performance - Why does loop alignment on 32 byte make code faster?

Question

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:

1. 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?
2. 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 false version should be faster at least a little bit. But no, exactly the same speed!
3. 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]

Answer 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.