小编Ben*_*tic的帖子

当我将一些fortran代码移植到c时,令我感到惊讶的是,使用ifort(intel fortran编译器)编译的fortran程序与使用gcc编译的c程序之间的大部分执行时间差异来自于对三角函数的评估(sin,cos).这令我感到惊讶,因为我曾经相信这个答案解释的是,正弦和余弦等函数是在微处理器内部的微码中实现的.

为了更明确地发现问题,我在fortran做了一个小测试程序

program ftest
  implicit none
  real(8) :: x
  integer :: i
  x = 0d0
  do i = 1, 10000000
    x = cos (2d0 * x)
  end do
  write (*,*) x
end program ftest

在intel Q6600处理器上,3.6.9-1-ARCH x86_64 Linux 我得到了ifort version 12.1.0

$ ifort -o ftest ftest.f90 
$ time ./ftest
  -0.211417093282753     

real    0m0.280s
user    0m0.273s
sys     0m0.003s

而与gcc version 4.7.2我得到

$ gfortran -o ftest ftest.f90 
$ time ./ftest …

我将使用以下示例程序演示此问题

{-# LANGUAGE BangPatterns #-}

data Point = Point !Double !Double

fmod :: Double -> Double -> Double
fmod a b | a < 0     = b - fmod (abs a) b 
         | otherwise = if a < b then a 
                       else let q = a / b 
                            in b * (q - fromIntegral (floor q :: Int))

standardMap :: Double -> Point -> Point
standardMap k (Point q p) = 
   Point (fmod (q + p) (2 * pi)) (fmod …

我的问题的描述几乎和这篇文章一样,但是虽然我认为我可以理解相应的解决方案,但我看不出它如何适用于我的问题,如果有的话.

这是我的示例程序

{-# LANGUAGE BangPatterns #-}

import System.Random (randoms, mkStdGen)
import Control.Parallel.Strategies
import Control.DeepSeq (NFData)
import Data.List

data Point = Point !Double !Double

fmod :: Double -> Double -> Double
fmod a b | a < 0     = b - fmod (abs a) b 
         | otherwise = if a < b then a 
                        else let q = a / b 
                             in b * (q - fromIntegral (floor q :: Int))

standardMap :: Double -> Point -> …

I am dealing with the computation which has as an intermediate result a list A=[B], which is a list of K lists of the length L. The time-complexity to compute an element of B is controlled by the parameter M and is theoretically linear in M. Theoretically I would expect the time-complexity for the computation of A to be O(K*L*M). However, this is not the case and I don't understand why?

Here is the simple complete sketch program which exhibits …