By: Travis Downs (, August 19, 2019 6:13 pm
Room: Moderated Discussions
I notice that when I interrupt a process on x86, the interrupt locations as measured by the return instruction (i.e., the instruction at which execution will resume at) line up exactly at the instruction after slow-to-retire instructions.

Naively, I would expect that when an interrupt comes in, all in-flight work is thrown away - that is, all unretired instructions. However, that would leave the interrupts pointing at the slow instruction, not one past the instruction.

Is there some reason it works like this? I seems like it would increase interrupt latency a bit.
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InterruptsTravis Downs2019/08/19 06:13 PM
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      InterruptsLinus Torvalds2019/08/22 10:58 AM
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          InterruptsTravis Downs2019/08/24 12:37 AM
        InterruptsTravis Downs2019/08/23 08:51 PM
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