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<think>
Okay, I need to figure out whether this bug report is user-mode or system-mode related. Let me read through the problem description again.
The issue is about the ADDV instruction in QEMU not being emulated properly. The documentation says that ADDV Rm, Rn adds Rm and Rn, stores the result in Rn, and sets T on overflow. But according to the user's test, QEMU is storing the result in Rm instead of Rn.
Looking at the example code provided: It uses inline assembly with three registers a (r8), b (r9), and c (r10). The assembly code is:
clrt
addv %2,%0
movt %1
The comments say that on real hardware, after running this, a becomes 0x80000000, which suggests that the addition overflowed. But in QEMU, a remains 0x7fffffff and b changes to 0x80000000.
So the ADDV instruction is supposed to add Rm (b) to Rn (a), store the result in Rn, and set T if there's an overflow. The test case shows that on real hardware, a overflows from 0x7fffffff +1 to 0x80000000. In QEMU, it seems like the result is stored in Rm instead of Rn.
The problem here is with how the ADDV instruction is being emulated. Since this is about the CPU's instruction emulation and registers, which are part of the user-mode execution (the program running on the CPU), it doesn't involve system calls, peripherals, or higher-privilege operations. The issue affects how programs execute their instructions.
Therefore, I think this bug is related to user-mode because it's about the CPU's instruction set and register handling during normal program execution.
</think>
user
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