add new results

1 files changed, 31 insertions, 0 deletions

diff --git a/results/classifier/qwen3:32b/reasoning/instruction/1915327 b/results/classifier/qwen3:32b/reasoning/instruction/1915327
new file mode 100644
index 000000000..24214a420
--- /dev/null
+++ b/results/classifier/qwen3:32b/reasoning/instruction/1915327
@@ -0,0 +1,31 @@
+<think>
+Okay, let's see. The user is reporting that the QEMU TCG (Tiny Code Generator) doesn't match the real x86_64 CPU behavior when executing the cmpxchg instruction. They provided a test program that shows different results between actual hardware and QEMU. 
+
+The problem is that cmpxchg is supposed to leave the RAX register unchanged when the comparison is successful. On real CPUs, the output is the original RAX value (0x1234567812345678), but in QEMU, RAX is zero-extended, resulting in just 0x12345678. 
+
+The key here is understanding how cmpxchg works. According to Intel's manual, when using cmpxchg with a 32-bit operand size, the EAX is compared with the memory operand. If it matches, then the EDX:EAX (for 32-bit) is written to the memory. However, when the operand size is 32 bits, the upper 32 bits of RAX (the R in RAX) should remain unchanged. Wait, no, actually, cmpxchg in 32-bit mode affects EAX, but in 64-bit mode, if you use the 32-bit operand size, then the upper 32 bits of RAX are zeroed. Wait, maybe the user's code is using a 32-bit cmpxchg in 64-bit mode?
+
+Looking at the code: the 'cmpxchg %[edi],%[mem]' instruction. The EDI is a 32-bit register (since edi is used, not rdi), so the operand size is 32 bits. The memory operand is an int (32 bits), and RAX is a 64-bit register. 
+
+In x86-64, when you perform a 32-bit cmpxchg (like cmpxchgl), the RAX register is treated as EAX, and the upper 32 bits (RAX's higher part) are zeroed. However, in the user's code, RAX is a 64-bit register. Wait, but in the code, the initial RAX is set to 0x1234567812345678 (a 64-bit value). The cmpxchg is using a 32-bit operand (EDI and the memory is an int). So the instruction is cmpxchgl (32-bit), which would compare EAX (the lower 32 bits of RAX) with the memory. If they are equal, then the EDI value (0x77777777) is written to memory. However, regardless of success or failure, the RAX register's value is updated. Wait, no: if the comparison is successful, the memory is updated, and the EAX is set to the value from the memory (which is 0x12345678?), wait no. Wait, cmpxchg works like this: compare the value in AL/AX/EAX/RAX with the memory operand. If equal, then the memory operand is replaced with the value in the second operand (in this case, EDI). If not, then the value in the memory is loaded into AL/AX/EAX/RAX.
+
+Wait, in the code, the instruction is "cmpxchg %[edi],%[mem]". The syntax here is a bit tricky. The AT&T syntax for cmpxchg is "cmpxchg src, dest", but I think in x86 syntax, it's "cmpxchg dest, src". Wait, no, in AT&T syntax, it's "cmpxchg %src, dest". So in this case, the instruction is cmpxchg edi, mem. Which means, compare the value in RAX (the implicit first operand) with the memory. If they are equal, then the memory is replaced with EDI's value. Otherwise, RAX is updated with the memory's value.
+
+But the operand size here is 32 bits. So when using cmpxchg with a 32-bit operand size, RAX's lower 32 bits (EAX) is compared with the memory. The initial RAX is 0x1234567812345678. So EAX is 0x12345678. The memory is 0x12345678. So the comparison is equal. Therefore, the memory is updated to EDI's value (0x77777777), and RAX should remain unchanged. But according to the Intel manual, when the comparison is successful, the EAX is not modified. Wait, no: when the comparison is successful, the memory is updated, but RAX is not changed. Wait, no, the cmpxchg instruction's behavior is: if the comparison is equal, then the memory is set to the value in the source operand (EDI here). The RAX is not modified in this case. If the comparison is not equal, then the memory's value is loaded into RAX.
