k1,k2 / jcc. Hopefully it will only be one uop, since mask registers are more like integer registers, and can be designed from the start to be interally "close" to the flags. ptest was only added in SSE4.1, after many generations of designs with no interaction between vectors and EFLAGS.

kmov does set you up for popcnt, bsf or bsr, though. (bsf/jcc doesn't macro-fuse, so in a search loop you're probably still going to want to test/jcc and only bsf when a non-zero is found. The extra byte to encode tzcnt doesn't buy you anything unless you're doing something branchless, because bsf still sets ZF on a zero input, even though the dest register is undefined. lzcnt gives 32 - bsr, though, so it can be useful even when you know the input is non-zero.)

We can also use vcmpEQps and combine our results differently:

VFPCLASSPS      k1,     zmm0, 0x81 ; k1 = set where there are NaNs in A
VFPCLASSPS      k2{k1}, zmm1, 0x81 ; k2 = set where there are NaNs in BOTH
;; where A doesn't have a NaN, k2 will be zero because of the zeromask
;; where B doesn't have a NaN, k2 will be zero because that's the FPCLASS result
vcmpEQ_OQps     k3, zmm0, zmm1
;; k3= 1 only where IEEE equal and ordered (cmpeqps normal operation)

;  k3   k2
;  1    0    ->  equal   (ieee equal)
;  1    1    ->  equal   (ieee equal and both-NaN (impossible))
;  0    0    ->  unequal (neither)
;  0    1    ->  equal   (both NaN)

KORTESTW        k3, k2  ; CF = set iff k3|k2 is all-ones.
jc .reflexive_equal

This way only works when there's a size of kortest that exactly matches the number of elements in our vectors. e.g. a 256b vector of double-precision elements only has 4 elements, but kortestb still sets CF according to the low 8 bits of the input mask registers.


Using only integer ops

Other than NaN, +/-0 is the only time when IEEE_equal is different from bitwise_equal. (Unless I'm missing something. Double-check this assumption before using!) +0 and -0 have all their bits zero, except that -0 has the sign bit set (the MSB).

If we ignore different NaN encodings, then bitwise_equal is the result we want, except in the the +/- 0 case. A OR B will be 0 everywhere except the sign bit iff A and B are +/- 0. A left-shift by one makes it all-zero or not-all-zero for depending on whether or not we need to override the bitwise-equal test.

This uses one more instruction than cmpneqps, because we're emulating the functionality we need from it with por / paddD. (or pslld by one, but that's one byte longer. It does run on a different port than pcmpeq, but you need to consider the the port distribution of the surrounding code to factor that into the decision.)

This algorithm might be useful on different SIMD architectures that don't provide the same vector FP tests for detecting NaN.

;inputs in xmm0:A  xmm1:B
movaps    xmm2, xmm0
pcmpeqd   xmm2, xmm1     ; xmm2=bitwise_equal.  (0:unequal -1:equal)

por       xmm0, xmm1
paddD     xmm0, xmm0     ; left-shift by 1 (one byte shorter than pslld xmm0, 1, and can run on more ports).

; xmm0=all-zero only in the +/- 0 case (where A and B are IEEE equal)

; xmm2     xmm0          desired result (0 means "no difference found")
;  -1       0        ->      0          ; bitwise equal and +/-0 equal
;  -1     non-zero   ->      0          ; just bitwise equal
;   0       0        ->      0          ; just +/-0 equal
;   0     non-zero   ->      non-zero   ; neither

ptest     xmm2, xmm0         ; CF = ( (not(xmm2) AND xmm0) == 0)
jc  reflexive_equal

The latency is lower than the cmpneqps version above, by one or two cycles.

We're really taking full advantage of PTEST here: Using its ANDN between two different operands, and using its compare-against-zero of the whole thing. We can't replace it with pandn / movmskps because we need to check all the bits, not just the sign bit of each element.

I haven't actually tested this, so it might be wrong even if my conclusion that +/-0 is the only time IEEE_equal is different from bitwise_equal (other than NaNs).


Handling non-bitwise-identical NaNs with integer-only ops is probably not worth it. The encoding is so similar to +/-Inf that I can't think of any simple checks that wouldn't take several instructions. Inf has all the exponent bits set, and an all-zero mantissa. NaN has all the exponent bits set, with a non-zero mantissa aka significand (so there are 23 bits of payload). The MSB of the mantissa is interpreted as an is_quiet flag to distinguish signalling / quiet NaNs. Also see Intel manual vol1, table 4-3 (Floating-Point Number and NaN Encodings).

If it wasn't for -Inf using the top-9-bits-set encoding, we could check for NaN with an unsigned compare for A > 0x7f800000. (0x7f800000 is single-precision +Inf). However, note that pcmpgtd / pcmpgtq are signed integer compares. AVX512F VPCMPUD is an unsigned compare (dest = a mask register).


The OP's idea: !(a<b) && !(b<a)

The OP's suggestion of !(a<b) && !(b<a) can't work, and neither can any variation of it. You can't tell the difference between one NaN and two NaNs just from two compares with reversed operands. Even mixing predicates can't help: No VCMPPS predicate differentiates one operand being NaN from both operands being NaN, or depends on whether it's the first or second operand that's NaN. Thus, it's impossible for a combination of them to have that information.

Paul R's solution of comparing a vector with itself does let us detect where there are NaNs and handle them "manually". No combination of results from VCMPPS between the two operands is sufficient, but using operands other than A and B does help. (Either a known-non-NaN vector or same operand twice).


Without the inversion, the bitwise-NaN code finds when at least one element is equal. (There's no inverse for pcmpeqd, so we can't use different logical operators and still get a test for all-equal):

; inputs in xmm0, xmm1
movaps   xmm2, xmm0
cmpeqps  xmm2, xmm1    ; -1:ieee_equal.  EQ_OQ predicate in the expanded notation for VEX encoding
pcmpeqd  xmm0, xmm1    ; -1:bitwise equal
orps     xmm0, xmm2
; xmm0 = -1:(where an element is bitwise or ieee equal)   0:elsewhere

movmskps eax, xmm0
test     eax, eax
jnz at_least_one_equal
; else  all different

PTEST isn't useful this way, since combining with OR is the only useful thing.


// UNFINISHED start of an idea
bitdiff = _mm_xor_si128(A, B);
signbitdiff = _mm_srai_epi32(bitdiff, 31);   // broadcast the diff in sign bit to the whole vector
signbitdiff = _mm_srli_epi32(bitdiff, 1);    // zero the sign bit
something = _mm_and_si128(bitdiff, signbitdiff);
like image 150
Peter Cordes Avatar answered Sep 30 '22 13:09

Peter Cordes


Here is one possible solution - it's not very efficient however, requiring 6 instructions:

__m128 v0, v1; // float vectors

__m128 v0nan = _mm_cmpeq_ps(v0, v0);                   // test v0 for NaNs
__m128 v1nan = _mm_cmpeq_ps(v1, v1);                   // test v1 for NaNs
__m128 vnan = _mm_or_si128(v0nan, v1nan);              // combine
__m128 vcmp = _mm_cmpneq_ps(v0, v1);                   // compare floats
vcmp = _mm_and_si128(vcmp, vnan);                      // combine NaN test
bool cmp = _mm_testz_si128(vcmp, vcmp);                // return true if all equal

Note that all the logic above is inverted, which may make the code a little hard to follow (ORs are effectively ANDs, and vice versa).

like image 35
Paul R Avatar answered Sep 30 '22 14:09

Paul R