編集履歴

質問編集履歴

メモリの解放について改善

2016/12/23 04:28

投稿

johejo

スコア12

test CHANGED Viewed

File without changes

test CHANGED Viewed

@@ -212,160 +212,166 @@
 	thread_id = ((struct ThreadArgs *) threadArgs) -> thread_id;
+	while ( *sum < *count_num) {
+		(*n)++;
+		printf("Thread%d %d\n", thread_id, *n);
+		for (i = 0; i < 50000000; i++) {}
+	}
 	free(threadArgs);
+}
+int main(int argc, char *argv[])
+{
+	int i, j, *n, *top, count_num, thread_num, sum;
+	clock_t time1, time0;
+	double diff;
+	struct ThreadArgs *threadArgs;
+	pthread_t mythread;
+	int thread_id;
+	if (argc != 3) {
+  		exit(1);
+  	}
+	count_num = atoi(argv[1]);
-	while ( *sum < *count_num) {
+	thread_num = atoi(argv[2]);
-		(*n)++;
-		printf("Thread%d %d\n", thread_id, *n);
+	printf("begin %d %d\n", count_num, thread_num);
+	n = (int *)malloc(sizeof(int) * thread_num);
+	top = n;
-		for (i = 0; i < 50000000; i++) {}
+	for (i = 0, n = top; i < thread_num; i++, n++) {
+		*n = 0;
 	}
+	sum = 0;
+	time0 = clock();
+	thread_id = 0;
+	for (i = 0, n = top; i < thread_num; i++, n++) {
+		if((threadArgs = (struct ThreadArgs *)malloc(sizeof(struct ThreadArgs))) == NULL){
+			fprintf(stderr, "malloc failed¥n"), exit(1);
+		}
+		threadArgs -> count_num = &count_num;
+		threadArgs -> n = n;
+		threadArgs -> sum = &sum;
+		threadArgs -> thread_id = thread_id;
+		if(pthread_create(&mythread, NULL, thread_func, (void *) threadArgs) != 0){
+			fprintf(stderr, "pthread_create() failed¥n"), exit(1);
+		}
+		thread_id++;
+	}
+	while (sum < count_num) {
+		pthread_mutex_lock(&MyMutex);
+		for (i = 0, sum = 0, n = top; i < thread_num; i++, n++) {
+			sum += *n;
+		}
+		pthread_mutex_unlock(&MyMutex);
+		//printf("%d %d\n", sum, count_num);
+	}
+	printf("sum %d\n", sum);
+	time1 = clock();
+	diff =(double)(time1 - time0) / CLOCKS_PER_SEC;
+	printf("%.3f[sec]\n", diff);
+	free(top);
+	return 0;
 }
-int main(int argc, char *argv[])
-{
-	int i, j, *n, *top, count_num, thread_num, sum;
-	clock_t time1, time0;
-	double diff;
-	struct ThreadArgs *threadArgs;
-	pthread_t mythread;
-	int thread_id;
-	if (argc != 3) {
-  		exit(1);
-  	}
-	count_num = atoi(argv[1]);
-	thread_num = atoi(argv[2]);
-	printf("begin %d %d\n", count_num, thread_num);
-	n = (int *)malloc(sizeof(int) * thread_num);
-	top = n;
-	for (i = 0, n = top; i < thread_num; i++, n++) {
-		*n = 0;
-	}
-	sum = 0;
-	time0 = clock();
-	thread_id = 0;
-	for (i = 0, n = top; i < thread_num; i++, n++) {
-		if((threadArgs = (struct ThreadArgs *)malloc(sizeof(struct ThreadArgs))) == NULL){
-			fprintf(stderr, "malloc failed¥n"), exit(1);
-		}
-		threadArgs -> count_num = &count_num;
-		threadArgs -> n = n;
-		threadArgs -> sum = &sum;
-		threadArgs -> thread_id = thread_id;
-		if(pthread_create(&mythread, NULL, thread_func, (void *) threadArgs) != 0){
-			fprintf(stderr, "pthread_create() failed¥n"), exit(1);
-		}
-		thread_id++;
-	}
-	while (sum < count_num) {
-		pthread_mutex_lock(&MyMutex);
-		for (i = 0, sum = 0, n = top; i < thread_num; i++, n++) {
-			sum += *n;
-		}
-		pthread_mutex_unlock(&MyMutex);
-		//printf("%d %d\n", sum, count_num);
-	}
-	printf("sum %d\n", sum);
-	time1 = clock();
-	diff =(double)(time1 - time0) / CLOCKS_PER_SEC;
-	printf("%.3f[sec]\n", diff);
-	return 0;
-}
 ```

ソースコードの誤りの訂正

2016/12/23 04:28

投稿

johejo

スコア12

test CHANGED Viewed

File without changes

test CHANGED Viewed

@@ -168,6 +168,14 @@
+static pthread_mutex_t MyMutex=PTHREAD_MUTEX_INITIALIZER;
+void *thread_func(void *threadArgs);
 struct ThreadArgs
 {
@@ -184,10 +192,6 @@
-void *thread_func(void *threadArgs);
 void *thread_func(void *threadArgs) {
 	int i, thread_id;

排他処理を追加しました。追加質問内容と実行例について加筆しました。

2016/12/23 04:24

投稿

johejo

スコア12

test CHANGED Viewed

File without changes

test CHANGED Viewed

@@ -32,6 +32,126 @@
+#実行例について
+スレッド数4で実行した場合
+$./a.out 100 4
+begin 100 4
+Thread0 1
+Thread1 1
+Thread2 1
+Thread3 1
+Thread3 2
+Thread1 2
+Thread2 2
+Thread0 2
+Thread1 3
+Thread3 3
+Thread2 3
+Thread0 3
+===中略===
+Thread3 24
+Thread1 24
+Thread2 24
+Thread0 24
+Thread3 25
+Thread1 25
+Thread0 25
+Thread2 25
+sum 100
+12.109[sec]
+スレッド数5で実行した場合
+$./a.out 100 5
+begin 100 5
+Thread0 1
+Thread1 1
+Thread2 1
+Thread3 1
+Thread1 2
+Thread3 2
+Thread0 2
+Thread2 2
+Thread1 3
+Thread3 3
+Thread0 3
+Thread2 3
+===中略===
+Thread1 95
+Thread3 95
+Thread2 96
+Thread0 97
+Thread1 96
+Thread3 96
+Thread2 97
+Thread0 98
+Thread1 97
+Thread4 1
+Thread3 97
+Thread2 98
+sum 391
+38.390[sec]
 ###該当のソースコード
 ```C
@@ -208,12 +328,16 @@
 	while (sum < count_num) {
+		pthread_mutex_lock(&MyMutex);
 		for (i = 0, sum = 0, n = top; i < thread_num; i++, n++) {
 			sum += *n;
 		}
+		pthread_mutex_unlock(&MyMutex);
 		//printf("%d %d\n", sum, count_num);
 	}
@@ -244,6 +368,16 @@
+##追加の質問内容
+なぜスレッド数5の実行例では、Thread 4(5つめのスレッド)がすぐに開始されないのでしょうか。
+##追記
+排他処理を追加しました。
+追加の質問内容と実行例について加筆しました。
 ###補足情報(言語/FW/ツール等のバージョンなど)