微优化c比较功能

我有一个Compare()函数,如下所示：

inline bool Compare(bool greater,int p1,int p2) {
  if (greater) return p1>=p2;
  else return p1<=p2;
}

我决定优化以避免分歧：

inline bool Compare2(bool greater,int p2) {
  bool ret[2] = {p1<=p2,p1>=p2};
  return ret[greater];
}

然后我通过这样做测试：

bool x = true;
int M = 100000;
int N = 100;

bool a[N];
int b[N];
int c[N];

for (int i=0;i<N; ++i) {
  a[i] = rand()%2;
  b[i] = rand()%128;
  c[i] = rand()%128;
}

// Timed the below loop with both Compare() and Compare2()
for (int j=0; j<M; ++j) {
  for (int i=0; i<N; ++i) {
    x ^= Compare(a[i],b[i],c[i]);
  }
}

结果：

Compare(): 3.14ns avg
Compare2(): 1.61ns avg

我会说个案封闭,避免分支FTW.但为了完整,我换了

a[i] = rand()%2;

有：

a[i] = true;

并得到〜3.14ns的完全相同的测量.大概当时没有分支,编译器实际上是重写Compare()来避免if语句.但是,为什么Compare2()更快？

不幸的是,我是汇编代码文盲,否则我本来会尝试回答这个问题.

编辑：下面是一些程序集：

_Z7Comparebii:
.LFB4:
    .cfi_startproc
    .cfi_personality 0x3,__gxx_personality_v0
    pushq   %rbp
    .cfi_def_cfa_offset 16
    movq    %rsp,%rbp
    .cfi_offset 6,-16
    .cfi_def_cfa_register 6
    movl    %edi,%eax
    movl    %esi,-8(%rbp)
    movl    %edx,-12(%rbp)
    movb    %al,-4(%rbp)
    cmpb    $0,-4(%rbp)
    je      .L2
    movl    -8(%rbp),%eax
    cmpl    -12(%rbp),%eax
    setge   %al
    jmp     .L3
.L2:
    movl    -8(%rbp),%eax
    setle   %al
.L3:
    leave
    ret
    .cfi_endproc
.LFE4:
    .size   _Z7Comparebii,.-_Z7Comparebii
    .section        .text._Z8Compare2bii,"axG",@progbits,_Z8Compare2bii,comdat
    .weak   _Z8Compare2bii
    .type   _Z8Compare2bii,@function
_Z8Compare2bii:
.LFB5:
    .cfi_startproc
    .cfi_personality 0x3,-24(%rbp)
    movl    %edx,-28(%rbp)
    movb    %al,-20(%rbp)
    movw    $0,-16(%rbp)
    movl    -24(%rbp),%eax
    cmpl    -28(%rbp),%eax
    setle   %al
    movb    %al,%eax
    setge   %al
    movb    %al,-15(%rbp)
    movzbl  -20(%rbp),%eax
    cltq
    movzbl  -16(%rbp,%rax),%eax
    leave
    ret
    .cfi_endproc
.LFE5:
    .size   _Z8Compare2bii,.-_Z8Compare2bii
    .text

现在,执行测试的实际代码可能是使用上述两个函数的内联版本,所以有可能这是错误的代码分析.就这样说,我在Compare()中看到一个jmp命令,所以我认为这意味着它是分支.如果是这样,我猜这个问题就变成了：为什么当我将一个[i]从rand()％2更改为true(或者对于该事情是false)时,为什么分支预测器不能提高Compare()的性能？

编辑2：我用“分支”代替“分支预测”,使我的帖子更加明智.

解决方法

我写了一个名为Celero的C库,旨在测试这样的优化和替代方案. (无耻自我推销： https://github.com/DigitalInBlue/Celero)

我使用以下代码运行您的案例：

class StackOverflowFixture : public celero::TestFixture
{
  public:
    StackOverflowFixture()
    {
    }

    inline bool NoOp(bool greater,int p2) 
    {
      return true;
    }

    inline bool Compare(bool greater,int p2) 
    {
      if(greater == true)
      {
        return p1>=p2;
      }

      return p1<=p2;
    }

    inline bool Compare2(bool greater,int p2)
    {
      bool ret[2] = {p1<=p2,p1>=p2};
      return ret[greater];
    }

    inline bool Compare3(bool greater,int p2) 
    {
      return (!greater != !(p1 <= p2)) | (p1 == p2);
    }

    inline bool Compare4(bool greater,int p2) 
    {
      return (greater ^ (p1 <= p2)) | (p1 == p2);
    }
};

BASELINE_F(StackOverflow,Baseline,StackOverflowFixture,100,5000000)
{
  celero::DoNotOptimizeAway(NoOp(rand()%2,rand(),rand()));
}

BENCHMARK_F(StackOverflow,Compare,5000000)
{
  celero::DoNotOptimizeAway(Compare(rand()%2,Compare2,5000000)
{
  celero::DoNotOptimizeAway(Compare2(rand()%2,Compare3,5000000)
{
  celero::DoNotOptimizeAway(Compare3(rand()%2,Compare4,5000000)
{
  celero::DoNotOptimizeAway(Compare4(rand()%2,rand()));
}

结果如下：

[==========]
[  CELERO  ]
[==========]
[ STAGE    ] Baselining
[==========]
[ RUN      ] StackOverflow.Baseline -- 100 samples,5000000 calls per run.
[     DONE ] StackOverflow.Baseline  (0.690499 sec) [5000000 calls in 690499 usec] [0.138100 us/call] [7241140.103027 calls/sec]
[==========]
[ STAGE    ] Benchmarking
[==========]
[ RUN      ] StackOverflow.Compare -- 100 samples,5000000 calls per run.
[     DONE ] StackOverflow.Compare  (0.782818 sec) [5000000 calls in 782818 usec] [0.156564 us/call] [6387180.672902 calls/sec]
[ BASELINE ] StackOverflow.Compare 1.133699
[ RUN      ] StackOverflow.Compare2 -- 100 samples,5000000 calls per run.
[     DONE ] StackOverflow.Compare2  (0.700767 sec) [5000000 calls in 700767 usec] [0.140153 us/call] [7135039.178500 calls/sec]
[ BASELINE ] StackOverflow.Compare2 1.014870
[ RUN      ] StackOverflow.Compare3 -- 100 samples,5000000 calls per run.
[     DONE ] StackOverflow.Compare3  (0.709471 sec) [5000000 calls in 709471 usec] [0.141894 us/call] [7047504.408214 calls/sec]
[ BASELINE ] StackOverflow.Compare3 1.027476
[ RUN      ] StackOverflow.Compare4 -- 100 samples,5000000 calls per run.
[     DONE ] StackOverflow.Compare4  (0.712940 sec) [5000000 calls in 712940 usec] [0.142588 us/call] [7013212.893091 calls/sec]
[ BASELINE ] StackOverflow.Compare4 1.032500
[==========]
[ COMPLETE ]
[==========]

考虑到这个测试,看起来Compare2是这个微型优化的最佳选择.

编辑：

Compare2装配(最好的情况)：

cmp r8d,r9d
movzx   eax,dl
setle   BYTE PTR ret$[rsp]
cmp r8d,r9d
setge   BYTE PTR ret$[rsp+1]
movzx   eax,BYTE PTR ret$[rsp+rax]

比较3装配(最好的情况)：

xor r11d,r11d
cmp r8d,r9d
mov r10d,r11d
setg    r10b
test    dl,dl
mov ecx,r11d
sete    cl
mov eax,r11d
cmp ecx,r10d
setne   al
cmp r8d,r9d
sete    r11b
or  eax,r11d

解决方法

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