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[英]Fastest way to sum dot product of vector of unsigned 64 bit integers using 192/256 bit integer?
[英]Vector of 64-bit double faster to dot-product than a vector of 32-bit unsigned int?
我有两种代码迭代大小为500的矢量设计。其中一个设计包含64位双精度数组,第二个设计使用包含32位整数的数组。 我期待32位设计更快,因为更多有用的数据可以打包在缓存中。
编译器MSVC,CPU Ivy Bridge,编译64位模式。
这是代码1,使用32位整数(在2600个 CPU周期中运行):
#include <vector>
#include <iostream>
int main(){
std::vector<unsigned int> x1;
std::vector<unsigned int> x2;
std::vector<unsigned int> x3;
x1.resize(500);
x2.resize(500);
x3.resize(500);
for(int i =0; i<500; i++){
x1[i] = i;
x2[i] = 2*i;
x3[i] = 4*i;
}
int counter = 0;
while(counter < 1000){
unsigned long long start = 0;
unsigned long long end = 0;
double m = 0;
double n = 0;
start = __rdtsc();
for(int i=0; i < 500; i++){
unsigned int a = x1[i];
unsigned int b = x2[i];
unsigned int g = x3[i];
m = m + (a * g);
n = n + (b * g);
}
end = __rdtscp();
std::cout << (end-start) << "\t\t"<<m << n << std::endl;
counter++;
}
}
产生这个asm(-Os):
start = __rdtscp(&p);
rdtscp
lea r8,[rbp+6Fh]
mov dword ptr [r8],ecx
shl rdx,20h
or rax,rdx
mov r10,rax
unsigned int p;
unsigned int q;
unsigned long long start = 0;
unsigned long long end = 0;
double m = 0;
mov r8,rbx
mov r9d,1F4h
unsigned int a = x1[i];
unsigned int b = x2[i];
unsigned int g = x3[i];
mov edx,dword ptr [r8+r15]
m = m + (a * g);
mov ecx,edx
imul ecx,dword ptr [r8+r14]
xorps xmm0,xmm0
cvtsi2sd xmm0,rcx
addsd xmm7,xmm0
n = n + (b * g);
imul edx,dword ptr [r8]
mov eax,edx
xorps xmm0,xmm0
cvtsi2sd xmm0,rax
addsd xmm8,xmm0
for(int i=0; i < 500; i++){
add r8,4
dec r9
jne main+0E5h (013F681261h)
}
end = __rdtscp(&q);
rdtscp
}
end = __rdtscp(&q);
lea r8,[rbp+6Fh]
mov dword ptr [r8],ecx
shl rdx,20h
or rdx,rax
这是代码2,使用64位双精度(代码在2000个 CPU周期中运行):
#include <vector>
#include <iostream>
int main(){
std::vector<double> x1;
std::vector<double> x2;
std::vector<unsigned long long> x3;
x1.resize(500);
x2.resize(500);
x3.resize(500);
for(int i =0; i<500; i++){
x1[i] = i;
x2[i] = 2*i;
x3[i] = 4*i;
}
int counter = 0;
while(counter < 1000){
unsigned int p;
unsigned int q;
unsigned long long start = 0;
unsigned long long end = 0;
double m = 0;
double n = 0;
start = __rdtscp(&p);
for(int i=0; i < 500; i++){
double a = x1[i];
double b = x2[i];
unsigned long long g = x3[i];
m = m + (a * g);
n = n + (b * g);
}
end = __rdtscp(&q);
std::cout << (end-start) << "\t\t"<<m << n << std::endl;
counter++;
}
}
这里是asm(-Os)产生的:
start = __rdtscp(&p);
rdtscp
lea r8,[rbp+6Fh]
mov dword ptr [r8],ecx
shl rdx,20h
or rax,rdx
mov r9,rax
unsigned int p;
unsigned int q;
unsigned long long start = 0;
unsigned long long end = 0;
double m = 0;
mov rdx,rbx
mov r8d,1F4h
double a = x1[i];
double b = x2[i];
unsigned long long g = x3[i];
mov rcx,qword ptr [rdx+r15]
xorps xmm1,xmm1
m = m + (a * g);
cvtsi2sd xmm1,rcx
test rcx,rcx
jns main+120h (013F32129Ch)
addsd xmm1,xmm9
movaps xmm0,xmm1
mulsd xmm0,mmword ptr [rdx+r14]
addsd xmm6,xmm0
n = n + (b * g);
mulsd xmm1,mmword ptr [rdx]
addsd xmm7,xmm1
for(int i=0; i < 500; i++){
add rdx,8
dec r8
jne main+10Ah (013F321286h)
}
end = __rdtscp(&q);
rdtscp
}
end = __rdtscp(&q);
lea r8,[rbp+6Fh]
mov dword ptr [r8],ecx
shl rdx,20h
or rdx,rax
区别在于第一个代码中整数到双精度的转换(向量包含unsigned int
,产品是整数运算,但积累使用double
,在汇编程序中,这会将cvtsi2sd
指令添加到代码中)。
在第二个代码中,您在任何地方都使用双打,因此您没有转换,代码运行得更快。
这种差异将是更加明显具有定点和浮点处理单元之间的严格区分在CPU上(在POWER平台是一个这样的例子)。 在这方面,X86平台非常宽容。
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