irq_set_chained_handler() called too early for hwirq to be initialized

[Roland Stigge]

On 28/10/12 18:34, Thomas Gleixner wrote:
> On Sun, 28 Oct 2012, Roland Stigge wrote:
>> consider arch/arm/mach-lpc32xx/irq.c: irq_set_chained_handler() is
>> called at a point where it accesses
>> irq_to_desc(IRQ_LPC32XX_SUB2IRQ)->irq_data.hwirq but which is not yet
>> initialized.
> 
> None of the functions which are called inside of
> irq_set_chained_handler() touches desc->irq_data.hwirq.
> 
> So what are you talking about?

Via the call trace:

irq_set_chained_handler()
-> __irq_set_handler()
-> irq_startup()
-> irq_enable()
-> desc->irq_data.chip->irq_unmask()

The code path comes back to irq.c's lpc32xx_unmask_irq() which reads the
above described ->hwirq which is only later initialized on
irq_domain_add_legacy(). Hope this explains my above short description.

> Of course are the interrupts preallocated, simply because
> machine_desc->nr_irqs is 0 and therefor the ARM core code allocates
> NR_IRQS irq descriptors in the early setup way before
> lpc32xx_init_irq() is called.

OK, will remove the call to  irq_alloc_descs() since it is superfluous.

Still, my question remains if I can move the irq_set_chained_handler()
calls to after of_irq_init() and irq_domain_add_legacy() since only the
latter initializes hwirq.

> If those interrupts would not be preallocated, then the code would
> fail to initialize any interrupt at all. And of course nothing would
> notice as all function calls to set_irq_* do not check the return
> value.

Do you mean mach-lpc32xx/irq.c's calls to set_irq_* not checking the
return values? Maybe because those are declared "void"?

static inline void
irq_set_chained_handler(unsigned int irq, irq_flow_handler_t handle);
void set_irq_flags(unsigned int irq, unsigned int iflags);
static inline void irq_set_chip_and_handler(unsigned int irq,
                                            struct irq_chip *chip,
                                            irq_flow_handler_t handle);

Or did I misunderstand sth.?

Thanks in advance,

Roland