Please note that I do not trust the “Suppress JIT optimization on module load” option, I spawn the process without debugging and attach my debugger after the JIT has run.
In the version where single-line runs faster, this is Main:
SingleLineTest();
00000000 push ebp
00000001 mov ebp,esp
00000003 call dword ptr ds:[0019380Ch]
MultiLineTest();
00000009 call dword ptr ds:[00193818h]
SingleLineTest();
0000000f call dword ptr ds:[0019380Ch]
MultiLineTest();
00000015 call dword ptr ds:[00193818h]
SingleLineTest();
0000001b call dword ptr ds:[0019380Ch]
MultiLineTest();
00000021 call dword ptr ds:[00193818h]
00000027 pop ebp
}
00000028 ret
Note that MultiLineTest has been placed on an 8 byte boundary, and SingleLineTest on a 4 byte boundary.
Here’s Main for the version where both run at the same speed:
MultiLineTest();
00000000 push ebp
00000001 mov ebp,esp
00000003 call dword ptr ds:[00153818h]
SingleLineTest();
00000009 call dword ptr ds:[0015380Ch]
MultiLineTest();
0000000f call dword ptr ds:[00153818h]
SingleLineTest();
00000015 call dword ptr ds:[0015380Ch]
MultiLineTest();
0000001b call dword ptr ds:[00153818h]
SingleLineTest();
00000021 call dword ptr ds:[0015380Ch]
MultiLineTest();
00000027 call dword ptr ds:[00153818h]
0000002d pop ebp
}
0000002e ret
Amazingly, the addresses chosen by the JIT are identical in the last 4 digits, even though it allegedly processed them in the opposite order. Not sure I believe that any more.
More digging is necessary. I think it was mentioned that the code before the loop wasn’t exactly the same in both versions? Going to investigate.
Here’s the “slow” version of SingleLineTest (and I checked, the last digits of the function address haven’t changed).
Stopwatch stopwatch = new Stopwatch();
00000000 push ebp
00000001 mov ebp,esp
00000003 push edi
00000004 push esi
00000005 push ebx
00000006 mov ecx,7A5A2C68h
0000000b call FFF91EA0
00000010 mov esi,eax
00000012 mov dword ptr [esi+4],0
00000019 mov dword ptr [esi+8],0
00000020 mov byte ptr [esi+14h],0
00000024 mov dword ptr [esi+0Ch],0
0000002b mov dword ptr [esi+10h],0
stopwatch.Start();
00000032 cmp byte ptr [esi+14h],0
00000036 jne 00000047
00000038 call 7A22B314
0000003d mov dword ptr [esi+0Ch],eax
00000040 mov dword ptr [esi+10h],edx
00000043 mov byte ptr [esi+14h],1
int count = 0;
00000047 xor edi,edi
for (uint i = 0; i < 1000000000; ++i) {
00000049 xor edx,edx
count += i % 16 == 0 ? 1 : 0;
0000004b mov eax,edx
0000004d and eax,0Fh
00000050 test eax,eax
00000052 je 00000058
00000054 xor eax,eax
00000056 jmp 0000005D
00000058 mov eax,1
0000005d add edi,eax
for (uint i = 0; i < 1000000000; ++i) {
0000005f inc edx
00000060 cmp edx,3B9ACA00h
00000066 jb 0000004B
}
stopwatch.Stop();
00000068 mov ecx,esi
0000006a call 7A23F2C0
Console.WriteLine("Single-line test --> Count: {0}, Time: {1}", count, stopwatch.ElapsedMilliseconds);
0000006f mov ecx,797C29B4h
00000074 call FFF91EA0
00000079 mov ecx,eax
0000007b mov dword ptr [ecx+4],edi
0000007e mov ebx,ecx
00000080 mov ecx,797BA240h
00000085 call FFF91EA0
0000008a mov edi,eax
0000008c mov ecx,esi
0000008e call 7A23ABE8
00000093 push edx
00000094 push eax
00000095 push 0
00000097 push 2710h
0000009c call 783247EC
000000a1 mov dword ptr [edi+4],eax
000000a4 mov dword ptr [edi+8],edx
000000a7 mov esi,edi
000000a9 call 793C6F40
000000ae push ebx
000000af push esi
000000b0 mov ecx,eax
000000b2 mov edx,dword ptr ds:[03392034h]
000000b8 mov eax,dword ptr [ecx]
000000ba mov eax,dword ptr [eax+3Ch]
000000bd call dword ptr [eax+1Ch]
000000c0 pop ebx
}
000000c1 pop esi
000000c2 pop edi
000000c3 pop ebp
000000c4 ret
And the “fast” version:
Stopwatch stopwatch = new Stopwatch();
00000000 push ebp
00000001 mov ebp,esp
00000003 push edi
00000004 push esi
00000005 push ebx
00000006 mov ecx,7A5A2C68h
0000000b call FFE11F70
00000010 mov esi,eax
00000012 mov ecx,esi
00000014 call 7A1068BC
stopwatch.Start();
00000019 cmp byte ptr [esi+14h],0
0000001d jne 0000002E
0000001f call 7A12B3E4
00000024 mov dword ptr [esi+0Ch],eax
00000027 mov dword ptr [esi+10h],edx
0000002a mov byte ptr [esi+14h],1
int count = 0;
0000002e xor edi,edi
for (uint i = 0; i < 1000000000; ++i) {
00000030 xor edx,edx
count += i % 16 == 0 ? 1 : 0;
00000032 mov eax,edx
00000034 and eax,0Fh
00000037 test eax,eax
00000039 je 0000003F
0000003b xor eax,eax
0000003d jmp 00000044
0000003f mov eax,1
00000044 add edi,eax
for (uint i = 0; i < 1000000000; ++i) {
00000046 inc edx
00000047 cmp edx,3B9ACA00h
0000004d jb 00000032
}
stopwatch.Stop();
0000004f mov ecx,esi
00000051 call 7A13F390
Console.WriteLine("Single-line test --> Count: {0}, Time: {1}", count, stopwatch.ElapsedMilliseconds);
00000056 mov ecx,797C29B4h
0000005b call FFE11F70
00000060 mov ecx,eax
00000062 mov dword ptr [ecx+4],edi
00000065 mov ebx,ecx
00000067 mov ecx,797BA240h
0000006c call FFE11F70
00000071 mov edi,eax
00000073 mov ecx,esi
00000075 call 7A13ACB8
0000007a push edx
0000007b push eax
0000007c push 0
0000007e push 2710h
00000083 call 782248BC
00000088 mov dword ptr [edi+4],eax
0000008b mov dword ptr [edi+8],edx
0000008e mov esi,edi
00000090 call 792C7010
00000095 push ebx
00000096 push esi
00000097 mov ecx,eax
00000099 mov edx,dword ptr ds:[03562030h]
0000009f mov eax,dword ptr [ecx]
000000a1 mov eax,dword ptr [eax+3Ch]
000000a4 call dword ptr [eax+1Ch]
000000a7 pop ebx
}
000000a8 pop esi
000000a9 pop edi
000000aa pop ebp
000000ab ret
Just the loops, fast on the left, slow on the right:
00000030 xor edx,edx 00000049 xor edx,edx
00000032 mov eax,edx 0000004b mov eax,edx
00000034 and eax,0Fh 0000004d and eax,0Fh
00000037 test eax,eax 00000050 test eax,eax
00000039 je 0000003F 00000052 je 00000058
0000003b xor eax,eax 00000054 xor eax,eax
0000003d jmp 00000044 00000056 jmp 0000005D
0000003f mov eax,1 00000058 mov eax,1
00000044 add edi,eax 0000005d add edi,eax
00000046 inc edx 0000005f inc edx
00000047 cmp edx,3B9ACA00h 00000060 cmp edx,3B9ACA00h
0000004d jb 00000032 00000066 jb 0000004B
The instructions are identical (being relative jumps, the machine code is identical even though the disassembly shows different addresses), but the alignment is different. There are three jumps. the je loading a constant 1 is aligned in the slow version and not in the fast version, but it hardly matters, since that jump is only taken 1/16 of the time. The other two jumps ( jmp after loading a constant zero, and jb repeating the entire loop) are taken millions more times, and are aligned in the “fast” version.
I think this is the smoking gun.