SSE intrinsics: Convert 32-bit floats to UNSIGNED 8-bit integers

问题

Using SSE intrinsics, I've gotten a vector of four 32-bit floats clamped to the range 0-255 and rounded to nearest integer. I'd now like to write those four out as bytes.

There is an intrinsic _mm_cvtps_pi8 that will convert 32-bit to 8-bit signed int, but the problem there is that any value over 127 gets clamped to 127. I can't find any instructions that will clamp to unsigned 8-bit values.

I have an intuition that what I may want to do is some combination of _mm_cvtps_pi16 and _mm_shuffle_pi8 followed by move instruction to get the four bytes I care about into memory. Is that the best way to do it? I'm going to see if I can figure out how to encode the shuffle control mask.

UPDATE: The following appears to do exactly what I want. Is there a better way?

#include <tmmintrin.h>
#include <stdio.h>

unsigned char out[8];
unsigned char shuf[8] = { 0, 2, 4, 6, 128, 128, 128, 128 };
float ins[4] = {500, 0, 120, 240};

int main()
{
    __m128 x = _mm_load_ps(ins);    // Load the floats
    __m64 y = _mm_cvtps_pi16(x);    // Convert them to 16-bit ints
    __m64 sh = *(__m64*)shuf;       // Get the shuffle mask into a register
    y = _mm_shuffle_pi8(y, sh);     // Shuffle the lower byte of each into the first four bytes
    *(int*)out = _mm_cvtsi64_si32(y); // Store the lower 32 bits

    printf("%d\n", out[0]);
    printf("%d\n", out[1]);
    printf("%d\n", out[2]);
    printf("%d\n", out[3]);
    return 0;
}

UPDATE2: Here's an even better solution based on Harold's answer:

#include <smmintrin.h>
#include <stdio.h>

unsigned char out[8];
float ins[4] = {10.4, 10.6, 120, 100000};

int main()
{   
    __m128 x = _mm_load_ps(ins);       // Load the floats
    __m128i y = _mm_cvtps_epi32(x);    // Convert them to 32-bit ints
    y = _mm_packus_epi32(y, y);        // Pack down to 16 bits
    y = _mm_packus_epi16(y, y);        // Pack down to 8 bits
    *(int*)out = _mm_cvtsi128_si32(y); // Store the lower 32 bits

    printf("%d\n", out[0]);
    printf("%d\n", out[1]);
    printf("%d\n", out[2]);
    printf("%d\n", out[3]);
    return 0;
}

回答1:

There is no direct conversion from float to byte, _mm_cvtps_pi8 is a composite. _mm_cvtps_pi16 is also a composite, and in this case it's just doing some pointless stuff that you undo with the shuffle. They also return annoying __m64's.

Anyway, we can convert to dwords (signed, but that doesn't matter), and then pack (unsigned) or shuffle them into bytes. _mm_shuffle_(e)pi8 generates a pshufb, Core2 45nm and AMD processors aren't too fond of it and you have to get a mask from somewhere.

Either way you don't have to round to the nearest integer first, the convert will do that. At least, if you haven't messed with the rounding mode.

Using packs 1: (not tested) -- probably not useful, packusdw already outputs unsigned words but then packuswb wants signed words again. Kept around because it is referred to elsewhere.

cvtps2dq xmm0, xmm0  
packusdw xmm0, xmm0     ; unsafe: saturates to a different range than packuswb accepts
packuswb xmm0, xmm0
movd somewhere, xmm0

Using different shuffles:

cvtps2dq xmm0, xmm0  
packssdw xmm0, xmm0     ; correct: signed saturation on first step to feed packuswb
packuswb xmm0, xmm0
movd somewhere, xmm0

Using shuffle: (not tested)

cvtps2dq xmm0, xmm0
pshufb xmm0, [shufmask]
movd somewhere, xmm0

shufmask: db 0, 4, 8, 12, 80h, 80h, 80h, 80h, 80h, 80h, 80h, 80h, 80h, 80h, 80h, 80h

回答2:

We can solve the unsigned clamping issue by doing the first stage of packing with signed saturation. [0-255] fits in a signed 16-bit int, so values in that range will remain unclamped. Values outside that range will stay on the same side of it. Thus, the signed16 -> unsigned8 step will clamp them correctly.

;; SSE2: good for arrays of inputs
cvtps2dq xmm0, [rsi]      ; 4 floats
cvtps2dq xmm1, [rsi+16]   ; 4 more floats
packssdw xmm0, xmm1       ; 8 int16_t

cvtps2dq xmm1, [rsi+32]
cvtps2dq xmm2, [rsi+48]
packssdw xmm1, xmm2       ; 8 more int16_t
                          ; signed because that's how packuswb treats its input
packuswb xmm0, xmm1       ; 16 uint8_t
movdqa   [rdi], xmm0

This only requires SSE2, not SSE4.1 for packusdw.

I assume this is the reason SSE2 only included signed pack from dword to word, but both signed and unsigned pack from word to byte. packuswd is only useful if your final goal is uint16_t, rather than further packing. (Since then you'd need to mask off the sign bit before feeding it to a further pack).

If you did use packusdw -> packuswb, you'd get bogus results when the first step saturated to a uint16_t > 0x7fff. packuswb would interpret that as a negative int16_t and saturate it to 0. packssdw would saturate such inputs to 0x7fff, the max int16_t.

(If your 32-bit inputs are always <= 0x7fff, you can use either, but SSE4.1 packusdw takes more instruction bytes than SSE2 packsswd, and never runs faster.)

If your source values can't be negative, and you only have one vector of 4 floats, not many, you can use harold's pshufb idea. If not, you need to clamp negative values to zero rather than truncate the by shuffling the low bytes into place.

Using

;; SSE4.1, good for a single vector.  Use the PACK version above for arrays
cvtps2dq   xmm0, xmm0
pmaxsd     xmm0, zeroed-register
pshufb     xmm0, [mask]
movd       [somewhere], xmm0

may be slightly more efficient than using two pack instructions, because pmax can run on port 1 or 5 (Intel Haswell). cvtps2dq is port 1 only, pshufb and pack* are port 5 only.

来源：https://stackoverflow.com/questions/29856006/sse-intrinsics-convert-32-bit-floats-to-unsigned-8-bit-integers

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