我一直在做一些测试，看看在循环中附加边界检查有多大区别。通过考虑访问数组时由C＃，Java等语言插入的隐式边界检查的成本来提示这一点。
更新：我在几台额外的计算机上尝试了相同的可执行程序，这可以为正在发生的事情提供更多的信息。我首先列出了原始计算机，其次是我的现代笔记本电脑。在我的现代笔记本电脑上，在循环中添加额外的检查只会增加1到4％的时间，相比之下原始硬件的3到30％。
Processor x86 Family 6 Model 30 Stepping 5 GenuineIntel ~2793 Mhz
Ratio 2 checks : 1 check = 1.0310
Ratio 3 checks : 1 check = 1.2769
Processor Intel(R) Core(TM) i7-3610QM CPU @ 2.30GHz, 2301 Mhz, 4 Core(s), 8 Logical Processor(s)
Ratio 2 checks : 1 check = 1.0090
Ratio 3 checks : 1 check = 1.0393
Processor Intel(R) Core(TM) i5-2500 CPU @ 3.30GHz, 4 Cores(s)
Ratio 2 checks : 1 check = 1.0035
Ratio 3 checks : 1 check = 1.0639
Processor Intel(R) Core(TM)2 Duo CPU T9300 @ 2.50GHz, 2501 Mhz, 2 Core(s), 2 Logical Processor(s)
Ratio 2 checks : 1 check = 1.1195
Ratio 3 checks : 1 check = 1.3597
Processor x86 Family 15 Model 43 Stepping 1 AuthenticAMD ~2010 Mhz
Ratio 2 checks : 1 check = 1.0776
Ratio 3 checks : 1 check = 1.1451
在下面的测试程序中，第一个函数只检查一个边界，第二个函数检查两个，第三个检查三个（在调用代码n1=n2=n3中）。我发现比率两个检查：一个约为1.03，比率三个检查：一个约为1.3。令我感到惊讶的是，再增加一张支票会对性能产生如此大的影响。我得到了interesting answer concerning the low cost of bounds checking on modern processors我原来的问题，这可能会对这里观察到的差异有所了解。
请注意，在未启用整个程序优化的情况下编译程序非常重要;否则编译器可以简单地删除附加边界检查。
// dotprod.cpp
#include "dotprod.h"
double SumProduct(const double* v1, const double* v2, int n)
{
double sum=0;
for(int i=0;
i<n;
++i)
sum += v1[i]*v2[i];
return sum;
}
double SumProduct(const double* v1, const double* v2, int n1, int n2)
{
double sum=0;
for(int i=0;
i<n1 && i <n2;
++i)
sum += v1[i]*v2[i];
return sum;
}
double SumProduct(const double* v1, const double* v2, int n1, int n2, int n3)
{
double sum=0;
for(int i=0;
i<n1 && i <n2 && i <n3;
++i)
sum += v1[i]*v2[i];
return sum;
}
此代码最初是使用Visual Studio 2010，Release，Win32构建的（我添加了'C'标记，因为速度差异背后的原因可能不是C ++特定的，可能不是Windows特定的）。谁能解释一下呢？
下面的其余代码，供参考。这里面有一些特定于C ++的东西。
标头文件
// dotprod.h
double SumProduct(const double*, const double*, int n);
double SumProduct(const double*, const double*, int n1, int n2);
double SumProduct(const double*, const double*, int n1, int n2, int n3);
测试工具
// main.cpp
#include <stdio.h>
#include <math.h>
#include <numeric>
#include <vector>
#include <windows.h>
#include "../dotprod/dotprod.h" // separate lib
typedef __int64 timecount_t;
inline timecount_t GetTimeCount()
{
LARGE_INTEGER li;
if (!QueryPerformanceCounter(&li)) {
exit(1);
}
return li.QuadPart;
}
int main()
{
typedef std::vector<double> dvec;
const int N = 100 * 1000;
// Initialize
dvec v1(N);
dvec v2(N);
dvec dp1(N);
dvec dp2(N);
dvec dp3(N);
for(int i=0; i<N; ++i) {
v1[i] = i;
v2[i] = log(static_cast<double>(i+1));
}
const timecount_t t0 = GetTimeCount();
// Check cost with one bound
for(int n=0; n<N; ++n) {
