图4为排序阶段CPU的使用率,可以看到只有一个核达到了100%的利用率。下面为一个多线程(线程的数量为核的数量)的排序版本,每个线程只对1G数据中的一部分进行快速排序,排序完成后再由另外一个线程进行归并,将结果写入文件。
多线程排序代码如下:
/*multi_thread_sort.c*/
- /*
- * Author: Chaos Lee
- * Date: 2012-06-30
- * Description: load, merge , store data with single core, but sorting data with all the cores provided by the SMP
- */
- #include
- #include
- #include
- #include
- #include
- #include
- #include
- #include
- #include "../error.h"
- #include "timer.h"
- uint64_t * buffer = NULL;
- pthread_mutex_t counter_mutex = PTHREAD_MUTEX_INITIALIZER;
- pthread_cond_t merge_start = PTHREAD_COND_INITIALIZER;
- int cores_number;
- int counter;
- int uint64_compare(const void * ptr1,const void * ptr2)
- {
- return *((uint64_t *)ptr1) > *((uint64_t *)ptr2) ? 1 : *((uint64_t *)ptr1) < *((uint64_t *)ptr2) ? -1 : 0;
- }
- typedef struct segment_tag
- {
- uint64_t start;
- uint64_t end;
- }segment_t,*segment_p;
- void barrier()
- {
- int status;
- status = pthread_mutex_lock(&counter_mutex);
- if(0 != status)
- err_abort("locking error.",status);
- counter++;
- if(cores_number == counter)
- {
- pthread_cond_signal(&merge_start);
- }
- status = pthread_mutex_unlock(&counter_mutex);
- if(0 != status)
- err_abort("unlocking error.",status);
- }
- void * sort_thread_routin(void * args)
- {
- DPRINTF(("%s","sorting thread start...\n"));
- segment_p seg = (segment_p) args;
- assert(buffer != NULL);
- DPRINTF(("%s","begin to sort...\n"));
- qsort(buffer+seg->start,seg->end-seg->start,sizeof(uint64_t),uint64_compare);
- DPRINTF(("%s","Entering barrier...\n"));
- barrier();
- pthread_exit((void *)0);
- }
- void * merge_thread_routin(void * args)
- {
- int status,i,finish_count,elapsed_seconds;
- FILE * fp_result;
- uint64_t tmp;
- restart_timer();
- DPRINTF(("%s","merging thread start...\n"));
- fp_result = fopen("multi-result.dat","wb");
- while(cores_number != counter)
- {
- status = pthread_cond_wait(&merge_start,&counter_mutex);
- if(0 != status)
- err_abort("waiting condition error.",status);
- }
- elapsed_seconds = get_elapsed_time();
- fprintf(stdout,"sorting cost %d seconds.\n",elapsed_seconds);
- status = pthread_mutex_unlock(&counter_mutex);
- if(0 != status)
- err_abort("unlocking error.",status);
- DPRINTF(("begin to merge...\n"));
- finish_count = 0;
- segment_p segs = (segment_p) args;
- restart_timer();
- while(finish_count
- {
- int i,first=0,j;
- for(i=0;i
- {
- if( 0 == first)
- {
- if(segs[i].start
end) - {
- tmp = buffer[segs[i].start];
- j = i;
- first = 1;
- }
- }
- else
- {
- if(segs[i].start
end && buffer[segs[i].start] - {
- tmp = buffer[segs[i].start];
- j = i;
- }
- }
- }
- segs[j].start++;
- if(segs[j].start >= segs[j].end)
- {
- finish_count++;
- }
- fwrite(&tmp,sizeof(uint64_t),1,fp_result);
- }
- elapsed_seconds = get_elapsed_time();
- fprintf(stdout,"merging cost %d seconds.\n",elapsed_seconds);
- DPRINTF(("merging is over\n"));
- fclose(fp_result);
- pthread_exit((void *)0);
- }
- int main(int argc,char *argv[])
- {
- int elapsed_seconds,status,i;
- segment_p segments;
- pthread_t * sort_threads;
- pthread_t * merge_thread;
- uint64_t size,length,seg_len;
- FILE * fp;
