Closed niscove closed 2 months ago
Hi @niscove, thank you for reporting the issue.
In fadd st(1)
, there is an implicit operand st(0)
, and the semantics is that it performs the addition of the value in st(1)
to the value in st(0)
and stores the result in st(0)
, right?
I can see that ddisasm produces fadd st(1)
for the instruction in the output asm, which seems to be correct.
When you say that Ddisasm interprets the binary as fadd st(1), st(0)
, do you mean that Ddisasm treats st(1) as the destination, and st(0) as the source?
The attached file has it all. Without getting into detail, ddisasm should read d8 c1 and output fadd ST(0),ST(1), not fadd ST(1),ST(0) and should read d8 c2 and output fadd ST(0),ST(2, not fadd ST(2),ST(0) all as in the attached file.
From: junghee @.> Sent: Thursday, May 23, 2024 4:37 PM To: GrammaTech/ddisasm @.> Cc: Iscove, Norman @.>; Mention @.> Subject: [External] Re: [GrammaTech/ddisasm] Incorrect disassembly of binary 'FADD ST(n1) ST(n2)' instructions (Issue #76)
Hi @niscovehttps://urldefense.com/v3/__https:/github.com/niscove__;!!CjcC7IQ!KXmEfJ47yD4Acjwffip25t1kc0BS2ls22ZMmEXlhyaueWNWXZSvhXfchS_Qlu7lee5ZqzQ63PUgBpeM4ysIa1GqvvxM$, thank you for reporting the issue.
In fadd st(1), there is an implicit operand st(0), and the semantics is that it performs the addition of the value in st(1) to the value in st(0) and stores the result in st(0), right?
I can see that ddisasm produces fadd st(1) for the instruction in the output asm, which seems to be correct.
When you say that Ddisasm interprets the binary as fadd st(1), st(0), do you mean that Ddisasm treats st(1) as the destination, and st(0) as the source?
—
Sorry, I can't find any attached file in this issue page.
This is the url for the page I sent you with detail on the FADD instruction
https://www.felixcloutier.com/x86/fadd:faddp:fiadd
From: junghee @.> Sent: Thursday, May 23, 2024 7:48 PM To: GrammaTech/ddisasm @.> Cc: Iscove, Norman @.>; Mention @.> Subject: [External] Re: [GrammaTech/ddisasm] Incorrect disassembly of binary 'FADD ST(n1) ST(n2)' instructions (Issue #76)
Sorry, I can't find any attached file in this issue page.
—
I assume you would have some binary that you observed the issue? If so, could you attach that?
I tried writing an assembly code with fadd instruction:
1 #include <stdio.h>
2
3 int main() {
4 float val1 = 3.14;
5 float val2 = 2.71;
6 float val3 = 1.62;
7 float result;
8
9 __asm__ (
10 "fld %1;" // Load val1 into ST(0)
11 "fld %2;" // Load val2 into ST(0), val1 is now in ST(1)
12 "fld %3;" // Load val3 into ST(0), val2 is in ST(1), val1 in ST(2)
13 "fadd %%st(1), %%st(0);" // ST(0) = ST(0) + ST(1) (val3 + val2)
14 //"fadd %%st(1);" // ST(0) = ST(0) + ST(1) (val3 + val2)
15 "fadd %%st(2);" // ST(0) = ST(0) + ST(2) (result of previous + val1)
16 "fstp %0;" // Store the result from ST(0) to result and pop ST(0)
17 : "=m" (result)
18 : "m" (val1), "m" (val2), "m" (val3)
19 : "st"
20 );
21
22 printf("Result: %f\n", result);
23 return 0;
24 }
Both fadd st(1)
and fadd st(1), st(0)
produces the same bytes: d8 c1
For both case, Ddisasm outputs fadd st(1)
.
" I assume you would have some binary that you observed the issue? If so, could you attach that?"
The binary code is
d8 c1
d8 c2
When Ddisasm is asked to disassemble this binary code, it outputs
fadd ST(1),ST(0)
fadd ST(2),ST(0)
The output is wrong.
It should be
fadd ST(0),ST(1)
fadd ST(0),ST(2)
The coding is exactly detailed in the link I sent you.
From: junghee @.> Sent: Thursday, May 23, 2024 8:28 PM To: GrammaTech/ddisasm @.> Cc: Iscove, Norman @.>; Mention @.> Subject: [External] Re: [GrammaTech/ddisasm] Incorrect disassembly of binary 'FADD ST(n1) ST(n2)' instructions (Issue #76)
I assume you would have some binary that you observed the issue? If so, could you attach that?
I tried writing an assembly code with fadd instruction:
1 #include
2
3 int main() {
4 float val1 = 3.14;
5 float val2 = 2.71;
6 float val3 = 1.62;
7 float result;
8
9 asm (
10 "fld %1;" // Load val1 into ST(0)
11 "fld %2;" // Load val2 into ST(0), val1 is now in ST(1)
12 "fld %3;" // Load val3 into ST(0), val2 is in ST(1), val1 in ST(2)
13 "fadd %%st(1), %%st(0);" // ST(0) = ST(0) + ST(1) (val3 + val2)
14 //"fadd %%st(1);" // ST(0) = ST(0) + ST(1) (val3 + val2)
15 "fadd %%st(2);" // ST(0) = ST(0) + ST(2) (result of previous + val1)
16 "fstp %0;" // Store the result from ST(0) to result and pop ST(0)
17 : "=m" (result)
18 : "m" (val1), "m" (val2), "m" (val3)
19 : "st"
20 );
21
22 printf("Result: %f\n", result);
23 return 0;
24 }
Both fadd st(1) and fadd st(1), st(0) produces the same bytes: d8 c1
For both case, Ddisasm outputs fadd st(1).
—
Can you give me your command line for running ddisasm on your binary? I don't quite understand how you create an input binary with your binary code to feed into ddisasm.
I wrote an assembly script, 'z.asm' which begins with the lines
.
.
fadd ST(0),ST(1)
fadd ST(0),ST(2)
pushd CF_TEXT
call IsClipboardFormatAvailable
'
.
I compiled z.asm with MASM generating the executable file z.exe . The first lines in z.exe disassembled by x64dbg are:
00401000 6A 00 push 0
00401002 E8 25010000 call <JMP.&GetModuleHandleA>
00401007 A3 00304000 mov dword ptr ds:[403000],eax
0040100C D8C1 **fadd st(0),st(1)** CORRECT
0040100E D8C2 **fadd st(0),st(2)** CORRECT
00401010 6A 01 push 1
00401012 E8 3F010000 call <JMP.&IsClipboardFormatAvailable>
.
.
Using ddisasm on z.exe with the command line
ddisasm --asm zDDout.asm z.exe
produces the disassembly file zDDout.asm. The first lines are
mov DWORD PTR [$L_403000],EAX
**fadd ST(1),ST(0)** WRONG
**fadd ST(2),ST(0)** WRONG
push 1
call FUN_401156
.
.
Thanks for the details. We were able to reproduce the problem, and are in process of resolving it. I will let you know when it is done.
I believe this is now solved with https://github.com/GrammaTech/gtirb-pprinter/commit/d9dc966176379e9a29e96558f28d589d00ff91a9
REVISION 1.2.840.113556.1.4.145 downloaded windows dev executable dated April 12 2024
Ddisasm interprets the binary sequences
d8 c1 as fadd ST(1), ST(0) d8 c2 as fadd ST(2), ST(0)
The correct outputs should instead be
d8 c1 fadd ST(0), ST(1) d8 c2 fadd ST(0), ST(2)
.