我是并发编程的新手,所以很好。我有一个基本的顺序程序(用于家庭作业),我正试图将它变成一个多线程程序。我不确定我的第二个共享变量是否需要锁。线程应该修改我的变量,但从不读取它们。应该读取的唯一时间计数是在产生我所有线程的循环完成分发密钥之后。
#define ARRAYSIZE 50000
#include <pthread.h>
#include <stdlib.h>
#include <stdio.h>
#include <sys/time.h>
void binary_search(int *array, int key, int min, int max);
int count = 0; // count of intersections
int l_array[ARRAYSIZE * 2]; //array to check for intersection
int main(void)
{
int r_array[ARRAYSIZE]; //array of keys
int ix = 0;
struct timeval start, stop;
double elapsed;
for(ix = 0; ix < ARRAYSIZE; ix++)
{
r_array[ix] = ix;
}
for(ix = 0; ix < ARRAYSIZE * 2; ix++)
{
l_array[ix] = ix + 500;
}
gettimeofday(&start, NULL);
for(ix = 0; ix < ARRAYSIZE; ix++)
{
//this is where I will spawn off separate threads
binary_search(l_array, r_array[ix], 0, ARRAYSIZE * 2);
}
//wait for all threads to finish computation, then proceed.
fprintf(stderr, "%d\n", count);
gettimeofday(&stop, NULL);
elapsed = ((stop.tv_sec - start.tv_sec) * 1000000+(stop.tv_usec-start.tv_usec))/1000000.0;
printf("time taken is %f seconds\n", elapsed);
return 0;
}
void binary_search(int *array, int key, int min, int max)
{
int mid = 0;
if (max < min) return;
else
{
mid = (min + max) / 2;
if (array[mid] > key) return binary_search(array, key, min, mid - 1);
else if (array[mid] < key) return binary_search(array, key, mid + 1, max);
else
{
//this is where I'm not sure if I need a lock or not
count++;
return;
}
}
}