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[kernel] Early-out earlier for ignored interrupts

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I was seeing more ignored interrupts when debugging, trying to shorten
their path more. Adding a separate ISR for ignored interrupts was the
shortest path, but that caused some strange instability that I'm not in
the mood to track down.
This commit is contained in:
Justin C. Miller
2021-08-01 17:18:23 -07:00
parent 2b16b69afa
commit 4dc9675a6c
2 changed files with 4 additions and 27 deletions
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1
src/kernel/interrupt_isrs.inc
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@@ -241,7 +241,6 @@ ISR (0xe0, 0, isrTimer)
ISR (0xe1, 0, isrLINT0)
ISR (0xe2, 0, isrLINT1)
ISR (0xe3, 0, isrAPICError)
ISR (0xe4, 0, isrAssert)
ISR (0xef, 0, isrSpurious)

30
src/kernel/interrupts.cpp
View File

@@ -83,8 +83,11 @@ disable_legacy_pic()
void
isr_handler(cpu_state *regs)
{
console *cons = console::get();
uint8_t vector = regs->interrupt & 0xff;
if ((vector & 0xf0) == 0xf0) {
*reinterpret_cast<uint32_t *>(apic_eoi_addr) = 0;
return;
}
// Clear out the IST for this vector so we just keep using
// this stack
@@ -169,31 +172,6 @@ isr_handler(cpu_state *regs)
// No EOI for the spurious interrupt
return;
case isr::isrIgnore0:
case isr::isrIgnore1:
case isr::isrIgnore2:
case isr::isrIgnore3:
case isr::isrIgnore4:
case isr::isrIgnore5:
case isr::isrIgnore6:
case isr::isrIgnore7:
//cons->printf("\nIGNORED: %02x\n", regs->interrupt);
outb(PIC1, 0x20);
break;
case isr::isrIgnore8:
case isr::isrIgnore9:
case isr::isrIgnoreA:
case isr::isrIgnoreB:
case isr::isrIgnoreC:
case isr::isrIgnoreD:
case isr::isrIgnoreE:
case isr::isrIgnoreF:
//cons->printf("\nIGNORED: %02x\n", regs->interrupt);
outb(PIC1, 0x20);
outb(PIC2, 0x20);
break;
default:
snprintf(message, sizeof(message), "Unknown interrupt 0x%x", regs->interrupt);
kassert(false, message);