dp1[n] = SumProduct(&(v1[0]),&(v2[0]),n);
}
const timecount_t t1 = GetTimeCount();
// Check cost with two bounds
for(int n=0; n<N; ++n) {
dp2[n] = SumProduct(&(v1[0]),&(v2[0]),n,n);
}
const timecount_t t2 = GetTimeCount();
// Check cost with three bounds
for(int n=0; n<N; ++n) {
dp3[n] = SumProduct(&(v1[0]),&(v2[0]),n,n,n);
}
const timecount_t t3 = GetTimeCount();
// Check results
const double sumSumProducts1 = std::accumulate(dp1.begin(), dp1.end(), 0.0);
const double sumSumProducts2 = std::accumulate(dp2.begin(), dp2.end(), 0.0);
const double sumSumProducts3 = std::accumulate(dp3.begin(), dp3.end(), 0.0);
printf("Sums of dot products: %.1f, %.1f, %.1f\n", sumSumProducts1, sumSumProducts2, sumSumProducts3);
// Output timings
const timecount_t elapsed1 = t1-t0;
const timecount_t elapsed2 = t2-t1;
const timecount_t elapsed3 = t3-t2;
printf("Elapsed: %.0f, %.0f, %.0f\n",
static_cast<double>(elapsed1),
static_cast<double>(elapsed2),
static_cast<double>(elapsed3));
const double ratio2to1 = elapsed2 / static_cast<double>(elapsed1);
const double ratio3to1 = elapsed3 / static_cast<double>(elapsed1);
printf("Ratio 2:1=%.2f\n", ratio2to1);
printf("Ratio 3:1=%.2f\n", ratio3to1);
return 0;
}
为了产生汇编，我接受了this answer中的建议（案例2，关闭整个程序优化），生成以下asm文件。
; Listing generated by Microsoft (R) Optimizing Compiler Version 16.00.40219.01
TITLE C:\dev\TestSpeed\dotprod\dotprod.cpp
.686P
.XMM
include listing.inc
.model flat
INCLUDELIB OLDNAMES
PUBLIC __real@0000000000000000
PUBLIC ?SumProduct@@YANPBN0HHH@Z ; SumProduct
EXTRN __fltused:DWORD
; COMDAT __real@0000000000000000
; File c:\dev\testspeed\dotprod\dotprod.cpp
CONST SEGMENT
__real@0000000000000000 DQ 00000000000000000r ; 0
; Function compile flags: /Ogtp
CONST ENDS
; COMDAT ?SumProduct@@YANPBN0HHH@Z
_TEXT SEGMENT
tv491 = -4 ; size = 4
_v1$ = 8 ; size = 4
_v2$ = 12 ; size = 4
_n1$ = 16 ; size = 4
_n2$ = 20 ; size = 4
_n3$ = 24 ; size = 4
?SumProduct@@YANPBN0HHH@Z PROC ; SumProduct, COMDAT
; 25 : {
push ebp
mov ebp, esp
push ecx
; 26 : double sum=0;
fldz
push ebx
mov ebx, DWORD PTR _v2$[ebp]
push esi
push edi
mov edi, DWORD PTR _n1$[ebp]
; 27 : for(int i=0;
xor ecx, ecx
; 28 : i<n1 && i <n2 && i <n3;
; 29 : ++i)
cmp edi, 4
jl $LC8@SumProduct
; 26 : double sum=0;
mov edi, DWORD PTR _v1$[ebp]
lea esi, DWORD PTR [edi+24]
; 30 : sum += v1[i]*v2[i];
sub edi, ebx
lea edx, DWORD PTR [ecx+2]
lea eax, DWORD PTR [ebx+8]
mov DWORD PTR tv491[ebp], edi
$LN15@SumProduct:
; 28 : i<n1 && i <n2 && i <n3;
; 29 : ++i)
mov ebx, DWORD PTR _n2$[ebp]
cmp ecx, ebx
jge $LN9@SumProduct
cmp ecx, DWORD PTR _n3$[ebp]
jge $LN9@SumProduct
; 30 : sum += v1[i]*v2[i];
fld QWORD PTR [eax-8]
lea edi, DWORD PTR [edx-1]
