Question

我编写了这段代码，将2个内核线程绑定到不同的CPU：

#include<linux/kthread.h>
#include<linux/sched.h>
#include<linux/delay.h>
#include<linux/slab.h>

struct task_struct *task1;
struct task_struct *task2;
int cpu, data;

int thread_function_one(void *data)
{
    int ret = 10;
    printk(KERN_INFO "IN THREAD FUNCTION 1 \n");

    while(!kthread_should_stop()){
       cpu = get_cpu();
       put_cpu();
       printk("t1 cpu = %d\n",cpu);
       msleep(5000);
    }
    printk(KERN_INFO "EXIT from thread function 1\n");
    return ret;
}

int thread_function_two(void *data)
{
    int ret = 10;
    printk(KERN_INFO "IN THREAD FUNCTION 2 \n");

    while(!kthread_should_stop()){
        cpu = get_cpu();
        put_cpu();
        printk("t2 cpu = %d\n",cpu);
        msleep(6000);
    }
    printk(KERN_INFO "EXIT from thread function 2\n");
    return ret;
}

static int kernel_init(void)
{
   printk(KERN_INFO "module_init\n");

   cpu = get_cpu();
   put_cpu();
   printk("main thread cpu = %d \n",cpu);

   task1 = kthread_create(&thread_function_one,(void *)&data,"one");
   kthread_bind(task1, cpu);
   wake_up_process(task1);

   cpu = 5;
   task2 = kthread_create(&thread_function_two,(void *)&data,"two");
   kthread_bind(task2, cpu);
   wake_up_process(task2);

   return 0;
}

static void kernel_exit(void)
{
   kthread_stop(task1);
   kthread_stop(task2);
   printk(KERN_INFO "module_exit\n");
}

module_init(kernel_init);
module_exit(kernel_exit);
MODULE_AUTHOR("K_K");
MODULE_LICENSE("GPL");

但是当我在thread_function_one和thread_function_two中打印cpu id时，两者都打印与主线程相同的CPU。

我做错了什么？

这个编程的输出：

kernel: [  956.816510] module_init
kernel: [  956.816515] main thread cpu = 8 
kernel: [  956.816743] IN THREAD FUNCTION 1 
kernel: [  956.816748] t1 cpu = 8
kernel: [  956.817057] IN THREAD FUNCTION 2 
kernel: [  956.817062] t2 cpu = 8
kernel: [  961.815160] t1 cpu = 8
kernel: [  962.814649] t2 cpu = 8
kernel: [  966.816760] t1 cpu = 8
kernel: [  968.815711] t2 cpu = 8
kernel: [  971.818307] EXIT from thread function 1
kernel: [  974.816813] EXIT from thread function 2
kernel: [  974.816872] module_exit

Answer 1

如果“int cpu”变量不是全局变量，则模块可以正常工作。线程1修改此全局，线程2将自身绑定到同一个CPU。这是主线程，线程1和线程2之间的竞争条件。

无法将内核线程绑定到CPU

1 个答案: