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Re: Spurious interrupt on ARM.
- From: Andrew Parlane <andrewp at carallon dot com>
- To: Nick Garnett <nickg at calivar dot com>, ecos-discuss at ecos dot sourceware dot org
- Date: Fri, 01 Nov 2013 17:20:46 +0000
- Subject: Re: Spurious interrupt on ARM.
- Authentication-results: sourceware.org; auth=none
- References: <52729021 dot 3080205 at carallon dot com> <5273DFAE dot 1080000 at calivar dot com>
Sorry, I should have been a bit more clear.
First we skip the ISR by jumping to the spurious_IRQ label, and then we
switch stacks if necessary, then we have (line numbers may vary):
941 // The return value from the handler (in r0) will indicate
whether a
942 // DSR is to be posted. Pass this together with a pointer to the
943 // interrupt object we have just used to the interrupt tidy
up routine.
944
945 // don't run this for spurious interrupts!
946 cmp v1,#CYGNUM_HAL_INTERRUPT_NONE
947 beq 17f
948 ldr r1,.hal_interrupt_objects
949 ldr r1,[r1,v1,lsl #2]
950 mov r2,v6 // register frame
951
952 THUMB_MODE(r3,10)
953
954 bl interrupt_end // post any bottom layer handler
955 // threads and call scheduler
956 ARM_MODE(r1,10)
957 17:
So it compares the result of hal_IRQ_handler (stored in v1) with
CYGNUM_HAL_INTERRUPT_NONE, and jumps forwards to label 17: which is
after interrupt_end. if it was a spurious IRQ.
Andrew
On 01/11/2013 17:06, Nick Garnett wrote:
On 31/10/13 17:15, Andrew Parlane wrote:
Looking at hal/arm/arch/current/src/vectors.S in IRQ:
We increment the scheduler lock and decrement it again in interrupt_end.
In the case of there being a spurious interrupt, we don't call
interrupt_end, and so the scheduler never gets decremented.
Am I missing something here?
interrupt_end() does get called. A spurious interrupt only causes the
code to skip calling an ISR by jumping to the spurious_IRQ label. From
there it follows the same code path and will call interrupt_end() as normal.
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