我刚刚开始学习汇编并制作一些自定义循环，以使用 C++ 的 asm{} 主体与 C-Free 5.0 中的 Digital-Mars 编译器交换两个变量
启用 -o（优化）
并得到了结果：
time of for-loop(cycles) 844
time of while-loop(cycles) 735
time of custom-loop-1(cycles) 562
time of custom-loop-2(cycles) 469
我找不到 Digital-Mars 编译器“asm 输出”选项进行比较。构建选项中没有其他优化选项。我应该改变我的编译器吗？如果是，是哪一个？你能看看下面的代码并告诉我为什么自定义循环更快吗？
这是for循环的标准：
t1=clock();
for(int i=0;i<200000000;i++)
{
temp=a;//instruction 1
a=b;//instruction 2
b=temp;//3 instructions total
}
t2=clock();
printf("\n time of for-loop(increasing) %i \n",(t2-t1));
这是标准的while循环：
t1=clock();
while(j<200000000)
{
temp=a;//again it is three instructions
a=b;
b=temp;
j++;
}
t2=clock();
printf("\n time of while-loop(cycles) %i \n",(t2-t1));
这是我的自定义循环 1：
t1=clock();
j=200000000;//setting the count
__asm
{
pushf //backup
push eax //backup
push ebx //backup
push ecx //backup
push edx //backup
mov ecx,0 //init of loop range(0 to 200000000)
mov edx,j
do_it_again: //begin to loop
mov eax,a //basic swap steps between cpu and mem(cache)
mov ebx,b
mov b,eax
mov a,ebx //four instructions total
inc ecx // j++
cmp ecx,edx //i<200000000 ?
jb do_it_again // end of loop block
pop edx //rolling back to history
pop ecx
pop ebx
pop eax
popf
}
t2=clock();
printf("\n time of custom-loop-1(cycles) %i \n",(t2-t1));
这是我的第二个自定义循环：
t1=clock();
j=200000000;//setting the count
__asm
{
pushf //backup
push eax
push ebx
push ecx
push edx
mov ecx,0 //init of loop range(0 to 200000000)
mov edx,j
mov eax,a //getting variables to registers
mov ebx,b
do_it_again2: //begin to loop
//swapping with using only 2 variables(only in cpu)
sub eax,ebx //a is now a-b
add ebx,eax //b is now a
sub eax,ebx //a is now -b
xor eax,80000000h //a is now b and four instructions total
inc ecx // j++
cmp ecx,edx //i<200000000 ?
jb do_it_again2 // end of loop block
pop edx //rollback
pop ecx
pop ebx
pop eax
popf
}
t2=clock();
printf("\n time of custom-loop-2(cycles) %i \n",(t2-t1));
完整代码：
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
int main()
{
int j=0;
int a=0,b=0,temp=0;
srand(time(0));
time_t t1=0;
time_t t2=0;
t1=clock();
for(int i=0;i<200000000;i++)
{
temp=a;//instruction 1
a=b;//instruction 2
b=temp;//3 instructions total
}
t2=clock();
printf("\n time of for-loop(cycles) %i \n",(t2-t1));
t1=clock();
while(j<200000000)
{
temp=a;//again it is three instructions
a=b;
b=temp;
j++;
}
t2=clock();
printf("\n time of while-loop(cycles) %i \n",(t2-t1));
t1=clock();
j=200000000;//setting the count
__asm
{
pushf //backup
push eax //backup
push ebx //backup
push ecx //backup
push edx //backup
mov ecx,0 //init of loop range(0 to 200000000)
mov edx,j
do_it_again: //begin to loop
mov eax,a //basic swap steps between cpu and mem(cache)
mov ebx,b
mov b,eax
mov a,ebx //four instructions total
inc ecx // j++
cmp ecx,edx //i<200000000 ?
jb do_it_again // end of loop block
pop edx //rolling back to history
pop ecx
pop ebx
pop eax
popf
}
t2=clock();
printf("\n time of custom-loop-1(cycles) %i \n",(t2-t1));
t1=clock();
j=200000000;//setting the count
__asm
{
pushf //backup
push eax
push ebx
push ecx
push edx
mov ecx,0 //init of loop range(0 to 200000000)
mov edx,j
mov eax,a //getting variables to registers
mov ebx,b
do_it_again2: //begin to loop
//swapping with using only 2 variables(only in cpu)
sub eax,ebx //a is now a-b
add ebx,eax //b is now a
sub eax,ebx //a is now -b
xor eax,80000000h //a is now b and four instructions total
inc ecx // j++
cmp ecx,edx //i<200000000 ?
jb do_it_again2 // end of loop block
pop edx //rollback
pop ecx
pop ebx
pop eax
popf
}
t2=clock();
printf("\n time of custom-loop-2(cycles) %i \n",(t2-t1));
return 0;
}
我只是在学习 c++ 和汇编，想知道事情是怎么回事。谢谢
windows xp, pentium 4 (2 GHz) Digital-Mars in C-Free