强制预先计算常量 [英] Force pre-computation of a constant

问题描述

我在Haskell中有一个常量声明 - 我能否提前对它进行评估？我看到一些代码看起来很粗糙，

  myList = [（a，b），（c，d）] 
 ... 
 map（f。fst）myList 
  

需要时间当我分析它时（它确实有168M的调用）时， fst 调用。 myList 的二进制表示非常小，例如可以复制到全局内存中（如果这是一个C程序）。我正在编译 -O3 -optc-O3 当然。

非常感谢！

为一个自定义类型生成Lift实例

赋予 lift call in sclv的答案必须是Lift的一个实例。有一个名为 th-lift 的库，它将为自定义数据类型生成Lift实例。查看该软件包的文档。

解决方案

使用Template Haskell生成编译时常量：

  { - ＃LANGUAGE TemplateHaskell＃ - } 
 import Language.Haskell.TH.Syntax（Lift（..））
 
 test = $（lift $（map（* 20）[0..100] :: [Int]））
  

lift 需要一个Haskell值并将其提升到TH Exp。 $（）运行封闭的引号，并在编译时将生成的exp拼接到代码中。

I have a constant declaration in Haskell -- can I force this to be evaluated ahead of time? I'm seeing some code that looks roughly like,

myList = [(a, b), (c, d)]
...
map (f . fst) myList

take time in the fst call when I profile it (it does have 168M calls). The binary representation of myList is quite small, and could be, for example, copied into global memory [if this were a C program]. I'm compiling with -O3 -optc-O3 of course.

Thanks very much!

Generating Lift instances for a custom type

Any expression given to the lift call in sclv's answer must be an instance of Lift. There's a library named th-lift which will generate Lift instances for custom data types. See that package's documentation.

解决方案

Generating a compile-time constant using Template Haskell:

{-# LANGUAGE TemplateHaskell #-}
import Language.Haskell.TH.Syntax(Lift(..))

test = $(lift $ (map (*20) [0..100] :: [Int]))

lift takes a Haskell value and lifts it into a TH Exp. The $() runs the enclosed quote, and splices the generated exp into the code at compile time.

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