Plain C ++代码比内联汇编器快10倍。为什么？ [英] Plain C++ Code 10 times faster than inline assembler. Why?
问题描述
这两个代码片段做同样的事情：添加两个float数组在一起并将结果存回他们。
These two code snippets do the same thing: Adding two float arrays together and storing the result back into them.
内联汇编：
void vecAdd_SSE(float* v1, float* v2) {
_asm {
mov esi, v1
mov edi, v2
movups xmm0, [esi]
movups xmm1, [edi]
addps xmm0, xmm1
movups [esi], xmm0
movups [edi], xmm0
}
}
普通C ++代码：
void vecAdd_Std(float* v1, float* v2) {
v1[0] = v1[0]+ v2[0];
v1[1] = v1[1]+ v2[1];
v1[2] = v1[2]+ v2[2];
v1[3] = v1[3]+ v2[3];
v2[0] = v1[0];
v2[1] = v1[1];
v2[2] = v1[2];
v2[3] = v1[3];
}
C ++代码的反汇编（在调试模式下进行反汇编，因为我无法查看反汇编在释放模式中）：
Disassembly for C++ Code (Disassembly made in Debug mode because i cannot view the Disassembly in Release mode for some reason):
void vecAdd_Std(float* v1, float* v2) {
push ebp
mov ebp,esp
sub esp,0C0h
push ebx
push esi
push edi
lea edi,[ebp-0C0h]
mov ecx,30h
mov eax,0CCCCCCCCh
rep stos dword ptr es:[edi]
v1[0] = v1[0]+ v2[0];
mov eax,4
imul ecx,eax,0
mov edx,4
imul eax,edx,0
mov edx,dword ptr [v1]
mov esi,dword ptr [v2]
movss xmm0,dword ptr [edx+ecx]
addss xmm0,dword ptr [esi+eax]
mov eax,4
imul ecx,eax,0
mov edx,dword ptr [v1]
movss dword ptr [edx+ecx],xmm0
v1[1] = v1[1]+ v2[1];
mov eax,4
shl eax,0
v1[1] = v1[1]+ v2[1];
mov ecx,4
shl ecx,0
mov edx,dword ptr [v1]
mov esi,dword ptr [v2]
movss xmm0,dword ptr [edx+eax]
addss xmm0,dword ptr [esi+ecx]
mov eax,4
shl eax,0
mov ecx,dword ptr [v1]
movss dword ptr [ecx+eax],xmm0
v1[2] = v1[2]+ v2[2];
mov eax,4
shl eax,1
mov ecx,4
shl ecx,1
mov edx,dword ptr [v1]
mov esi,dword ptr [v2]
movss xmm0,dword ptr [edx+eax]
addss xmm0,dword ptr [esi+ecx]
mov eax,4
shl eax,1
mov ecx,dword ptr [v1]
movss dword ptr [ecx+eax],xmm0
v1[3] = v1[3]+ v2[3];
mov eax,4
imul ecx,eax,3
mov edx,4
imul eax,edx,3
mov edx,dword ptr [v1]
mov esi,dword ptr [v2]
movss xmm0,dword ptr [edx+ecx]
addss xmm0,dword ptr [esi+eax]
mov eax,4
imul ecx,eax,3
mov edx,dword ptr [v1]
movss dword ptr [edx+ecx],xmm0
v2[0] = v1[0];
mov eax,4
imul ecx,eax,0
mov edx,4
imul eax,edx,0
mov edx,dword ptr [v2]
mov esi,dword ptr [v1]
mov ecx,dword ptr [esi+ecx]
mov dword ptr [edx+eax],ecx
v2[1] = v1[1];
mov eax,4
shl eax,0
mov ecx,4
shl ecx,0
mov edx,dword ptr [v2]
mov esi,dword ptr [v1]
mov eax,dword ptr [esi+eax]
mov dword ptr [edx+ecx],eax
v2[2] = v1[2];
mov eax,4
shl eax,1
mov ecx,4
shl ecx,1
mov edx,dword ptr [v2]
mov esi,dword ptr [v1]
mov eax,dword ptr [esi+eax]
mov dword ptr [edx+ecx],eax
v2[3] = v1[3];
mov eax,4
imul ecx,eax,3
mov edx,4
imul eax,edx,3
mov edx,dword ptr [v2]
mov esi,dword ptr [v1]
mov ecx,dword ptr [esi+ecx]
mov dword ptr [edx+eax],ecx
}
现在我做的那些功能的时间测量，发现内联汇编代码需要更长的时间大约为10倍（以释放模式）。
有人知道为什么吗？
Now I made a time measurement on those to functions and noticed that the inline assembler code takes approximately 10 times longer (in Release mode). Does anybody know why?
提前感谢。
推荐答案
在我的机器（VS2015 64位模式）中，编译器内嵌 vecAdd_Std 并产生
On my machine (VS2015 64-bit mode), the compiler inlines vecAdd_Std and produces
00007FF625921C8F vmovups xmm1,xmmword ptr [__xmm@4100000040c000004080000040000000 (07FF625929D60h)]
00007FF625921C97 vmovups xmm4,xmm1
00007FF625921C9B vcvtss2sd xmm1,xmm1,xmm4
测试代码
int main() {
float x[4] = {1.0, 2.0, 3.0, 4.0};
float y[4] = {1.0, 2.0, 3.0, 4.0};
vecAdd_Std(x, y);
std::cout << x[0];
}
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