删除D元编程/编译时数组生成的突变

问题描述

我的计划是用D语言编写一个无变异的代码，以便我的值在运行时可用.有人跟我谈到了循环展开和编译时代码生成的问题，但我不清楚如何做到这一点.我在下面做了D模板，但是不能保证在编译时会进行评估，因为这是在两个赋值语句(突变)上进行的.建议将不胜感激.建议最好是在没有宏的D或C ++中使用.

My plan is to write a mutation-less code in D-language so that my values are available by runtime. Someone spoke to me about loop-unrolling and compile time code generation but I have no clear idea how to do that. I have made the D-template below but it has no guarantee to be evaluated at compile-time because on the two assignment statements(mutations) . Advice would be greatly appreciated. Suggestions could be preferably in D or C++ without macros.

import std.stdio;
import std.string;
import std.conv;

const char[] ALPHABET="ABFCDFRGHDSTHFG";
const string pattern="ABA";


I[C]  computeAtCompileTime(S ,C,I)( const S  pattern ){
  I[C] table1;

  const int size = to!int(pattern.length) ;//Length of the pattern to be matched

  foreach( c; ALPHABET){   //Initialise array
          table1[c] = size;
    }

  foreach(i; 0..size-1){
             table1[pattern[i]] = size -i-1;
  }
  return table1;
}
enum TableFromCompiler  = computeAtCompileTime!(const string ,char, int)(pattern);
void main(){

    // enum TableFromCompiler  = computeAtCompileTime!(const string ,char, int)(pattern);

     writeln(TableFromCompiler);

 }

推荐答案

import std.stdio;

immutable char[] ALPHABET = "ABFCDFRGHDSTHFG";

int[char] compute(in string pattern)
{
    int[char] table;

    foreach (c; ALPHABET) {
        table[c] = cast(int)pattern.length;
    }

    foreach (i, c; pattern) {
        table[c] = cast(int)(pattern.length - i - 1);
    }

    return table;
}

void main()
{
    enum table = compute("ABA");
    writeln(table);
}

输出:

['A':0, 'B':1, 'C':3, 'D':3, 'F':3, 'G':3, 'H':3, 'R':3, 'S':3, 'T':3]

dpaste上的代码.

这篇关于删除D元编程/编译时数组生成的突变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！