如何解决在C中使用移位运算符的乘法和除法实际上更快吗？
| 例如，可以使用位运算符来实现乘法和除法i*2 = i<<1
i*3 = (i<<1) + i;
i*10 = (i<<3) + (i<<1)
解决方法
简短答案：不太可能。
长答案：
您的编译器中包含一个优化器，该优化器知道如何以目标处理器体系结构所能达到的速度快速进行乘法。最好的选择是清楚地告诉编译器您的意图（即i * 2而不是i << 1），然后让它决定最快的汇编/机器码序列。处理器本身甚至可能已将乘法指令实现为一系列移位和微码加法运算。
最重要的是-不要花很多时间担心这个问题。如果您要转移，那就转移。如果您打算乘，乘。做语义上最清晰的事情-您的同事稍后会感谢您。或者更有可能的是，如果不这样做，稍后再诅咒您。
, 只是一个具体的测量点：很多年前，我以两个为基准
我的哈希算法的版本：
unsigned
hash( char const* s )
{
unsigned h = 0;
while ( *s != \'\\0\' ) {
h = 127 * h + (unsigned char)*s;
++ s;
}
return h;
}
unsigned
hash( char const* s )
{
unsigned h = 0;
while ( *s != \'\\0\' ) {
h = (h << 7) - h + (unsigned char)*s;
++ s;
}
return h;
}
std::string
x * 2 + z
x << 1 + z
+ - * / %
& ^ | >>
*
<<
source literal approach optimised approach
#define N 0
int x; .word x xor registerA,registerA
x *= N; move x -> registerA
move x -> registerB
A = B * immediate(0)
store registerA -> x
...............do something more with x...............
xor
...main(...) { return (argc << 4) + (argc << 2) + argc; }imull $21,8(%ebp),%eax
x
y
int
int x;
x >> 1; // divide by 2?
x
/
>>
int a = ...;
int b = a * 10;
MOV EAX,DWORD PTR SS:[ESP+1C] ; Move a into EAX
LEA EAX,DWORD PTR DS:[EAX+EAX*4] ; Multiply by 5 without shift !
SHL EAX,1 ; Multiply by 2 using shift
include ConstDiv.inc
...
mov eax,9999999
; divide eax by 100000
cdiv 100000
; edx = quotient
mov eax,9999999
mov edx,0A7C5AC47h
add eax,1
.if !CARRY?
mul edx
.endif
shr edx,16
#include <stdio.h>
int main(void)
{
int i;
for (i = 5; i >= -5; --i)
{
printf(\"%d / 2 = %d,%d >> 1 = %d\\n\",i,i / 2,i >> 1);
}
return 0;
}
5 / 2 = 2,5 >> 1 = 2
4 / 2 = 2,4 >> 1 = 2
3 / 2 = 1,3 >> 1 = 1
2 / 2 = 1,2 >> 1 = 1
1 / 2 = 0,1 >> 1 = 0
0 / 2 = 0,0 >> 1 = 0
-1 / 2 = 0,-1 >> 1 = -1
-2 / 2 = -1,-2 >> 1 = -1
-3 / 2 = -1,-3 >> 1 = -2
-4 / 2 = -2,-4 >> 1 = -2
-5 / 2 = -2,-5 >> 1 = -3
>>> from timeit import timeit
>>> setup_str = \'import scipy; from scipy import random; scipy.random.seed(0)\'
>>> N = 10*1000*1000
>>> timeit(\'x=random.randint(65536);\',setup=setup_str,number=N)
1.894096851348877 # Time from generating the random #s and no opperati
>>> timeit(\'x=random.randint(65536); x*2\',number=N)
2.2799630165100098
>>> timeit(\'x=random.randint(65536); x << 1\',number=N)
2.2616429328918457
>>> timeit(\'x=random.randint(65536); x*10\',number=N)
2.2799630165100098
>>> timeit(\'x=random.randint(65536); (x << 3) + (x<<1)\',number=N)
2.9485139846801758
>>> timeit(\'x=random.randint(65536); x // 2\',number=N)
2.490908145904541
>>> timeit(\'x=random.randint(65536); x / 2\',number=N)
2.4757170677185059
>>> timeit(\'x=random.randint(65536); x >> 1\',number=N)
2.2316000461578369
