我如何同步这 9 个线程，以便它们在主线程之前执行？
我想检查大小为 9 的二维数组中行的有效性。每行应包含值（1 到 9）。为此，我在主线程中创建了一个名为“void* checksRows(void* arg)”的线程，并将其与 main 连接。然后线程checkingRows 又创建了9 个线程来检查每一行的有效性。
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Pthread_t id1;
pthread_mutex_t mut1;
int arr[9][9] = {
{6,2,4,5,3,9,1,8,7},
{6,6,9,7,2,8,6,3,4},
{8,3,7,6,1,4,2,9,5},
{1,4,3,8,6,5,7,2,9},
{9,5,8,2,4,7,3,6,1},
{7,6,2,3,9,1,4,5,8},
{3,7,1,9,5,6,8,4,2},
{4,9,6,1,8,2,5,7,3},
{2,8,5,4,7,3,9,1,6}
};
````````````````````````````````````
void* rowCheck(void* arg){
int* argument = (int*) arg;
int idx = *argument;
int count = 0;
for(int i = 0; i < 9; i++){
int temp = arr[idx][i];
count = 0;
for(int j = i; j < 9; j++){
if(arr[idx][j] == temp || arr[idx][j] <= 0 || arr[idx][j] >= 10){
count++;
}
if(count > 1){
pthread_mutex_lock(&mut1);
count = 0;
cout<<"ERROR at"<<arr[idx][j]<<endl;
pthread_mutex_unlock(&mut1);
break;
}
}
}
pthread_exit(NULL);
}
````````````````````````````````````
void* checkingRows(void* arg){
int *row = new int;
*row = 0;
for(int i = 0; i<gridSize; i++){
pthread_create(&workerIdRow[i], NULL, &rowCheck, row);
*row = *row + 1;
}
pthread_exit(NULL);
}
`````````````````````````````````
int main(){
pthread_mutex_init(&mut1, NULL);
pthread_create(&id1, NULL, &checkingRows, NULL);
pthread_join(id1,NULL);
retrun 0;
}
````````````````````````````````````
ERROR at 6
ERROR at 6
第一个创建的线程没用，因为你创建它并立即等待它的结束
相反，您不会等待其他 9 个线程的结束，因此您会在它们结束之前退出
当您启动子线程时，您没有同步以确保它们在为下一个线程修改它之前有时间读取行号，几个可以检查同一行。为此，您可以使用条件（仍然使用 C 线程库），或者您可以在每次给定索引的情况下提供一个新创建的int。
你有内存泄漏，因为你没有删除int，实际上在你的情况下在堆中分配它是没用的
来自你的代码的建议，如你所见，我直接将相关行的地址提供给每个线程，我删除了中间线程并等待9个线程结束，更多一些其他更改
#include <pthread.h>
#include <iostream>
pthread_mutex_t mut1;
int arr[9][9] = {
{6,2,4,5,3,9,1,8,7},
{6,6,9,7,2,8,6,3,4},
{8,3,7,6,1,4,2,9,5},
{1,4,3,8,6,5,7,2,9},
{9,5,8,2,4,7,3,6,1},
{7,6,2,3,9,1,4,5,8},
{3,7,1,9,5,6,8,4,2},
{4,9,6,1,8,2,5,7,3},
{2,8,5,4,7,3,9,1,6}
};
void * rowCheck(void* arg)
{
int * row = (int *) arg;
const char * result = "ok";
for(int i = 0; i < 9; i++){
int temp = row[i];
int count = 0;
for(int j = i; j < 9; j++){
if(row[j] == temp || row[j] <= 0 || row[j] >= 10){
count++;
}
if(count > 1){
result = "ko";
pthread_mutex_lock(&mut1);
std::cout<<"ERROR at "<< row[j]
<< " (row " << (row - arr[0])/9
<< " column " << j << ")" << std::endl;
pthread_mutex_unlock(&mut1);
break;
}
}
}
pthread_exit((void *) result);
}
int main()
{
pthread_t id[9];
pthread_mutex_init(&mut1, NULL);
for (int i = 0; i != 9; ++i) {
if (pthread_create(&id[i], NULL, &rowCheck, arr[i]) != 0) {
std::cerr << "error when creating thread " << i << std::endl;
return -1;
}
}
for (int i = 0; i != 9; ++i) {
void * retval;
pthread_join(id[i], &retval);
pthread_mutex_lock(&mut1);
std::cout << i << ":" << (char *) retval << std::endl;
pthread_mutex_unlock(&mut1);
}
return 0;
}
编译和执行：
/tmp % g++ -pedantic -Wextra -Wall c.cc -lpthread
/tmp % ./a.out
ERROR at 6 (row 1 column 1)
ERROR at 6 (row 1 column 6)
0:ok
1:ko
2:ok
3:ok
4:ok
5:ok
6:ok
7:ok
8:ok
另一种方法是通过专用于每个线程的已分配int提供索引（并在线程中删除它）：
#include <pthread.h>
#include <iostream>
pthread_mutex_t mut1;
int arr[9][9] = {
{6,2,4,5,3,9,1,8,7},
{6,6,9,7,2,8,6,3,4},
{8,3,7,6,1,4,2,9,5},
{1,4,3,8,6,5,7,2,9},
{9,5,8,2,4,7,3,6,1},
{7,6,2,3,9,1,4,5,8},
{3,7,1,9,5,6,8,4,2},
{4,9,6,1,8,2,5,7,3},
{2,8,5,4,7,3,9,1,6}
};
void * rowCheck(void* arg)
{
int rowNum = *((int*) arg);
delete (int*) arg;
int * row = arr[rowNum];
const char * result = "ok";
for(int i = 0; i < 9; i++){
int temp = row[i];
int count = 0;
for(int j = i; j < 9; j++){
if(row[j] == temp || row[j] <= 0 || row[j] >= 10){
count++;
}
if(count > 1){
result = "ko";
pthread_mutex_lock(&mut1);
std::cout<<"ERROR at "<< row[j]
<< " (row " << rowNum
<< " column " << j << ")" << std::endl;
pthread_mutex_unlock(&mut1);
break;
}
}
}
pthread_exit((void *) result);
}
int main()
{
pthread_t id[9];
pthread_mutex_init(&mut1, NULL);
for (int i = 0; i != 9; ++i) {
if (pthread_create(&id[i], NULL, &rowCheck, new int(i)) != 0) {
std::cerr << "error when creating thread " << i << std::endl;
return -1;
}
}
for (int i = 0; i != 9; ++i) {
void * retval;
pthread_join(id[i], &retval);
pthread_mutex_lock(&mut1);
std::cout << i << ":" << (char *) retval << std::endl;
pthread_mutex_unlock(&mut1);
}
return 0;
}
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