为什么这个函数的递归版本更快?

时间:2012-06-03 00:03:49

标签: c++ c++11 compiler-optimization

这是一个用于迭代多维数值范围的简单类:

#include <array>
#include <limits>

template <int N>
class NumericRange
{
public:
  //  typedef std::vector<double>::const_iterator const_iterator;
  NumericRange() {
    _lower.fill(std::numeric_limits<double>::quiet_NaN());
    _upper.fill(std::numeric_limits<double>::quiet_NaN());
    _delta.fill(std::numeric_limits<double>::quiet_NaN());
  }
  NumericRange(const std::array<double, N> & lower, const std::array<double, N> & upper, const std::array<double, N> & delta):
    _lower(lower), _upper(upper), _delta(delta) {
    _state.fill(std::numeric_limits<double>::quiet_NaN());
    _next_index_to_advance = 0;
  }

  const std::array<double, N> & get_state() const {
    return _state;
  }

  void start() {
    _state = _lower;
  }

  bool in_range(int index_to_advance = N-1) const {
    return ( _state[ index_to_advance ] - _upper[ index_to_advance ] ) < _delta[ index_to_advance ];
  }

  void advance(int index_to_advance = 0) {
    _state[ index_to_advance ] += _delta[ index_to_advance ];
    if ( ! in_range(index_to_advance) ) {
      if (index_to_advance < N-1) {
    // restart index_to_advance
    _state[index_to_advance] = _lower[index_to_advance];

    // carry
    index_to_advance;
    advance(index_to_advance+1);
      }
    }
  }

private:
  std::array<double, N> _lower, _upper, _delta, _state;
  int _next_index_to_advance;
};

int main() {
  std::array<double, 7> lower{0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0};
  std::array<double, 7> upper{1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0};
  std::array<double, 7> delta{0.03, 0.06, 0.03, 0.06, 0.03, 0.06, 0.03};

  NumericRange<7> nr(lower, upper, delta);
  int c = 0;
  for (nr.start(); nr.in_range(); nr.advance()) {
    const std::array<double, 7> & st = nr.get_state();
    ++c;
  }
  std::cout << "took " << c << " steps" << std::endl;

  return 0;
}

当我用非递归变量替换advance函数时,运行时增加

void advance(int index_to_advance = 0) {
  bool carry;
  do {
    carry = false;
    _state[ index_to_advance ] += _delta[ index_to_advance ];
    if ( ! in_range(index_to_advance) ) {
      if (index_to_advance < N-1) {
    // restart index_to_advance
    _state[index_to_advance] = _lower[index_to_advance];

    // carry
    ++index_to_advance;
    carry = true;
    //    advance(index_to_advance);
      }
    }
  } while (carry);
}

通过命令time使用unix用户时间获取运行时。代码是使用带有选项-std=c++11 -O3的gcc-4.7编译的(但我认为它应该与gcc-4.6上的c++0x一起使用)。递归版本需要13秒,迭代版本需要30秒。两者都需要相同数量的advance调用才能终止(如果在nr.get_state()循环中打印for(ns.start()...)数组,则两者都做同样的事情。)

这是一个有趣的谜语!帮我弄清楚为什么递归会更有效/更可优化。

2 个答案:

答案 0 :(得分:13)

递归版本是 tail-recursion 的一个示例,这意味着编译器可以将递归转换为迭代。现在,一旦执行了转换,递归函数看起来就像这样:

void advance(int index_to_advance = 0) {
    _state[ index_to_advance ] += _delta[ index_to_advance ];
    while ( !in_range(index_to_advance) && index_to_advance < N-1 ) {
        // restart index_to_advance
        _state[index_to_advance] = _lower[index_to_advance];

        // carry
        ++index_to_advance;
        _state[ index_to_advance ] += _delta[ index_to_advance ];
    }
  }

如您所见,您的版本包含一个额外的测试和条件变量。如果仔细观察,循环等同于

for( ; index_to_advance < N-1 && !in_range(index_to_advance);++index_to_advance)

(删除末尾的++index_to_advance),优化器可能更有可能展开它。

话虽如此,我不认为这解释了巨大的时差,尽管它确实解释了为什么递归版本比迭代版本慢得多。检查生成的程序集以查看编译器实际执行的操作。

答案 1 :(得分:3)

只是为David Rodriguez所说的更多细节添加:

使用尾递归优化,函数变为:

 void advance(int index_to_advance = 0) {
  top:
  _state[ index_to_advance ] += _delta[ index_to_advance ];
  if ( ! in_range(index_to_advance) ) {
    if (index_to_advance < N-1) {
      // restart index_to_advance
      _state[index_to_advance] = _lower[index_to_advance];

      // carry
      ++index_to_advance;
      goto top;
    }
  }
}

这确实与我系统上的递归版本具有相同的性能(g ++ 4.6.3 -std = c ++ 0x)