fmul QWORD PTR [esi-24]
faddp ST(1), ST(0)
cmp edi, ebx
jge SHORT $LN9@SumProduct
; 28 : i<n1 && i <n2 && i <n3;
; 29 : ++i)
cmp edi, DWORD PTR _n3$[ebp]
jge SHORT $LN9@SumProduct
; 30 : sum += v1[i]*v2[i];
mov edi, DWORD PTR tv491[ebp]
fld QWORD PTR [edi+eax]
fmul QWORD PTR [eax]
faddp ST(1), ST(0)
cmp edx, ebx
jge SHORT $LN9@SumProduct
; 28 : i<n1 && i <n2 && i <n3;
; 29 : ++i)
cmp edx, DWORD PTR _n3$[ebp]
jge SHORT $LN9@SumProduct
; 30 : sum += v1[i]*v2[i];
fld QWORD PTR [eax+8]
lea edi, DWORD PTR [edx+1]
fmul QWORD PTR [esi-8]
faddp ST(1), ST(0)
cmp edi, ebx
jge SHORT $LN9@SumProduct
; 28 : i<n1 && i <n2 && i <n3;
; 29 : ++i)
cmp edi, DWORD PTR _n3$[ebp]
jge SHORT $LN9@SumProduct
; 30 : sum += v1[i]*v2[i];
fld QWORD PTR [eax+16]
mov edi, DWORD PTR _n1$[ebp]
fmul QWORD PTR [esi]
add ecx, 4
lea ebx, DWORD PTR [edi-3]
add eax, 32 ; 00000020H
add esi, 32 ; 00000020H
faddp ST(1), ST(0)
add edx, 4
cmp ecx, ebx
jl SHORT $LN15@SumProduct
mov ebx, DWORD PTR _v2$[ebp]
$LC8@SumProduct:
; 28 : i<n1 && i <n2 && i <n3;
; 29 : ++i)
cmp ecx, edi
jge SHORT $LN9@SumProduct
mov edx, DWORD PTR _v1$[ebp]
lea eax, DWORD PTR [ebx+ecx*8]
sub edx, ebx
$LC3@SumProduct:
cmp ecx, DWORD PTR _n2$[ebp]
jge SHORT $LN9@SumProduct
cmp ecx, DWORD PTR _n3$[ebp]
jge SHORT $LN9@SumProduct
; 30 : sum += v1[i]*v2[i];
fld QWORD PTR [eax+edx]
inc ecx
fmul QWORD PTR [eax]
add eax, 8
faddp ST(1), ST(0)
cmp ecx, edi
jl SHORT $LC3@SumProduct
$LN9@SumProduct:
; 31 : return sum;
; 32 : }
pop edi
pop esi
pop ebx
mov esp, ebp
pop ebp
ret 0
?SumProduct@@YANPBN0HHH@Z ENDP ; SumProduct
_TEXT ENDS
PUBLIC ?SumProduct@@YANPBN0HH@Z ; SumProduct
; Function compile flags: /Ogtp
; COMDAT ?SumProduct@@YANPBN0HH@Z
_TEXT SEGMENT
tv448 = -4 ; size = 4
_v1$ = 8 ; size = 4
_v2$ = 12 ; size = 4
_n1$ = 16 ; size = 4
_n2$ = 20 ; size = 4
?SumProduct@@YANPBN0HH@Z PROC ; SumProduct, COMDAT
; 15 : {
push ebp
mov ebp, esp
push ecx
; 16 : double sum=0;
fldz
push ebx
mov ebx, DWORD PTR _v2$[ebp]
push esi
push edi
mov edi, DWORD PTR _n1$[ebp]
; 17 : for(int i=0;
xor ecx, ecx
; 18 : i<n1 && i <n2;
; 19 : ++i)
cmp edi, 4
jl SHORT $LC8@SumProduct@2
; 16 : double sum=0;
mov edi, DWORD PTR _v1$[ebp]
lea edx, DWORD PTR [edi+24]
; 20 : sum += v1[i]*v2[i];
sub edi, ebx
lea esi, DWORD PTR [ecx+2]
lea eax, DWORD PTR [ebx+8]
mov DWORD PTR tv448[ebp], edi
$LN19@SumProduct@2:
mov edi, DWORD PTR _n2$[ebp]
cmp ecx, edi
jge SHORT $LN9@SumProduct@2
fld QWORD PTR [eax-8]
lea ebx, DWORD PTR [esi-1]
fmul QWORD PTR [edx-24]
faddp ST(1), ST(0)
cmp ebx, edi
jge SHORT $LN9@SumProduct@2
mov ebx, DWORD PTR tv448[ebp]
fld QWORD PTR [ebx+eax]
fmul QWORD PTR [eax]
faddp ST(1), ST(0)
cmp esi, edi
jge SHORT $LN9@SumProduct@2
