我试图通过将数组拆分为大小为10的段,将1000个元素的整数数组(其中每个元素为1(与pthread库相加。因此,有效地使用了100个线程来实现这一点。此并行操作的结果与预期的一样(1000(。但有趣的是,在第一次调用pthread_join()
之后,我在创建线程之前计算的顺序和被设置为零。不确定我是否遗漏了什么。有人能在这里发现虫子吗?
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#define SEGMENT_SIZE 10
#define NUM_THREADS 100
int *array = NULL;
void* segment_sum(void *args)
{
int index = (int)args;
int sum = 0;
for (int i = index * SEGMENT_SIZE; i < (index + 1) * SEGMENT_SIZE; i++) {
sum += array[i];
}
return (void *)sum;
}
int main()
{
pthread_t thread[NUM_THREADS];
int res = 0;
int seq_res = 0;
int par_res = 0;
array = calloc(1, sizeof(int) * NUM_THREADS * SEGMENT_SIZE);
for (int i = 0; i < NUM_THREADS * SEGMENT_SIZE; i++) {
array[i] = 1;
seq_res += 1;
}
for (int i = 0; i < NUM_THREADS; i++) {
res = pthread_create(&thread[i], NULL, segment_sum, (void *)i);
if (res != 0) {
printf("nError creating new thread");
}
}
printf("nindex = %d", seq_res); // the sequential sum here is 1000
for (int i = 0; i < NUM_THREADS; i++) {
int sum = 0;
res = pthread_join(thread[i], (void **)&sum);
if (res != 0) {
printf("nError creating new thread");
}
printf("nindex = %d", seq_res); // Here it is becoming zero!!!
par_res += sum;
}
printf("nmultithreaded sum: %d single threaded sum: %dn", par_res, seq_res);
}
编译程序时,请尽可能消除警告,因为它们经常指出不可移植的行为或隐藏错误。这里的汇编指出了以下几点:
pte.c: In function 'segment_sum':
pte.c:11:21: warning: cast from pointer to integer of different size [-Wpointer-to-int-cast]
11 | int index = (int)args;
| ^
pte.c:18:16: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
18 | return (void *)sum;
| ^
pte.c: In function 'main':
pte.c:36:69: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]
36 | res = pthread_create(&thread[i], NULL, segment_sum, (void *)i);
| ^
传递给线程的参数是将指针强制转换为"指针";int";。是的建议通过";int";。因此,您可以定义每个线程上下文:
struct thd_ctx {
pthread_t thread;
int index;
int sum;
};
pthread_join((传递一个指针的地址,该指针将获得地址线程将其结果存储到的内存位置。螺纹必须返回此内存位置的地址,而不是存储在其中的值。此外,线程不应返回自动变量的地址(即在其堆栈中(,因为它是未指定的。结果必须是地址直接或通过pthread_exit((返回的全局变量(或从连接线程可见的"某物"(。在该程序的增强中;sum";线程上下文中的字段:
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <errno.h>
#define SEGMENT_SIZE 10
#define NUM_THREADS 100
int *array = NULL;
struct thd_ctx {
pthread_t thread;
int index;
int sum;
};
void *segment_sum(void *args)
{
int i;
struct thd_ctx *ctx = (struct thd_ctx *)args;
ctx->sum = 0;
for (i = ctx->index * SEGMENT_SIZE; i < (ctx->index + 1) * SEGMENT_SIZE; i++) {
ctx->sum += array[i];
}
return (void *)&(ctx->sum);
}
int main(void)
{
struct thd_ctx thd_ctx[NUM_THREADS];
int res = 0;
int seq_res = 0;
int par_res = 0;
int i;
array = calloc(1, sizeof(int) * NUM_THREADS * SEGMENT_SIZE);
if (!array) {
fprintf(stderr, "calloc(): error %dn", errno);
return 1;
}
for (i = 0; i < NUM_THREADS * SEGMENT_SIZE; i++) {
array[i] = 1;
seq_res += 1;
}
for (i = 0; i < NUM_THREADS; i++) {
thd_ctx[i].index = i;
res = pthread_create(&(thd_ctx[i].thread), NULL, segment_sum, (void *)&(thd_ctx[i]));
if (res != 0) {
fprintf(stderr, "Error %d creating new thread#%dn", res, i);
free(array);
return 1;
}
}
printf("Index = %dn", seq_res); // the sequential sum here is 1000
for (i = 0; i < NUM_THREADS; i++) {
int *sum = 0;
res = pthread_join(thd_ctx[i].thread, (void **)&(sum));
if (res != 0) {
printf("Error %d joining thread#%d", res, i);
free(array);
return 1;
}
par_res += *sum;
printf("sum = %dn", par_res);
}
printf("nMultithreaded sum: %d single threaded sum: %dn", par_res, seq_res);
free(array);
return 0;
}