#include <stdio.h>
#include <chrono>
static const unsigned s_bc = 32;
static const unsigned long long s_p = 1ULL << s_bc;
static const unsigned long long s_hp = s_p / 2;
static unsigned long long s_f;
static unsigned long long s_fr;
static void fastDivInitialize(const unsigned d)
{
s_f = s_p / d;
s_fr = s_f * (s_p - (s_f * d));
}
static unsigned fastDiv(const unsigned n)
{
return (s_f * n + ((s_fr * n + s_hp) >> s_bc)) >> s_bc;
}
static bool fastDivCheck(const unsigned n,const unsigned d)
{
// 32 to 64 cycles latency on modern cpus
const unsigned expected = n / d;
// At least 10 cycles latency on modern cpus
const unsigned result = fastDiv(n);
if (result != expected)
{
printf(\"Failed for: %u/%u != %u\\n\",n,d,expected);
return false;
}
return true;
}
int main()
{
unsigned result = 0;
// Make sure to verify it works for your expected set of inputs
const unsigned MAX_N = 65535;
const unsigned MAX_D = 40000;
const double ONE_SECOND_COUNT = 1000000000.0;
auto t0 = std::chrono::steady_clock::now();
unsigned count = 0;
printf(\"Verifying...\\n\");
for (unsigned d = 1; d <= MAX_D; ++d)
{
fastDivInitialize(d);
for (unsigned n = 0; n <= MAX_N; ++n)
{
count += !fastDivCheck(n,d);
}
}
auto t1 = std::chrono::steady_clock::now();
printf(\"Errors: %u / %u (%.4fs)\\n\",count,MAX_D * (MAX_N + 1),(t1 - t0).count() / ONE_SECOND_COUNT);
t0 = t1;
for (unsigned d = 1; d <= MAX_D; ++d)
{
fastDivInitialize(d);
for (unsigned n = 0; n <= MAX_N; ++n)
{
result += fastDiv(n);
}
}
t1 = std::chrono::steady_clock::now();
printf(\"Fast division time: %.4fs\\n\",(t1 - t0).count() / ONE_SECOND_COUNT);
t0 = t1;
count = 0;
for (unsigned d = 1; d <= MAX_D; ++d)
{
for (unsigned n = 0; n <= MAX_N; ++n)
{
result += n / d;
}
}
t1 = std::chrono::steady_clock::now();
printf(\"Normal division time: %.4fs\\n\",(t1 - t0).count() / ONE_SECOND_COUNT);
getchar();
return result;
}
push rbp
mov rbp,rsp
mov DWORD PTR [rbp-4],edi
mov eax,DWORD PTR [rbp-4]
add eax,eax
pop rbp
ret
#include <stdio.h>
#include <time.h>
#define MULTIPLYINTBYMINUS(X,Y) (-((X >> 30) & 1)&(Y<<30))+(-((X >> 29) & 1)&(Y<<29))+(-((X >> 28) & 1)&(Y<<28))+(-((X >> 27) & 1)&(Y<<27))+(-((X >> 26) & 1)&(Y<<26))+(-((X >> 25) & 1)&(Y<<25))+(-((X >> 24) & 1)&(Y<<24))+(-((X >> 23) & 1)&(Y<<23))+(-((X >> 22) & 1)&(Y<<22))+(-((X >> 21) & 1)&(Y<<21))+(-((X >> 20) & 1)&(Y<<20))+(-((X >> 19) & 1)&(Y<<19))+(-((X >> 18) & 1)&(Y<<18))+(-((X >> 17) & 1)&(Y<<17))+(-((X >> 16) & 1)&(Y<<16))+(-((X >> 15) & 1)&(Y<<15))+(-((X >> 14) & 1)&(Y<<14))+(-((X >> 13) & 1)&(Y<<13))+(-((X >> 12) & 1)&(Y<<12))+(-((X >> 11) & 1)&(Y<<11))+(-((X >> 10) & 1)&(Y<<10))+(-((X >> 9) & 1)&(Y<<9))+(-((X >> 8) & 1)&(Y<<8))+(-((X >> 7) & 1)&(Y<<7))+(-((X >> 6) & 1)&(Y<<6))+(-((X >> 5) & 1)&(Y<<5))+(-((X >> 4) & 1)&(Y<<4))+(-((X >> 3) & 1)&(Y<<3))+(-((X >> 2) & 1)&(Y<<2))+(-((X >> 1) & 1)&(Y<<1))+(-((X >> 0) & 1)&(Y<<0))
