我的老师给了我应该编译的代码。应该使用线程来计算PI。然后,我们应该对程序进行计时和/或在代码中添加适当的计时系统调用。运行不同的线程并创建一个比较时间和线程数的图表...问题是,我不知道如何解决此问题。我从来没有处理过类似的事情,我自己也没有构建代码,所以我对如何解决它感到困惑。
编译器说问题出在:
rand_no_y = (double) (rand_r( &seed ))/(double)RAND_MAX;
^
和
rand_no_x = (double) (rand_r( &seed ))/(double)RAND_MAX;
整个代码如下:
#include <pthread.h>
#include <stdlib.h>
#include <stdio.h>
#define MAX_THREADS 512
void *compute_pi( void * );
//int sample_points;
int total_hits;
int total_misses;
int hits[ MAX_THREADS ];
int sample_points;
int sample_points_per_thread;
int num_threads;
int main( int argc, char *argv[] )
{
/* local variables */
int ii;
int retval;
pthread_t p_threads[MAX_THREADS];
pthread_attr_t attr;
double computed_pi;
/* initialize local variables */
retval = 0;
pthread_attr_init( &attr );
pthread_attr_setscope( &attr, PTHREAD_SCOPE_SYSTEM );
/* parse command line arguments into sample points and number of threads */
/* there is no error checking here!!!!! */
sample_points = atoi(argv[1]);
num_threads = atoi(argv[2]);
/* uncomment this block if you want interactive input!!!! */
/* if so...comment out the two statements above */
/*
printf( "Enter number of sample points: " );
scanf( "%d", &sample_points );
printf( "Enter number of threads: " );
scanf( "%d%", &num_threads );
*/
total_hits = 0;
sample_points_per_thread = sample_points / num_threads;
for( ii=0; ii<num_threads; ii++ )
{
hits[ii] = ii;
pthread_create( &p_threads[ ii ], &attr, compute_pi, (void *) &hits[ii] );
}
for( ii=0; ii<num_threads; ii++ )
{
pthread_join( p_threads[ ii ], NULL );
total_hits += hits[ ii ];
}
computed_pi = 4.0 * (double) total_hits / ((double) (sample_points));
printf( "Computed PI = %lf\n", computed_pi );
/* return to calling environment */
return( retval );
}
void *compute_pi( void *s )
{
int seed;
int ii;
int *hit_pointer;
int local_hits;
double rand_no_x;
double rand_no_y;
hit_pointer = (int *) s;
seed = *hit_pointer;
local_hits = 0;
for( ii=0; ii < sample_points_per_thread; ii++ )
{
rand_no_x = (double) (rand_r( &seed ))/(double)RAND_MAX;
rand_no_y = (double) (rand_r( &seed ))/(double)RAND_MAX;
if(((rand_no_x - 0.5) * (rand_no_x - 0.5) +
(rand_no_y - 0.5) * (rand_no_y - 0.5)) < 0.25)
local_hits++;
seed *= ii;
}
*hit_pointer = local_hits;
pthread_exit(0);
}
答案 0 :(得分:0)