+
+In the test case, the initial RAX is 0x1234567812345678 (so EAX is 0x12345678). The memory is 0x12345678. So the comparison is equal. Therefore, RAX should remain as 0x1234567812345678. However, in QEMU, the user is seeing RAX being set to 0x12345678. That suggests that QEMU is zeroing the upper 32 bits of RAX when the operand size is 32 bits. On real hardware, when using a 32-bit cmpxchg, the upper 32 bits of RAX (when in 64-bit mode) should be zeroed. Wait, no. Wait, in 64-bit mode, when you perform a 32-bit operation (like cmpxchgl), the upper 32 bits of RAX are cleared. For example, if you do a 32-bit mov to EAX, the upper 32 bits of RAX become zero. Similarly, cmpxchg with 32-bit operand size would cause the RAX to be treated as EAX. So when the instruction is executed, even if the comparison is successful, the upper 32 bits of RAX (the part beyond EAX) are zeroed. Wait, but in the test case, the user's initial RAX is 0x1234567812345678. If the cmpxchg is 32-bit, then after the instruction, RAX should be 0x0000000012345678. But the user says that on real hardware, RAX remains as 0x1234567812345678. That's conflicting with what I thought. 
+
+Wait, maybe I'm mixing up the behavior. Let me check the Intel manual. 
+
+In the Intel Software Developer's Manual, Volume 2, instruction CMPXCHG. For the 32-bit operand size in 64-bit mode: when the comparison is equal, the source operand (EDI) is stored in the destination (memory). The RAX register's lower 32 bits (EAX) are not changed. However, in 64-bit mode, when a 32-bit instruction is executed, the upper 32 bits of the RAX register are zeroed. Wait, yes! For example, if you do a 32-bit operation on EAX, the upper 32 bits of RAX are cleared. So if you have RAX = 0x1234567887654321 and you perform a 32-bit operation on EAX, then EAX is modified and the upper 32 bits become zero. However, in the case of the cmpxchg instruction where the comparison is equal, the RAX is not modified. Wait, no. If the comparison is equal, the RAX is not modified. But the operand size is 32 bits. Wait, but the instruction is cmpxchg with a 32-bit operand size. So, in 64-bit mode, if you use cmpxchgl (32-bit operand size), then during execution, the comparison is between EAX and the memory. If equal, the memory is updated. However, the RAX register is not modified. However, in 64-bit mode, when you perform a 32-bit operation, the upper 32 bits of RAX are zeroed. Wait, but this zeroing only happens when the 32-bit operation writes to EAX. For example, if you do a mov eax, 0x1234, then RAX becomes 0x000000001234. However, if the instruction does not write to EAX, then the upper bits remain. 
+
+In the case of cmpxchg with 32-bit operand size: when the comparison is equal, the instruction does not write to EAX. Therefore, the upper 32 bits of RAX should remain as they were. But according to the user's test, on real hardware, RAX remains as 0x1234567812345678. However, in QEMU, RAX becomes 0x12345678. That suggests that QEMU is zeroing the upper 32 bits of RAX when the operand size is 32 bits, even though the instruction didn't modify EAX. 
+
+But according to the Intel manual, when a 32-bit instruction is executed in 64-bit mode, and the instruction does not modify the lower 32 bits (EAX), then the upper 32 bits are not zeroed. For example, if you have an instruction that doesn't touch EAX, like a 32-bit add that doesn't affect EAX (but that's not possible?), but in the case of cmpxchg where the comparison is equal, the EAX is not modified. Therefore, the upper 32 bits of RAX should remain as they were. Therefore, the correct RAX after the instruction would still be 0x1234567812345678. But QEMU is zeroing the upper part, leading to RAX being 0x0000000012345678. 
+
+So this is a discrepancy in QEMU's TCG implementation. The problem is that QEMU's TCG is not preserving the upper 32 bits of RAX when a 32-bit cmpxchg is executed and the comparison is successful. This suggests that the TCG code for cmpxchg is incorrectly zeroing RAX when it shouldn't. 
+
+The category here is an instruction-related issue because it's about the incorrect emulation of a specific instruction (cmpxchg) in QEMU. The problem is not with syscalls or runtime environment, but with how the instruction is translated or executed in TCG. So the correct category is "instruction".
+</think>
+
+instruction
\ No newline at end of file