[Question][Hw5] memory in Analysis.check_instr

[nubi12336]

In example1.c, there is if (x<0) and it is %div3 = sdiv i32 10, %call, !dbg !23 in example1.ll, if.then2.
I printed out a table and I got a result %div3 = sdiv i32 10, %call, !dbg !23 -> %call -> Neg.
Then, I thought, in Analysis.check_instr, as it matches with Llvm.Opcode.SDiv and v is Neg(as e is %call)
Memory.Value.order zero v must be false.
However, it prints out Potential Division-by-zero @ example1.c:main:9:14, %call = Top.

What is wrong?

[nubi12336]

I printed out in Memory.pp Format.std_formatter memory Analysis.check_instr.
Below is result.

I think second row is weird..

[RiceBiscuits]

I can't find the bug due to the lack of information.
However, I guess analysis.run function ortransfer_*may have a problem.
Please debug your run and transfer_* function carefully.

[nubi12336]

How come it can happen that, in the resulting table there is %div3 = sdiv i32 10, %call, !dbg !23 -> %call -> Neg, but in the result div3, %call become top?(above picture)
%div3 ~ is 10/x in if (x < 0) {z = 10 / x}.