Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Why does adding extra check in loop make big difference on some machines, and small difference on others?

Tags:

c++

c

I have been doing some testing to see how much of a difference additional bounds checking makes in loops. This is prompted by thinking about the cost of implicit bounds checking inserted by languages such as C#, Java etc, when you access arrays.

Update: I have tried the same executable program out on several additional computers, which throws a lot more light onto what is happening. I've listed the original computer first, and second my modern laptop. On my modern laptop, adding additional checks in the loop adds only between 1 and 4% to the time taken, compared to between 3 and 30% for the original hardware.

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

In the test program, below, the first function checks just one bound, the second function checks two, and the third checks three (in the calling code, n1=n2=n3). I found that the ratio two checks:one was about 1.03, and the ratio three checks:one was about 1.3. I was surprised by that adding one more check made such a difference to performance. I got an interesting answer concerning the low cost of bounds checking on modern processors to my original question, which may throw some light on the differences observed here.

Note that it's important to compile the program without whole program optimization turned on; otherwise the compiler can simply remove the additional bounds checking.

// 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;
}

This code was originally built using Visual Studio 2010, Release, Win32 (I've added the 'C' tag because the reasoning behind the difference in speed is not likely to be C++ specific, and may not be Windows specific). Can anyone explain it?

Rest of the code below, for information. This has some C++ specific stuff in it.

Header file

// 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);

Test harness

// 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;
}

In order to produce assembly, I took the advice in this answer (case 2, turning off whole program optimization), producing the following asm file.

; 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
like image 640
TooTone Avatar asked Jun 12 '13 16:06

TooTone


1 Answers

One big difference between CPUs is the pipeline optimization

The CPU can execute in parallel several instructions until reaches a conditional branch. From this point instead of waiting until all the instructions are executed, the CPU can continue with a branch in parallel until the condition is available and ready to be evaluated. If the assumption was correct, then we have a gain. Otherwise the CPU will go with the other branch.

So the tricky part for a CPU is to find the best assumptions and to execute as many instructions in parallel as possible.

like image 163
Ovidiu Andoniu Avatar answered Oct 27 '22 23:10

Ovidiu Andoniu