fld QWORD PTR [eax+8]
lea ebx, DWORD PTR [esi+1]
fmul QWORD PTR [edx-8]
faddp ST(1), ST(0)
cmp ebx, edi
jge SHORT $LN9@SumProduct@2
fld QWORD PTR [eax+16]
mov edi, DWORD PTR _n1$[ebp]
fmul QWORD PTR [edx]
add ecx, 4
lea ebx, DWORD PTR [edi-3]
add eax, 32 ; 00000020H
add edx, 32 ; 00000020H
faddp ST(1), ST(0)
add esi, 4
cmp ecx, ebx
jl SHORT $LN19@SumProduct@2
mov ebx, DWORD PTR _v2$[ebp]
$LC8@SumProduct@2:
; 18 : i<n1 && i <n2;
; 19 : ++i)
cmp ecx, edi
jge SHORT $LN9@SumProduct@2
mov edx, DWORD PTR _v1$[ebp]
lea eax, DWORD PTR [ebx+ecx*8]
sub edx, ebx
$LC3@SumProduct@2:
cmp ecx, DWORD PTR _n2$[ebp]
jge SHORT $LN9@SumProduct@2
; 20 : sum += v1[i]*v2[i];
fld QWORD PTR [eax+edx]
inc ecx
fmul QWORD PTR [eax]
add eax, 8
faddp ST(1), ST(0)
cmp ecx, edi
jl SHORT $LC3@SumProduct@2
$LN9@SumProduct@2:
; 21 : return sum;
; 22 : }
pop edi
pop esi
pop ebx
mov esp, ebp
pop ebp
ret 0
?SumProduct@@YANPBN0HH@Z ENDP ; SumProduct
_TEXT ENDS
PUBLIC ?SumProduct@@YANPBN0H@Z ; SumProduct
; Function compile flags: /Ogtp
; COMDAT ?SumProduct@@YANPBN0H@Z
_TEXT SEGMENT
_v1$ = 8 ; size = 4
_v2$ = 12 ; size = 4
?SumProduct@@YANPBN0H@Z PROC ; SumProduct, COMDAT
; _n$ = eax
; 5 : {
push ebp
mov ebp, esp
mov edx, DWORD PTR _v2$[ebp]
; 6 : double sum=0;
fldz
push ebx
push esi
mov esi, eax
; 7 : for(int i=0;
xor ebx, ebx
push edi
mov edi, DWORD PTR _v1$[ebp]
; 8 : i<n;
; 9 : ++i)
cmp esi, 4
jl SHORT $LC9@SumProduct@3
; 6 : double sum=0;
lea eax, DWORD PTR [edx+8]
lea ecx, DWORD PTR [edi+24]
; 10 : sum += v1[i]*v2[i];
sub edi, edx
lea edx, DWORD PTR [esi-4]
shr edx, 2
inc edx
lea ebx, DWORD PTR [edx*4]
$LN10@SumProduct@3:
fld QWORD PTR [eax-8]
add eax, 32 ; 00000020H
fmul QWORD PTR [ecx-24]
add ecx, 32 ; 00000020H
dec edx
faddp ST(1), ST(0)
fld QWORD PTR [edi+eax-32]
fmul QWORD PTR [eax-32]
faddp ST(1), ST(0)
fld QWORD PTR [eax-24]
fmul QWORD PTR [ecx-40]
faddp ST(1), ST(0)
fld QWORD PTR [eax-16]
fmul QWORD PTR [ecx-32]
faddp ST(1), ST(0)
jne SHORT $LN10@SumProduct@3
; 6 : double sum=0;
mov edx, DWORD PTR _v2$[ebp]
mov edi, DWORD PTR _v1$[ebp]
$LC9@SumProduct@3:
; 8 : i<n;
; 9 : ++i)
cmp ebx, esi
jge SHORT $LN8@SumProduct@3
sub edi, edx
lea eax, DWORD PTR [edx+ebx*8]
sub esi, ebx
$LC3@SumProduct@3:
; 10 : sum += v1[i]*v2[i];
fld QWORD PTR [eax+edi]
add eax, 8
dec esi
fmul QWORD PTR [eax-8]
faddp ST(1), ST(0)
jne SHORT $LC3@SumProduct@3
$LN8@SumProduct@3:
; 11 : return sum;
; 12 : }
pop edi
pop esi
pop ebx
pop ebp
ret 0
?SumProduct@@YANPBN0H@Z ENDP ; SumProduct
_TEXT ENDS
END
答案 0 :(得分：2)
CPU之间的一个重要区别是管道优化
CPU可以并行执行多条指令，直到到达条件分支。从这一点开始，CPU不再等待所有指令执行，而是可以并行继续分支，直到条件可用并准备好进行评估。如果假设是正确的，那么我们就有了收益。否则CPU将与另一个分支一起使用。
因此，CPU的棘手部分是找到最佳假设并尽可能并行执行尽可能多的指令。