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我在程序中創建了5個線程,並分別爲它們分配了ID 1,2,3,4,5。每個線程都會嘗試訪問Next_ID。當一個線程獲取Next_ID時,它會將其ID與Next_ID進行比較。如果它匹配我打印輪到我打開而其他打印不輪到我。並遞增1 NEXT_ID如果NEXT_ID達到6將其重置回1。但是我的代碼打印如下:編程線程
My Turn: 1
Not My Turn: 3
My Turn: 4
My Turn: 5
Not My Turn: 2
Not My Turn: 1
Not My Turn: 3
My Turn: 4
My Turn: 5
Not My Turn: 2
Not My Turn: 1
Not My Turn: 3
My Turn: 4
My Turn: 5
Not My Turn: 2
Not My Turn: 1
Not My Turn: 3
My Turn: 4
My Turn: 5
Not My Turn: 2
Not My Turn: 1
Not My Turn: 3
My Turn: 4
My Turn: 5
預期的輸出應該是:
My Turn: 1
not my Turn: 2
not my Turn: 3
not my Turn: 4
not my Turn: 5
not my Turn: 1
My Turn: 2
not my Turn: 3
not my Turn: 4
not my Turn: 5
我的代碼:
#include <stdio.h>
#include <pthread.h>
#include <stdlib.h>
#include <unistd.h>
#define NUM_THREADS 5
int Next_ID = 1;
pthread_t threads[NUM_THREADS];
pthread_mutex_t mutex;
typedef struct threadArgs {
int threadId;
int numOfCalls;
} ThreadArgs;
ThreadArgs threadArgsArray[NUM_THREADS];
void * printThread(void *pThreadArgs) {
ThreadArgs *threadArgs = (ThreadArgs *) pThreadArgs;
int *threadId = &threadArgs->threadId;
for(int i = 0; i < 20; i++) {
pthread_mutex_lock(&mutex);
if (Next_ID == *threadID) {
printf("My Turn: %d\n", *threadId);
} else {
printf("Not My Turn: %d\n", *threadId);
}
Next_ID = (*threadId == 5) ? 1 : *threadId + 1;
pthread_mutex_unlock (&mutex);
}
pthread_exit(NULL);
}
void * printThread(void *);
int main(int argc, char const *argv[]) {
pthread_attr_t attr;
void *status;
pthread_mutex_init(&mutex, NULL);
pthread_attr_init(&attr);
pthread_attr_setdetachstate(&attr, PTHREAD_CREATE_JOINABLE);
for (int i = 0; i < NUM_THREADS; i++) {
threadArgsArray[i].threadId = i + 1;
int rc = pthread_create(&threads[i], &attr, printThread,
&threadArgsArray[i]);
}
for (int i = 0; i < NUM_THREADS; i++) {
int rc = pthread_join(threads[i], &status);
}
pthread_attr_destroy(&attr);
pthread_mutex_destroy(&mutex);
pthread_exit(NULL);
}
你肯定是你的真實和精確的輸出?它看起來太經常而不真實,並且看起來不符合您所看到的代碼的行爲方式。 – kaylum
這樣比較好。原因是因爲你的假設是錯誤的。你的代碼中沒有任何東西可以保證線程2會在線程1之後立即執行,等等。也就是說,你有互斥,但你沒有排序。 – kaylum
我該怎麼做? – user456