The open_softirq function cannot be recognized in the module.

The open_softirq function cannot be recognized in the module-general Linux technology-Linux programming and kernel information. The following is a detailed description. My problem is that in the following code, the softirq function is called in softirq. c is defined in interrupt. the expression in the h function is extern, but during fc6 compilation, open_softirq is always prompted to be unrecognizable.
However, when the make command is successful, the module is loaded again and the error is still caused by open_softirq being an unrecognizable symbol.

At the beginning, I thought it was the fc6 source code package. Only the. h file does not have the. c file. So I downloaded the 2.6.18.1 source code package and compiled it under fc6, and then the make loaded module still encountered the same problem.
However, in the 2.6.15.5 kernel environment, you can compile it, load it normally, and print the output hello, world !!
I 'd like to ask you what is going on.

# Include
# Include
# Include

# Include

MODULE_DESCRIPTION ("My kernel module ");
MODULE_AUTHOR ("root (root@localhost.localdomain )");
MODULE_LICENSE ("$ LICENSE $ ");

Static void timer_func (struct softirq_action * h)
{
Printk (KERN_ALERT "hello, world !! \ N ");
}

Void fastcall raise_softirq (unsigned int nr)
{
Unsigned long flags;

Local_irq_save (flags );
Raise_softirq_irqoff (nr );
Local_irq_restore (flags );
}

Static int test1_init_module (void)
{
Printk (KERN_DEBUG "Module test1 init \ n ");
Open_softirq (TIMER_SOFTIRQ, timer_func, NULL );
Raise_softirq (TIMER_SOFTIRQ );
Return 0;
}

Static void test1_exit_module (void)
{
Printk (KERN_DEBUG "Module test1 exit \ n ");
}

Module_init (test1_init_module );
Module_exit (test1_exit_module );