[英]How to make a memcpy inside of an thread of pthread?
我正在嘗試使用c ++中的pthreads做2個矩陣的總和。 我一直試圖將線程內部計算的總和的結果傳遞給我的主函數。
要添加的2個值在結構內部:
struct sum{
int value1;
int value2;
int result;
}typedef struct_sum;
並將包含值的結構作為參數傳遞給pthread_create()
以便在線程內部執行操作。
這是我的例程:
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個cout
我檢查我的值是否正確到達線程。 在最后一個cout
(退出線程之前),我檢查了結構的result元素的內存地址(因此,我可以看到它在main函數中具有相同的地址)。
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);
}
我有兩個問題:
即使結果在主線程和線程中具有相同的地址,當我將i
的值復制到main
函數中的((struct_sum *)sum)->result
, ((struct_sum *)sum)->result
result->result
仍為0。
當我取消對memcpy()
行的注釋時,線程根本不會運行,所以我不知道我做錯了什么。
我期望在我的main
功能中,語句std::cout << (result->result) <<std::endl
將返回操作結果,但當前值為0。
那么,如何在線程中正確執行memcpy()
?
您必須加入線程。 這意味着,等待他們完成。 您的工作方式基本上是啟動線程,而不給它保證的時間做任何事情。 除此之外,對API進行了一些重要更改,請檢查以下注釋:
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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