- struct stat data_stat;
- cores_number = get_nprocs();
- status = stat("data.dat",&data_stat);
- if(0 != status)
- error_abort("stat file error.\n");
- size = data_stat.st_size;
- length = size / sizeof(uint64_t);
- seg_len = length / cores_number;
- buffer = (uint64_t *) malloc(size);
- if(NULL == buffer)
- {
- fprintf(stderr,"mallocing error.\n");
- exit(1);
- }
- fp = fopen("data.dat","rb");
- if(NULL == fp)
- {
- fprintf(stderr,"file open error.\n");
- exit(1);
- }
- start_timer();
- fread(buffer,size,1,fp);
- elapsed_seconds = get_elapsed_time();
- fprintf(stdout,"loading cost %d seconds\n",elapsed_seconds);
- segments = (segment_p)malloc(sizeof(segment_t)*cores_number);
- if(NULL == segments)
- {
- fprintf(stderr,"at %s:%d : %s",__FILE__,__LINE__,"malloc error.\n");
- exit(1);
- }
- for(i=0;i
- {
- segments[i].start = i * seg_len;
- if(i != cores_number-1)
- segments[i].end = (i + 1 ) * seg_len;
- else
- segments[i].end = length;
- }
- sort_threads = (pthread_t *)malloc(sizeof(pthread_t) * cores_number);
- if(NULL == sort_threads)
- {
- fprintf(stderr,"at %s:%d :%s",__FILE__,__LINE__,"malloc failuer.\n");
- exit(1);
- }
- merge_thread = (pthread_t *)malloc(sizeof(pthread_t));
- if(NULL == merge_thread)
- {
- fprintf(stderr,"at %s:%d :%s",__FILE__,__LINE__,"malloc failuer.\n");
- exit(1);
- }
- for(i=0;i
- {
- status = pthread_create(&sort_threads[i],NULL,sort_thread_routin,(void *)&segments[i]);
- if(0 != status)
- err_abort("creating threads faulire.\n",status);
- }
- status = pthread_create(merge_thread,NULL,merge_thread_routin,(void *)segments);
- if(0 != status)
- err_abort("creating thread faulier.\n",status);
- for(i=0;i
- {
- status = pthread_join(sort_threads[i],NULL);
- if(0 != status)
- err_abort("joining threads error.\n",status);
- }
- status = pthread_join(*merge_thread,NULL);
- if(0 != status)
- err_abort("joining thread error.\n",status);
- free(buffer);
- fclose(fp);
- return 0;
- }
再编译运行下,以下为测试结果:
- [lichao@sg01 thread_power]$ gcc multi_thread_sort.c -o multi_thread_sort timer.o -lpthread
- [lichao@sg01 thread_power]$ ./multi_thread_sort
- loading cost 14 seconds
- sorting cost 22 seconds.
- merging cost 44 seconds.
下图5为多线程排序时CPU的利用率,可以看到CPU的四个核都已经达到100%的利用率,即:硬件没有白投资:D。当然排序的时间效果也很好,几乎达到了之前的4倍的加速比。另外可以看到文件的加载速度和回写速度也有所提高,这点也是让我比较疑惑的。下面再次运行单线程排序版本。
图5 排序阶段CPU的利用率
- [lichao@sg01 thread_power]$ ./single_thread_sort
- loading cost 17 seconds
- sorting cost 81 seconds
- writing results cost 12 seconds
可以看到加载速度和回写速度有了显著的提升,虽然排序时间还是没有多大变化。
再次运行多线程排序版本试试:
- [lichao@sg01 thread_power]$ ./multi_thread_sort
- loading cost 31 seconds
- sorting cost 22 seconds.
- merging cost 23 seconds.
加载速度又延长了,排序速度几乎不变,回写速度也提高了不少。我想这主要是因为文件系统本身提供了缓冲的作用,即上次用过的文件可以放在交换区,便于迅速载入内存吧。这样第二次使用的时候,由于这些文件还存放在交换区中,所以以很高的速度传入内存中。回写的原理应该也一样。对于1G的文件回写到内存,只用了23s,大致的回写速度为50MB/s
假设文件系统一直起作用,并能达到第二次实验的效果,即分块排序22s,归并排序并回写文件系统23s,那么计算和归并回写是能够重合的。对于200G的文件A来说,分块排序的处理时间大致为:200*22s =~1.2h,就扩大为1小时15分钟吧。这样对文件B来说也差不多为1小时15分钟,一共需要2个半小时,接下来开始归并比较了,假设文件的缓冲系统能够启作用,即速度能达到50MB/s,这样,对于2个200G的文件都需要在内存中过一遍,大致时间应该为400*10^3/50 = 8000s,大致为2小时15分钟,所以加上前面的2个半小时,对于2个200G的文件寻找相同值共需要的时间为 5个小时左右,至少比300万年好点。
PS: =~这个符号表示约等于。