提交 624db9ea 编写于 作者: T Thomas Gleixner

x86: Select CONFIG_HAVE_IRQ_EXIT_ON_IRQ_STACK

Now that all invocations of irq_exit_rcu() happen on the irq stack, turn on
CONFIG_HAVE_IRQ_EXIT_ON_IRQ_STACK which causes the core code to invoke
__do_softirq() directly without going through do_softirq_own_stack().

That means do_softirq_own_stack() is only invoked from task context which
means it can't be on the irq stack. Remove the conditional from
run_softirq_on_irqstack_cond() and rename the function accordingly.
Signed-off-by: NThomas Gleixner <tglx@linutronix.de>
Reviewed-by: NKees Cook <keescook@chromium.org>
Link: https://lore.kernel.org/r/20210210002513.068033456@linutronix.de
上级 52d743f3
......@@ -187,6 +187,7 @@ config X86
select HAVE_HW_BREAKPOINT
select HAVE_IDE
select HAVE_IOREMAP_PROT
select HAVE_IRQ_EXIT_ON_IRQ_STACK if X86_64
select HAVE_IRQ_TIME_ACCOUNTING
select HAVE_KERNEL_BZIP2
select HAVE_KERNEL_GZIP
......
......@@ -189,19 +189,16 @@
"call %P[__func] \n"
/*
* Macro to invoke __do_softirq on the irq stack. Contrary to the above
* the only check which is necessary is whether the interrupt stack is
* in use already.
* Macro to invoke __do_softirq on the irq stack. This is only called from
* task context when bottom halfs are about to be reenabled and soft
* interrupts are pending to be processed. The interrupt stack cannot be in
* use here.
*/
#define run_softirq_on_irqstack_cond() \
#define run_softirq_on_irqstack() \
{ \
if (__this_cpu_read(hardirq_stack_inuse)) { \
__do_softirq(); \
} else { \
__this_cpu_write(hardirq_stack_inuse, true); \
call_on_irqstack(__do_softirq, ASM_CALL_SOFTIRQ); \
__this_cpu_write(hardirq_stack_inuse, false); \
} \
__this_cpu_write(hardirq_stack_inuse, true); \
call_on_irqstack(__do_softirq, ASM_CALL_SOFTIRQ); \
__this_cpu_write(hardirq_stack_inuse, false); \
}
#else /* CONFIG_X86_64 */
......
......@@ -76,5 +76,5 @@ int irq_init_percpu_irqstack(unsigned int cpu)
void do_softirq_own_stack(void)
{
run_softirq_on_irqstack_cond();
run_softirq_on_irqstack();
}
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