在模板函数中包含不变假设 [英] Including an invariant assumption in a template function

问题描述

考虑一个典型的有限差分应用程序:

Consider a typical finite difference application:

// assuming T_size > 2
void process_T(double *T0, double *T, const int &T_size, bool periodic) {
  for (int i = 0; i < T_size; ++i) {
    double sum = 0;
    double base = T0[i];
    if (i > 0) sum += (T0[i-1]-base);
    if (i < 0) sum += (T0[i+1]-base);
    if (periodic) { 
       if (i == 0) sum += (T0[T_size-1]-base);
       if (i == T_size-1) sum += (T0[0]-base);
    } else {
      if (i == 1 || i == T_size-1) sum += 0.5*(T0[i-1]-base);
      if (i == 0 || i == T_size-2) sum += 0.5*(T0[i+1]-base);
    }
    T[i] = T0[i] + sum * 0.08; // where 0.08 is some magic number
  }
}

periodic的检查是循环不变的，但是由于仅在运行时才知道，因此每次都会产生条件检查费用.我可以创建一个假设其中一种情况的专用函数，但是要维护通用基数会很麻烦，尤其是在三维问题中，它会增长到8个函数(周期性:none，x，y，z， xy，xz，yz，xyz)来考虑所有组合.

The check for periodic is loop-invariant, but since is only known at run-time, the conditional check cost is incurred everytime. I could create a specialized function which assumes one of the cases, but it would be cumbersome to maintain the common base, especially in case of three-dimensional problem where it would grow to 8 functions (periodicity: none, x, y, z, xy, xz, yz, xyz) to consider all combinations.

是否可以通过元编程解决此问题?

Is it possible to solve this problem via metaprogramming?

P/S:分支预测器可以相应地对其进行优化吗?

P/S: can the branch predictor optimize this accordingly?

推荐答案

是的，您可以拥有

enum Periodicity
{
    PERIODICITY_NONE,
    PERIODICITY_X,
    PERIODICITY_Y
    // etc
};

然后

template <Periodicity P>
void process_T(double* T0, double* T, const int& T_size)
{
    if (P == PERIODICITY_NONE) // ... do something
    if (P == PERIODICITY_X) // ... do something else

    // Common code
}

任何体面的优化编译器都可以在编译时执行检查，并且可以消除任何无效代码(即使在-O0上，g ++似乎也可以这样做).

Any decent optimising compiler would be able to perform the check at compile time, and would eliminate any dead code (g++ appears to do this even at -O0).

这篇关于在模板函数中包含不变假设的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！