I am migrating some parts of old C++ code, originally compiled with CodeGear C++Builder® 2009 Version 12.0.3170.16989
The following code - minimal version of a bigger piece - outputs -34
with any modern compiler. Although, in the original platform it outputs 84
:
char Key[4];
Key[0] = 0x1F;
Key[1] = 0x01;
Key[2] = 0x8B;
Key[3] = 0x55;
for(int i = 0; i < 2; i++) {
Key[i] = Key[2*i] ^ Key[2*i + 1];
}
std::cout << (int) Key[1] << std::endl;
The following code outputs
-34
with both old and new compilers:
for(int i = 0; i < 2; i++) {
char a = Key[2*i];
char b = Key[2*i + 1];
char c = a ^ b;
Key[i] = c;
}
Also, manually unrolling the loop seems to work with both compilers:
Key[0] = Key[0] ^ Key[1];
Key[1] = Key[2] ^ Key[3];
It is important that I match the behavior of the old code. Can anyone please help me understand why the original compiler produces those results?
This seems to be a bug:
The line
Key[i] = Key[2*i] ^ Key[2*i + 1];
generates the following code:
00401184 8B55F8 mov edx,[ebp-$08]
00401187 8A4C55FD mov cl,[ebp+edx*2-$03]
0040118B 8B5DF8 mov ebx,[ebp-$08]
0040118E 304C1DFC xor [ebp+ebx-$04],cl
That does not make sense. This is something like:
Key[i] ^= Key[i*2 + 1];
And that explains how the result came to be: 0x01 ^ 0x55
is indeed 0x54
, or 84
.
It should be something like:
mov edx,[ebp-$08]
mov cl,[ebp+edx*2-$04]
xor cl,[ebp+edx*2-$03]
mov [ebp+ebx-$04],cl
So this is definitely a code generation bug. It seems to persist until now, C++Builder 10.2 Tokyo, for the "classic" (Borland) compiler.
But if I use the "new" (clang) compiler, it produces 222
. The code produced is:
File7.cpp.12: Key[i] = Key[2*i] ^ Key[2*i + 1];
004013F5 8B45EC mov eax,[ebp-$14]
004013F8 C1E001 shl eax,$01
004013FB 0FB64405F0 movzx eax,[ebp+eax-$10]
00401400 8B4DEC mov ecx,[ebp-$14]
00401403 C1E101 shl ecx,$01
00401406 0FB64C0DF1 movzx ecx,[ebp+ecx-$0f]
0040140B 31C8 xor eax,ecx
0040140D 88C2 mov dl,al
0040140F 8B45EC mov eax,[ebp-$14]
00401412 885405F0 mov [ebp+eax-$10],dl
That doesn't look optimal to me (I used O2 and O3 with the same result), but it produces the right result.