Return true after queueing a guest syscall override - #211
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dispatchSyscallOverride returns bool, and its caller uses that to decide whether the built-in handler still needs to run. The success path -- the one that actually queues the guest's handler as an invocation -- fell off the end of the function without returning. That is undefined behaviour, and the practical failure mode is bad: whatever happened to be in the return register decided whether the built-in syscall ran in addition to the guest's override, so a game that overrides a syscall could get the effect applied twice, or not at all, depending on the build. (cherry picked from commit 6bf1feff8bd38892f1a464322f455fe016814c5c)
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| parent.r[2] = completed.r[2]; | ||
| }; | ||
| scheduler.invokeCurrent(std::move(invocation)); | ||
| // The invocation is queued and the caller must not fall through to the |
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Nice, I think I remove the return when I has moving the code, can you remove this comments
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waitttt. I remenber now.
look at invokeCurrent its a [[noreturn]] so your return true; is unreachable
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dispatchSyscallOverridereturnsbool, and its caller uses that to decide whether the built-in handler still needs to run. The success path, the one that actually queues the guest's handler as an invocation falls off the end of the function without returning.That is undefined behaviour, and the practical failure mode is unpleasant: whatever happens to be in the return register decides whether the built-in syscall runs in addition to the guest's override. A game that overrides a syscall can get the effect applied twice, or not at all, depending on the build.
One-line fix.