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我想用定時器多次調用一個函數。 爲了實現這個，我使用了this的例子。它的默認工作沒有任何錯誤。內核模塊定時器殺死系統
當我將此代碼片段集成到我的源代碼中時，它導致內核恐慌（？）。發生長時間錯誤，虛擬機停止工作 。
印在控制檯上的錯誤：
[ 1130.520474] [<c12bf6ae>] ? no_context+0x14b/0x155
[ 1130.520474] [<c12bf7ca>] ? bad_area_nosemaphore+0xa/0xc
[ 1130.520474] [<c12c7370>] ? do_page_fault+0x1bf/0x35e
[ 1130.520474] [<c100d6bf>] ? show_trace_log_lvl+0x37/0x3d
[ 1130.520474] [<c12c71b1>] ? vmalloc_fault+0x87/0x87
[ 1130.520474] [<c12c5377>] ? error_code+0x67/0x6c
[ 1130.520474] [<c12c306e>] ? __schedule+0x31/0x5a4
[ 1130.520474] [<c100fa07>] ? sched_clock+0x9/0xd
[ 1130.520474] [<c1051896>] ? sched_clock_local+0x10/0x14b
[ 1130.520474] [<c102abe2>] ? test_tsk_need_resched+0xa/0x13
[ 1130.520474] [<c102abe2>] ? test_tsk_need_resched+0xa/0x13
[ 1130.520474] [<c102e59c>] ? check_preempt_curr+0x48/0x58
[ 1130.520474] [<c121f462>] ? __napi_complete+0x1b/0x22
[ 1130.520474] [<c12213f0>] ? napi_complete+0x25/0x31
[ 1130.520474] [<f825d2d6>] ? e1000_clean+0x33e/0x35c [e1000]
[ 1130.520474] [<c105daf5>] ? arch_local_irq_save+0xf/0x14
[ 1130.520474] [<c12c45a2>] ? _raw_spin_lock_irq+0x9/0x12
[ 1130.520474] [<c1042307>] ? run_timer_softirq+0x1eb/0x1f3
[ 1130.520474] [<c107b86c>] ? __rcu_process_callbacks+0x5f/0x252
[ 1130.520474] [<c103cfdb>] ? local_bh_enable+0x2/0x2
[ 1130.520474] [<c103d0dc>] ? __do_softirq+0x101/0x12f
[ 1130.520474] [<c103c6e8>] ? __local_bh_enable+0x37/0x67
[ 1130.520474] <IRQ> [<c10cd712>] ? sys_write+0x58/0x61
[ 1130.520474] [<c12c8e5f>] ? sysenter_do_call+0x12/0x28
我想我的問題是返回先前的前一個新的呼叫被觸發。
爲了測試這一點，我修改了上面提到的定時器玩具例子：
int i;
void timer1_routine(unsigned long data)
{
int no;
no=i;
i++;
printk(KERN_INFO "process %d starts",no);
msleep(5000);
printk(KERN_INFO "process %d ends",no);
mod_timer(&timer1, jiffies + HZ); /* restarting timer */
}
這導致了同樣的錯誤，所以我想，我的猜測是正確的：有兩個電話之間的衝突。我的問題是如何避免這個問題，如果我不知道函數的執行需要多少時間？