我的编译器在做什么?(优化memcpy)
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26-09-2019 - |
题
我正在使用 VC++2010 中的以下设置编译一些代码:/O2 /Ob2 /Oi /Ot
但是,我在理解生成的程序集的某些部分时遇到了一些困难,我在代码中添加了一些问题作为注释。
另外,现代 cpu 上通常建议的预取距离是多少?我可以在自己的 cpu 上进行 ofc 测试,但我希望得到一些能够在更广泛的 cpu 上正常工作的值。也许可以使用动态预取距离?
<--编辑:
另一件令我惊讶的事情是编译器不会以某种形式交错 movdqa 和 movntdq 指令?因为根据我的理解,这些指令在某种意义上是异步的。
此代码还假设预取时有 32 字节缓存行,但高端 cpu 似乎有 64 字节缓存行,因此可以删除其中 2 个预取。
-->
void memcpy_aligned_x86(void* dest, const void* source, size_t size)
{
0052AC20 push ebp
0052AC21 mov ebp,esp
const __m128i* source_128 = reinterpret_cast<const __m128i*>(source);
for(size_t n = 0; n < size/16; n += 8)
0052AC23 mov edx,dword ptr [size]
0052AC26 mov ecx,dword ptr [dest]
0052AC29 mov eax,dword ptr [source]
0052AC2C shr edx,4
0052AC2F test edx,edx
0052AC31 je copy+9Eh (52ACBEh)
__m128i xmm0 = _mm_setzero_si128();
__m128i xmm1 = _mm_setzero_si128();
__m128i xmm2 = _mm_setzero_si128();
__m128i xmm3 = _mm_setzero_si128();
__m128i xmm4 = _mm_setzero_si128();
__m128i xmm5 = _mm_setzero_si128();
__m128i xmm6 = _mm_setzero_si128();
__m128i xmm7 = _mm_setzero_si128();
__m128i* dest_128 = reinterpret_cast<__m128i*>(dest);
0052AC37 push esi
0052AC38 push edi
0052AC39 lea edi,[edx-1]
0052AC3C shr edi,3
0052AC3F inc edi
{
_mm_prefetch(reinterpret_cast<const char*>(source_128+8), _MM_HINT_NTA);
_mm_prefetch(reinterpret_cast<const char*>(source_128+10), _MM_HINT_NTA);
_mm_prefetch(reinterpret_cast<const char*>(source_128+12), _MM_HINT_NTA);
_mm_prefetch(reinterpret_cast<const char*>(source_128+14), _MM_HINT_NTA);
xmm0 = _mm_load_si128(source_128++);
xmm1 = _mm_load_si128(source_128++);
xmm2 = _mm_load_si128(source_128++);
xmm3 = _mm_load_si128(source_128++);
xmm4 = _mm_load_si128(source_128++);
xmm5 = _mm_load_si128(source_128++);
xmm6 = _mm_load_si128(source_128++);
xmm7 = _mm_load_si128(source_128++);
0052AC40 movdqa xmm6,xmmword ptr [eax+70h] // 1. Why is this moved before the pretecthes?
0052AC45 prefetchnta [eax+80h]
0052AC4C prefetchnta [eax+0A0h]
0052AC53 prefetchnta [eax+0C0h]
0052AC5A prefetchnta [eax+0E0h]
0052AC61 movdqa xmm0,xmmword ptr [eax+10h]
0052AC66 movdqa xmm1,xmmword ptr [eax+20h]
0052AC6B movdqa xmm2,xmmword ptr [eax+30h]
0052AC70 movdqa xmm3,xmmword ptr [eax+40h]
0052AC75 movdqa xmm4,xmmword ptr [eax+50h]
0052AC7A movdqa xmm5,xmmword ptr [eax+60h]
0052AC7F lea esi,[eax+70h] // 2. What is happening in these 2 lines?
0052AC82 mov edx,eax //
0052AC84 movdqa xmm7,xmmword ptr [edx] // 3. Why edx? and not simply eax?
_mm_stream_si128(dest_128++, xmm0);
0052AC88 mov esi,ecx // 4. Is esi never used?
0052AC8A movntdq xmmword ptr [esi],xmm7
_mm_stream_si128(dest_128++, xmm1);
0052AC8E movntdq xmmword ptr [ecx+10h],xmm0
_mm_stream_si128(dest_128++, xmm2);
0052AC93 movntdq xmmword ptr [ecx+20h],xmm1
_mm_stream_si128(dest_128++, xmm3);
0052AC98 movntdq xmmword ptr [ecx+30h],xmm2
_mm_stream_si128(dest_128++, xmm4);
0052AC9D movntdq xmmword ptr [ecx+40h],xmm3
_mm_stream_si128(dest_128++, xmm5);
0052ACA2 movntdq xmmword ptr [ecx+50h],xmm4
_mm_stream_si128(dest_128++, xmm6);
0052ACA7 movntdq xmmword ptr [ecx+60h],xmm5
_mm_stream_si128(dest_128++, xmm7);
0052ACAC lea edx,[ecx+70h]
0052ACAF sub eax,0FFFFFF80h
0052ACB2 sub ecx,0FFFFFF80h
0052ACB5 dec edi
0052ACB6 movntdq xmmword ptr [edx],xmm6 // 5. Why not simply ecx?
0052ACBA jne copy+20h (52AC40h)
0052ACBC pop edi
0052ACBD pop esi
}
}
原始代码:
void memcpy_aligned_x86(void* dest, const void* source, size_t size)
{
assert(dest != nullptr);
assert(source != nullptr);
assert(source != dest);
assert(size % 128 == 0);
__m128i xmm0 = _mm_setzero_si128();
__m128i xmm1 = _mm_setzero_si128();
__m128i xmm2 = _mm_setzero_si128();
__m128i xmm3 = _mm_setzero_si128();
__m128i xmm4 = _mm_setzero_si128();
__m128i xmm5 = _mm_setzero_si128();
__m128i xmm6 = _mm_setzero_si128();
__m128i xmm7 = _mm_setzero_si128();
__m128i* dest_128 = reinterpret_cast<__m128i*>(dest);
const __m128i* source_128 = reinterpret_cast<const __m128i*>(source);
for(size_t n = 0; n < size/16; n += 8)
{
_mm_prefetch(reinterpret_cast<const char*>(source_128+8), _MM_HINT_NTA);
_mm_prefetch(reinterpret_cast<const char*>(source_128+10), _MM_HINT_NTA);
_mm_prefetch(reinterpret_cast<const char*>(source_128+12), _MM_HINT_NTA);
_mm_prefetch(reinterpret_cast<const char*>(source_128+14), _MM_HINT_NTA);
xmm0 = _mm_load_si128(source_128++);
xmm1 = _mm_load_si128(source_128++);
xmm2 = _mm_load_si128(source_128++);
xmm3 = _mm_load_si128(source_128++);
xmm4 = _mm_load_si128(source_128++);
xmm5 = _mm_load_si128(source_128++);
xmm6 = _mm_load_si128(source_128++);
xmm7 = _mm_load_si128(source_128++);
_mm_stream_si128(dest_128++, xmm0);
_mm_stream_si128(dest_128++, xmm1);
_mm_stream_si128(dest_128++, xmm2);
_mm_stream_si128(dest_128++, xmm3);
_mm_stream_si128(dest_128++, xmm4);
_mm_stream_si128(dest_128++, xmm5);
_mm_stream_si128(dest_128++, xmm6);
_mm_stream_si128(dest_128++, xmm7);
}
}
解决方案
eax+70h 读取会向上移动,因为 eax+70h 与 eax 位于不同的缓存行中,并且编译器可能希望硬件预取器尽快忙于获取该行。
它不进行交错,要么是因为它希望通过避免加载到存储的依赖关系来最大化性能(即使 AMD 优化指南明确指出要交错),要么只是因为它不确定存储不会覆盖加载。如果将 __restrict 关键字添加到源和目标,它会改变行为吗?
其余部分的目的也让我无法理解。对于 AMD 或 Intel 来说,可能是一些模糊的指令解码或硬件预取器考虑因素,但我找不到任何理由。我想知道当你删除这些指令时代码会变得更快还是更慢?
建议的预取距离取决于循环大小。需要足够远,以便数据在需要时有时间从内存到达。我认为你通常需要给它至少 100 个时钟周期。
其他提示
我还没有想出编译器做什么,但是我虽然我会分享我的一些测试结果。我已经重写的函数在组件。
系统:至强W3520
4.55 Gb / s的:常规的memcpy
5.52 GB / S:的memcpy所讨论
5.58 GB / S:下面的memcpy
7.48 GB / S:低于多线程
的memcpyvoid* memcpy(void* dest, const void* source, size_t num)
{
__asm
{
mov esi, source;
mov edi, dest;
mov ebx, num;
shr ebx, 7;
cpy:
prefetchnta [esi+80h];
prefetchnta [esi+0C0h];
movdqa xmm0, [esi+00h];
movdqa xmm1, [esi+10h];
movdqa xmm2, [esi+20h];
movdqa xmm3, [esi+30h];
movntdq [edi+00h], xmm0;
movntdq [edi+10h], xmm1;
movntdq [edi+20h], xmm2;
movntdq [edi+30h], xmm3;
movdqa xmm4, [esi+40h];
movdqa xmm5, [esi+50h];
movdqa xmm6, [esi+60h];
movdqa xmm7, [esi+70h];
movntdq [edi+40h], xmm4;
movntdq [edi+50h], xmm5;
movntdq [edi+60h], xmm6;
movntdq [edi+70h], xmm7;
lea edi, [edi+80h];
lea esi, [esi+80h];
dec ebx;
jnz cpy;
}
return dest;
}
void* memcpy_tbb(void* dest, const void* source, size_t num)
{
tbb::parallel_for(tbb::blocked_range<size_t>(0, num/128), [&](const tbb::blocked_range<size_t>& r)
{
memcpy_SSE2_3(reinterpret_cast<char*>(dest) + r.begin()*128, reinterpret_cast<const char*>(source) + r.begin()*128, r.size()*128);
}, tbb::affinity_partitioner());
return dest;
}
0052AC82 mov edx,eax //
0052AC84 movdqa xmm7,xmmword ptr [edx] // 3. Why edx? and not simply eax? <--
因为它要propably分裂数据通路所以该指令
0052ACAF sub eax,0FFFFFF80h
可以被并行执行。
点号 4 可能是预取的提示... propably(因为否则它没有任何意义,也可能是一个编译器/优化器的bug /怪癖)。
我没有关于任何点想法的 5 强>