[英]How to make a memcpy inside of an thread of pthread?
我正在尝试使用c ++中的pthreads做2个矩阵的总和。 我一直试图将线程内部计算的总和的结果传递给我的主函数。
要添加的2个值在结构内部:
struct sum{
int value1;
int value2;
int result;
}typedef struct_sum;
并将包含值的结构作为参数传递给pthread_create()
以便在线程内部执行操作。
这是我的例程:
void * routine(void * sum) {
std::cout<<((struct_sum *)sum)->value1 + ((struct_sum *)sum)->value2<<std::endl;
std::cout<<((struct_sum *)sum)->value1<<std::endl;
std::cout<<((struct_sum *)sum)->value2<<std::endl;
int i = (((struct_sum *) sum)->value1 + ((struct_sum *) sum)->value2);
// memcpy(&(((struct_sum *)sum)->result), reinterpret_cast<const void *>(i), sizeof(i));
((struct_sum *)sum)->result = i;
std::cout<<&(((struct_sum *)sum)->result)<<std::endl;
pthread_exit(nullptr);
}
在前3个cout
我检查我的值是否正确到达线程。 在最后一个cout
(退出线程之前),我检查了结构的result元素的内存地址(因此,我可以看到它在main函数中具有相同的地址)。
主要功能如下:
int main(int argc, char * argv[]) {
int mat_1[ROW_SIZE][COLUMN_SIZE] = {{1, 2},
{6, 7}};
int mat_2[ROW_SIZE][COLUMN_SIZE] = {{3, 15},
{9, 14}};
int mat_result[ROW_SIZE][COLUMN_SIZE];
int mat_size = sizeof(mat_1) / sizeof(int);
int row_size = sizeof(mat_1) / sizeof(mat_1[0]);
int column_size = sizeof(mat_1[0]) / sizeof(int);
pthread_t threads[mat_size];
int thread_number = 0;
int thread_handler;
for (int row = 0; row < row_size; row++) {
for (int column = 0; column < column_size; column++) {
struct_sum *result;
result = static_cast<struct_sum *>(malloc(sizeof(struct_sum)));
result->value1 = mat_1[row][column];
result->value2 = mat_2[row][column];
result->result = 0;
thread_handler = pthread_create(&threads[thread_number], nullptr, routine, result);
if(thread_handler) return(-1);
std::cout << &(result->result)<<std::endl;
thread_number++;
mat_result[row][column] = result->result;
// free(result);
}
}
pthread_exit(nullptr);
}
我有两个问题:
即使结果在主线程和线程中具有相同的地址,当我将i
的值复制到main
函数中的((struct_sum *)sum)->result
, ((struct_sum *)sum)->result
result->result
仍为0。
当我取消对memcpy()
行的注释时,线程根本不会运行,所以我不知道我做错了什么。
我期望在我的main
功能中,语句std::cout << (result->result) <<std::endl
将返回操作结果,但当前值为0。
那么,如何在线程中正确执行memcpy()
?
您必须加入线程。 这意味着,等待他们完成。 您的工作方式基本上是启动线程,而不给它保证的时间做任何事情。 除此之外,对API进行了一些重要更改,请检查以下注释:
void * routine(void * sum) {
int i = (((struct_sum *) sum)->value1 + ((struct_sum *) sum)->value2);
((struct_sum *)sum)->result = i;
// notice you don't need memcpy(), in fact...
// but you could use it here if you want... it won't fail.
// you have to use this function so it return your result to the main thread.
pthread_exit(sum);
}
int main(int argc, char * argv[]) {
int mat_1[ROW_SIZE][COLUMN_SIZE] = {{1, 2},
{6, 7}};
int mat_2[ROW_SIZE][COLUMN_SIZE] = {{3, 15},
{9, 14}};
int mat_result[ROW_SIZE][COLUMN_SIZE];
int mat_size = sizeof(mat_1) / sizeof(int);
int row_size = sizeof(mat_1) / sizeof(mat_1[0]);
int column_size = sizeof(mat_1[0]) / sizeof(int);
pthread_t threads[mat_size];
int thread_number = 0;
int thread_handler;
for (int row = 0; row < row_size; row++) {
for (int column = 0; column < column_size; column++) {
struct_sum *result;
result = static_cast<struct_sum *>(malloc(sizeof(struct_sum)));
result->value1 = mat_1[row][column];
result->value2 = mat_2[row][column];
result->result = 0;
thread_handler = pthread_create(&threads[thread_number], nullptr, routine, result);
if(thread_handler) return(-1);
// std::cout << &(result->result)<<std::endl;
thread_number++;
// forget this line below
// mat_result[row][column] = result->result;
}
}
// here you wait for the threads to JOIN
// here it means they actually "finished" their job
for (int i = 0; i < mat_size; i++)
{
struct_sum *result;
// here you wait for the threads to finish their job
// add something to your struct to "identify" the thread, so
// you can figure out where in the final matrix you put the result
pthread_join(threads[i], (void**)&result);
std::cout << result->result << "\n";
// this will print the correct sums: 4, 17, 15 and 21
// notice: it will be printed in ANY order, once you
// don't know which thread will finish first
// but result->result has the... result you need!
// you have to figure out how to fit this result in your matrix.
// but this is out of scope of the question
// and you can do yourself. have fun! :-)
// here you can free the result, you already got the value!
free(result);
}
// you don't need this line below... this goes to routine()
// pthread_exit(nullptr);
return 0;
}
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