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STM8 port: Update IAR assembler comments. The IAR compiler is saving ?b0:?b7 in interrupt handlers as expected.

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Kelvin Lawson
2010-05-25 00:57:14 +01:00
parent 619c8c3e4a
commit c0c953b476
1 changed files with 11 additions and 7 deletions
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18
ports/stm8/atomport-asm-iar.s
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@@ -103,13 +103,17 @@ archContextSwitch:
; (IAR) Compiler will have already saved any scratch registers
; (A, X, Y, CC, and ?b0 to ?b7) which it needs before calling here
; for cooperative switches. So these will already be on the stack
; and do not need to be context-switched. This assumes that
; __interrupt funcs (although EWSTM8 docs say they don't save
; vregs) do actually save ?b0 to ?b7 if they call out to another
; function. Can't verify this with Kickstart edition. If this
; assumption is incorrect then we actually need to save all vregs
; here, and this would be inefficient if called for a cooperative
; switch where we know ?b0 to ?b7 do not need to be saved.
; and do not need to be context-switched. The same goes for
; __interrupt functions (i.e. preemptive switches): the IAR
; compiler saves A, X, Y and CC for interrupt handlers, and
; (because we call out from interrupt handlers to C kernel code
; (e.g. atomIntExit()) before calling here for a context-switch,
; the compiler also has to save ?b0 to ?b7. This is because those
; called C functions cannot know they were called from an interrupt
; and will assume that they have ?b0 to ?b7 available as scratch
; registers. Either way (cooperative or interrupt/preemptive) we
; know that the only registers which must be preserved that are not
; already on the stack-frame are ?b8 to ?b15.
PUSH ?b8
PUSH ?b9
PUSH ?b10