#define MULTIPLYINTBYSHIFT(X,Y) (((((X >> 30) & 1)<<31)>>31)&(Y<<30))+(((((X >> 29) & 1)<<31)>>31)&(Y<<29))+(((((X >> 28) & 1)<<31)>>31)&(Y<<28))+(((((X >> 27) & 1)<<31)>>31)&(Y<<27))+(((((X >> 26) & 1)<<31)>>31)&(Y<<26))+(((((X >> 25) & 1)<<31)>>31)&(Y<<25))+(((((X >> 24) & 1)<<31)>>31)&(Y<<24))+(((((X >> 23) & 1)<<31)>>31)&(Y<<23))+(((((X >> 22) & 1)<<31)>>31)&(Y<<22))+(((((X >> 21) & 1)<<31)>>31)&(Y<<21))+(((((X >> 20) & 1)<<31)>>31)&(Y<<20))+(((((X >> 19) & 1)<<31)>>31)&(Y<<19))+(((((X >> 18) & 1)<<31)>>31)&(Y<<18))+(((((X >> 17) & 1)<<31)>>31)&(Y<<17))+(((((X >> 16) & 1)<<31)>>31)&(Y<<16))+(((((X >> 15) & 1)<<31)>>31)&(Y<<15))+(((((X >> 14) & 1)<<31)>>31)&(Y<<14))+(((((X >> 13) & 1)<<31)>>31)&(Y<<13))+(((((X >> 12) & 1)<<31)>>31)&(Y<<12))+(((((X >> 11) & 1)<<31)>>31)&(Y<<11))+(((((X >> 10) & 1)<<31)>>31)&(Y<<10))+(((((X >> 9) & 1)<<31)>>31)&(Y<<9))+(((((X >> 8) & 1)<<31)>>31)&(Y<<8))+(((((X >> 7) & 1)<<31)>>31)&(Y<<7))+(((((X >> 6) & 1)<<31)>>31)&(Y<<6))+(((((X >> 5) & 1)<<31)>>31)&(Y<<5))+(((((X >> 4) & 1)<<31)>>31)&(Y<<4))+(((((X >> 3) & 1)<<31)>>31)&(Y<<3))+(((((X >> 2) & 1)<<31)>>31)&(Y<<2))+(((((X >> 1) & 1)<<31)>>31)&(Y<<1))+(((((X >> 0) & 1)<<31)>>31)&(Y<<0))
int main()
{
int randomnumber=23;
int randomnumber2=23;
int checknum=23;
clock_t start,diff;
srand(time(0));
start = clock();
for(int i=0;i<1000000;i++)
{
randomnumber = rand() % 10000;
randomnumber2 = rand() % 10000;
checknum=MULTIPLYINTBYMINUS(randomnumber,randomnumber2);
if (checknum!=randomnumber*randomnumber2)
{
printf(\"s %i and %i and %i\",checknum,randomnumber,randomnumber2);
}
}
diff = clock() - start;
int msec = diff * 1000 / CLOCKS_PER_SEC;
printf(\"MULTIPLYINTBYMINUS Time %d milliseconds\",msec);
start = clock();
for(int i=0;i<1000000;i++)
{
randomnumber = rand() % 10000;
randomnumber2 = rand() % 10000;
checknum=MULTIPLYINTBYSHIFT(randomnumber,randomnumber2);
}
}
diff = clock() - start;
msec = diff * 1000 / CLOCKS_PER_SEC;
printf(\"MULTIPLYINTBYSHIFT Time %d milliseconds\",msec);
start = clock();
for(int i=0;i<1000000;i++)
{
randomnumber = rand() % 10000;
randomnumber2 = rand() % 10000;
checknum= randomnumber*randomnumber2;
if (checknum!=randomnumber*randomnumber2)
{
printf(\"s %i and %i and %i\",randomnumber2);
}
}
diff = clock() - start;
msec = diff * 1000 / CLOCKS_PER_SEC;
printf(\"normal * Time %d milliseconds\",msec);
return 0;
}
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