Question

所以，我大约99％肯定我已经实施了一些错误的东西，但这是交易。

我一直在玩Grand Central Dispatch，并组织了一个计算MD5哈希的实验。我正在使用i5运行macbook air，因此可以使用4个内核。这将使我相信使用Grand Central Dispatch来计算哈希值将大约快4倍。但是，出于某种原因，它似乎更慢。

以下代码

使用GCD

#include <stdio.h>
#include <time.h>
#import <CommonCrypto/CommonDigest.h>
#import <dispatch/dispatch.h>

int main (int argc, const char * argv[])
{

    int i,j,k,l,a;
    int num_chars = 4, total;
    clock_t start, end;
    double elap;

    printf("Calculating hashes for %d chars\n", num_chars);

    total = num_chars ^ 64;

    printf("Calculating %d hashes\n", total);

    dispatch_queue_t queue = dispatch_get_global_queue(0,0);
    dispatch_queue_t main = dispatch_get_main_queue();
    dispatch_group_t group = dispatch_group_create();

    start = clock();

   printf("Starting calculation queue\n");
   for(i=0;i<62;i++) {
        for(j=0;j<62;j++) {
            for(k=0;k<62;k++) {
                for(l=0;l<62;l++) {

                    dispatch_group_async(group, queue, ^{

                        char *letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
                        char buffer[10];
                        char out[100];
                        unsigned char hash[16];

                        sprintf(buffer, "%c%c%c%c", letters[i], letters[j], letters[k], letters[l]);
                        CC_MD5(buffer, strlen(buffer), hash); 

                    });


                }
            }
        }
    }

   printf("Finished calculation queue\n");
   dispatch_group_wait(group, DISPATCH_TIME_FOREVER);

    end = clock();

    elap = ((double) end - start) / CLOCKS_PER_SEC;

    printf("Time taken  %2f\n", elap);
    return 0;
}

编译并运行......

gcc -o a.out main.c
./a.out
Calculating hashes for 4 chars
Calculating 68 hashes
Starting calculation queue
Finished calculation queue
Time taken  35.193133

查看Activity Monitor，我可以看到脚本运行时所有4个核心最大化。

现在，注释掉调度......

#include <stdio.h>
#include <time.h>
#import <CommonCrypto/CommonDigest.h>
#import <dispatch/dispatch.h>

int main (int argc, const char * argv[])
{

    int i,j,k,l,a;
    int num_chars = 4, total;
    clock_t start, end;
    double elap;

    printf("Calculating hashes for %d chars\n", num_chars);

    total = num_chars ^ 64;

    printf("Calculating %d hashes\n", total);

    dispatch_queue_t queue = dispatch_get_global_queue(0,0);
    dispatch_queue_t main = dispatch_get_main_queue();
    dispatch_group_t group = dispatch_group_create();

    start = clock();

   printf("Starting calculation queue\n");
   for(i=0;i<62;i++) {
        for(j=0;j<62;j++) {
            for(k=0;k<62;k++) {
                for(l=0;l<62;l++) {

                    //dispatch_group_async(group, queue, ^{

                        char *letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
                        char buffer[10];
                        char out[100];
                        unsigned char hash[16];

                        sprintf(buffer, "%c%c%c%c", letters[i], letters[j], letters[k], letters[l]);
                        CC_MD5(buffer, strlen(buffer), hash); 

                    //});




   }
        }
    }
}

printf("Finished calculation queue\n");
//dispatch_group_wait(group, DISPATCH_TIME_FOREVER);

end = clock();

elap = ((double) end - start) / CLOCKS_PER_SEC;

printf("Time taken  %2f\n", elap);
return 0;
}

编译并运行

gcc -o b.out main.c
./b.out
Calculating hashes for 4 chars
Calculating 68 hashes
Starting calculation queue
Finished calculation queue
Time taken  7.511273

查看Activity Monitor，它只显示脚本运行时的1个核心活动。

Answer 1

在调度中可能完成的工作太少，以至于它不会使调度的开销变得有价值。我会尝试增加每次调度的工作量。我不知道这是否会有所帮助，但请尝试：

将调度移动几个循环，或许将k或j循环包裹在调度块中，以使其执行更多工作。

移除电话sprintf和strlen。实际上，该块可以简化为：

static const char *letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
unsigned char hash[16];
char buffer[4] = { letters[i], letters[j], letters[k], letters[l] };

CC_MD5(buffer, sizeof buffer, hash);

Grand Central dispatch给我一个较慢的执行时间

1 个答案: