问题
Look at this code:
one.cpp:
bool test(int a, int b, int c, int d);
int main() {
volatile int va = 1;
volatile int vb = 2;
volatile int vc = 3;
volatile int vd = 4;
int a = va;
int b = vb;
int c = vc;
int d = vd;
int s = 0;
__asm__("nop"); __asm__("nop"); __asm__("nop"); __asm__("nop");
__asm__("nop"); __asm__("nop"); __asm__("nop"); __asm__("nop");
__asm__("nop"); __asm__("nop"); __asm__("nop"); __asm__("nop");
__asm__("nop"); __asm__("nop"); __asm__("nop"); __asm__("nop");
for (int i=0; i<2000000000; i++) {
s += test(a, b, c, d);
}
return s;
}
two.cpp:
bool test(int a, int b, int c, int d) {
// return a == d || b == d || c == d;
return false;
}
There are 16 nops in one.cpp. You can comment/decomment them to change alignment of the loop's entry point between 16 and 32. I've compiled them with g++ one.cpp two.cpp -O3 -mtune=native.
Here are my questions:
- the 32-aligned version is faster than the 16-aligned version. On Sandy Bridge, the difference is 20%; on Haswell, 8%. Why is the difference?
- with the 32-aligned version, the code runs the same speed on Sandy Bridge, doesn't matter which return statement is in two.cpp. I thought the
return falseversion should be faster at least a little bit. But no, exactly the same speed! - If I remove
volatiles from one.cpp, code becomes slower (Haswell: before: ~2.17 sec, after: ~2.38 sec). Why is that? But this only happens, when the loop aligned to 32.
The fact that 32-aligned version is faster, is strange to me, because Intel® 64 and IA-32 Architectures Optimization Reference Manual says (page 3-9):
Assembly/Compiler Coding Rule 12. (M impact, H generality) All branch targets should be 16- byte aligned.
Another little question: is there any tricks to make only this loop 32-aligned (so rest of the code could keep using 16-byte alignment)?
Note: I've tried compilers gcc 6, gcc 7 and clang 3.9, same results.
Here's the code with volatile (the code is the same for 16/32 aligned, just the address differ):
0000000000000560 <main>:
560: 41 57 push r15
562: 41 56 push r14
564: 41 55 push r13
566: 41 54 push r12
568: 55 push rbp
569: 31 ed xor ebp,ebp
56b: 53 push rbx
56c: bb 00 94 35 77 mov ebx,0x77359400
571: 48 83 ec 18 sub rsp,0x18
575: c7 04 24 01 00 00 00 mov DWORD PTR [rsp],0x1
57c: c7 44 24 04 02 00 00 mov DWORD PTR [rsp+0x4],0x2
583: 00
584: c7 44 24 08 03 00 00 mov DWORD PTR [rsp+0x8],0x3
58b: 00
58c: c7 44 24 0c 04 00 00 mov DWORD PTR [rsp+0xc],0x4
593: 00
594: 44 8b 3c 24 mov r15d,DWORD PTR [rsp]
598: 44 8b 74 24 04 mov r14d,DWORD PTR [rsp+0x4]
59d: 44 8b 6c 24 08 mov r13d,DWORD PTR [rsp+0x8]
5a2: 44 8b 64 24 0c mov r12d,DWORD PTR [rsp+0xc]
5a7: 0f 1f 44 00 00 nop DWORD PTR [rax+rax*1+0x0]
5ac: 66 2e 0f 1f 84 00 00 nop WORD PTR cs:[rax+rax*1+0x0]
5b3: 00 00 00
5b6: 66 2e 0f 1f 84 00 00 nop WORD PTR cs:[rax+rax*1+0x0]
5bd: 00 00 00
5c0: 44 89 e1 mov ecx,r12d
5c3: 44 89 ea mov edx,r13d
5c6: 44 89 f6 mov esi,r14d
5c9: 44 89 ff mov edi,r15d
5cc: e8 4f 01 00 00 call 720 <test(int, int, int, int)>
5d1: 0f b6 c0 movzx eax,al
5d4: 01 c5 add ebp,eax
5d6: 83 eb 01 sub ebx,0x1
5d9: 75 e5 jne 5c0 <main+0x60>
5db: 48 83 c4 18 add rsp,0x18
5df: 89 e8 mov eax,ebp
5e1: 5b pop rbx
5e2: 5d pop rbp
5e3: 41 5c pop r12
5e5: 41 5d pop r13
5e7: 41 5e pop r14
5e9: 41 5f pop r15
5eb: c3 ret
5ec: 0f 1f 40 00 nop DWORD PTR [rax+0x0]
Without volatile:
0000000000000560 <main>:
560: 55 push rbp
561: 31 ed xor ebp,ebp
563: 53 push rbx
564: bb 00 94 35 77 mov ebx,0x77359400
569: 48 83 ec 08 sub rsp,0x8
56d: 66 0f 1f 84 00 00 00 nop WORD PTR [rax+rax*1+0x0]
574: 00 00
576: 66 2e 0f 1f 84 00 00 nop WORD PTR cs:[rax+rax*1+0x0]
57d: 00 00 00
580: b9 04 00 00 00 mov ecx,0x4
585: ba 03 00 00 00 mov edx,0x3
58a: be 02 00 00 00 mov esi,0x2
58f: bf 01 00 00 00 mov edi,0x1
594: e8 47 01 00 00 call 6e0 <test(int, int, int, int)>
599: 0f b6 c0 movzx eax,al
59c: 01 c5 add ebp,eax
59e: 83 eb 01 sub ebx,0x1
5a1: 75 dd jne 580 <main+0x20>
5a3: 48 83 c4 08 add rsp,0x8
5a7: 89 e8 mov eax,ebp
5a9: 5b pop rbx
5aa: 5d pop rbp
5ab: c3 ret
5ac: 0f 1f 40 00 nop DWORD PTR [rax+0x0]
回答1:
This doesn't answer point 2 (return a == d || b == d || c == d; being the same speed as return false). That's still a maybe-interesting question, since that must compile multiple uop-cache lines of instructions.
The fact that 32-aligned version is faster, is strange to me, because [Intel's manual says to align to 32]
That optimization-guide advice is a very general guideline, and definitely doesn't mean that larger never helps. Usually it doesn't, and padding to 32 would be more likely to hurt than help. (I-cache misses, ITLB misses, and more code bytes to load from disk).
In fact, 16B alignment is rarely necessary, especially on CPUs with a uop cache. For a small loop that can run from the loop buffer, it alignment is usually totally irrelevant.
16B is still not bad as a broad recommendation, but it doesn't tell you everything you need to know to understand one specific case on a couple of specific CPUs.
Compilers usually default to aligning loop branches and function entry-points, but usually don't align other branch targets. The cost of executing a NOP (and code bloat) is often larger than the likely cost of an unaligned non-loop branch target.
Code alignment has some direct and some indirect effects. The direct effects include the uop cache on Intel SnB-family. For example, see Branch alignment for loops involving micro-coded instructions on Intel SnB-family CPUs.
Another section of Intel's optimization manual goes into some detail about how the uop cache works:
2.3.2.2 Decoded ICache:
- All micro-ops in a Way (uop cache line) represent instructions which are statically contiguous in the code and have their EIPs within the same aligned 32-byte region. (I think this means an instruction that extends past the boundary goes in the uop cache for the block containing its start, rather than end. Spanning instructions have to go somewhere, and the branch target address that would run the instruction is the start of the insn, so it's most useful to put it in a line for that block).
- A multi micro-op instruction cannot be split across Ways.
- An instruction which turns on the MSROM consumes an entire Way.
- Up to two branches are allowed per Way.
- A pair of macro-fused instructions is kept as one micro-op.
See also Agner Fog's microarch guide. He adds:
- An unconditional jump or call always ends a μop cache line
- lots of other stuff that that probably isn't relevant here.
Also, that if your code doesn't fit in the uop cache, it can't run from the loop buffer.
The indirect effects of alignment include:
- larger/smaller code-size (L1I cache misses, TLB). Not relevant for your test
- which branches alias each other in the BTB (Branch Target Buffer).
If I remove
volatiles from one.cpp, code becomes slower. Why is that?
The larger instructions push the last instruction into the loop across a 32B boundary:
59e: 83 eb 01 sub ebx,0x1
5a1: 75 dd jne 580 <main+0x20>
So if you aren't running from the loop buffer (LSD), then without volatile one of the uop-cache fetch cycles gets only 1 uop.
If sub/jne macro-fuses, this might not apply. And I think only crossing a 64B boundary would break macro-fusion.
Also, those aren't real addresses. Have you checked what the addresses are after linking? There could be a 64B boundary there after linking, if the text section has less than 64B alignment.
Sorry I haven't actually tested this to say more about this specific case. The point is, when you bottleneck on the front-end from stuff like having a call/ret inside a tight loop, alignment becomes important and can get is extremely complex. Boundary-crossing or not for all future instructions is affected. Do not expect it to be simple. If you've read my other answers, you'll know I'm not usually the kind of person to say "it's too complicated to fully explain", but alignment can be that way.
See also Code alignment in one object file is affecting the performance of a function in another object file
In your case, make sure tiny functions inline. Use link-time optimization if your code-base has any important tiny functions in separate .c files instead of in a .h where they can inline. Or change your code to put them in a .h.
来源：https://stackoverflow.com/questions/45298870/why-does-loop-alignment-on-32-byte-make-code-faster