同一指令具有不同的CPU周期

时间:2018-10-18 08:20:12

标签: c++ performance intel avx instructions

我正在调试一个c ++程序,两个不同的输入导致几乎相同的数字或指令,但是cpu周期之间存在明显的差异。

下面是perf stat信息,这确实很奇怪。有人知道什么可能导致这种时差吗?

./sai2 100000,C,98.700,60,98.695,0.2,0.0的性能计数器统计信息:

   2684.371940      task-clock (msec)         #    1.000 CPUs utilized          
            17      context-switches          #    0.006 K/sec                  
             1      cpu-migrations            #    0.000 K/sec                  
           921      page-faults               #    0.343 K/sec                  
 7,292,413,665      cycles                    #    2.717 GHz                    
11,267,827,416      instructions              #    1.55  insn per cycle         
   940,483,779      branches                  #  350.355 M/sec                  
       132,437      branch-misses             #    0.01% of all branches        

   2.685433574 seconds time elapsed

./sai2 100000,P,98.700,60,98.695,0.2,0.0的性能计数器统计信息

  25698.831411      task-clock (msec)         #    1.000 CPUs utilized          
            72      context-switches          #    0.003 K/sec                  
            19      cpu-migrations            #    0.001 K/sec                  
           921      page-faults               #    0.036 K/sec                  
70,402,935,601      cycles                    #    2.740 GHz                    
10,418,026,021      instructions              #    0.15  insn per cycle         
   956,483,724      branches                  #   37.219 M/sec                  
       323,444      branch-misses             #    0.03% of all branches        

  25.702346518 seconds time elapsed

真正让我感到困惑的是,大多数时间成本代码几乎相同。

// const double vector definitions
__m256d vec_1_p = _mm256_set1_pd(v_1_p_discount);
__m256d vec_p = _mm256_set1_pd(v_p_discount);
__m256d vec_tricky = _mm256_set_pd(tricky_3, tricky_2, tricky, 1);
__m256d vec_strike = _mm256_set1_pd(strike);

for (int m = m_stepNumber - 1; m > 0; m--) {
  double m_s_pointer = power_d / d;
  power_d = m_s_pointer;
  for (int n = 0; n < m; n += 4) {
    double *val = known + n;
    double *dest = to_calc + n;

    __m256d hold_0 = _mm256_load_pd(val);
    __m256d hold_1 = _mm256_loadu_pd(val+1);
    hold_0 = _mm256_mul_pd(hold_0, vec_1_p);
    hold_1 = _mm256_mul_pd(hold_1, vec_p);
    hold_0 = _mm256_add_pd(hold_0, hold_1);

    __m256d vec_tmp = _mm256_set1_pd(m_s_pointer);
    vec_tmp = _mm256_mul_pd(vec_tmp, vec_tricky);

    //vec_tmp = _mm256_sub_pd(vec_tmp, vec_strike);   this run in quick case
    //vec_tmp = _mm256_sub_pd(vec_strike, vec_tmp);   this run in slow case

    hold_0 = _mm256_max_pd(hold_0, vec_tmp);
    _mm256_store_pd(dest, hold_0);
    m_s_pointer *= tricky_4;
  }
  std::swap(known, to_calc);
}

我已经使用perf record来查看如下所示的较慢情况的热点

    0.00 │520:┌─→vmulpd -0x8(%rdx),%ymm1,%ymm3
   41.48 │    │  add    $0x4,%ecx
    1.94 │    │  add    $0x20,%rdx
    0.52 │    │  vmovup -0x20(%rdx),%ymm0
    3.85 │    │  add    $0x20,%rsi
    0.31 │    │  vfmadd %ymm2,%ymm3,%ymm0
   42.73 │    │  vmovup -0x20(%rsi),%ymm3
    5.53 │    │  vmaxpd %ymm3,%ymm0,%ymm0
    2.77 │    │  vmovap %ymm0,-0x28(%rdx)
    0.86 │    │  cmp    %edi,%ecx 
         │    └──jle    520

更快的情况

      3.43 │398:┌─→vmovup -0x8(%rdx),%ymm3
      9.18 │    │  add    $0x4,%ecx
      1.64 │    │  add    $0x20,%rdx
      5.44 │    │  vmovup -0x20(%rdx),%ymm0
     12.03 │    │  add    $0x20,%rsi
      2.32 │    │  vmulpd %ymm1,%ymm3,%ymm3
      7.28 │    │  vfmadd %ymm2,%ymm3,%ymm0
     16.77 │    │  vmovup -0x20(%rsi),%ymm3
     19.33 │    │  vmaxpd %ymm3,%ymm0,%ymm0
     15.83 │    │  vmovup %ymm0,-0x28(%rdx)
      6.70 │    │  cmp    %edi,%ecx
           │    └──jle    398

0 个答案:

没有答案
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