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javascript - 在 javascript 解释器中解析循环

问题描述

我一直在探索用 javascript 编写一个非常基本/有限的解释器作为练习。在我介绍 LOOP 的概念之前，一切都进行得很顺利。

给定以下脚本：

LOOP 2
  A
  LOOP 3
    B
  END
  LOOP 4
    C
    LOOP 5
      D
    END
    E
  END
  F
END

该算法应按以下顺序访问内部令牌：

ABBBCDDDDDECDDDDDECDDDDDECDDDDDEFABBBCDDDDDECDDDDDECDDDDDECDDDDDEF

以下是诀窍，但它需要对令牌进行大量迭代。这是对我之前使用的手动扩展循环的切片方法的改进，但远非最佳。

/**
 * In practice, we'll grab each token as we read the script,
 * but to keep this simple and focus on the loop algorithm,
 * we can cheat and make an array of all the tokens.
 */

const getTokens = (s) => s.replace(/[\W_]+/g, " ").split(" ").filter(Boolean);

/* Temp vars - ideally, I'd like to solve this with arrays. */
const start = []; // Loop start indices
const end = []; // Loop end indices
const counts = []; // Times to loop
const completed = []; // Loops completed

for (let i = 0; i < tokens.length; i++) {
  const token = tokens[i];

  if (token === "LOOP") {
    if (start.length == 0 || i > start[start.length - 1]) {
      // Add new loop index if we haven't seen it before
      start.push(i); // Store the loop index
      counts.push(Number(tokens[i + 1])); // The loop count is always next LOOP token
      completed.push(0); // Initialize with 0 completed at index

      // Find the end index for the loop
      // Note: This is the slowest part.
      let skip = 0;
      for (let j = i + 2; j < tokens.length; j++) {
        if (tokens[j] == "LOOP") {
          skip++; // Increase nest depth
        } else if (tokens[j] == "END") {
          if (skip == 0) {
            end.push(j); // Found matching loop close
            break;
          }
          skip--;
        }
      }
    }

    i++; // Skip over the loop count
    continue;
  } else if (token === "END") {
    let j;
    for (j = 0; j < end.length; j++) {
      if (end[j] == i) break; // Found matching end index
    }
    const isCompleted = completed[j] == counts[j] - 1;
    if (!isCompleted) {
      i = start[j] + 1;
      completed[j]++;
      for (let k = j + 1; k < start.length; k++) {
        completed[k] = 0; // Reset nested loops in between
      }
    }
    continue;
  }

  console.log(tokens[i]);
}

https://jsfiddle.net/5wpa8t4n/

使用单遍脚本或最坏的 2 遍而不是 N-LOOP 遍来完成这种基于数组的方法的更好方法是什么？

标签： javascriptloopsecmascript-6computer-scienceinterpreter

解决方案

end在开始解释它时，您不需要知道循环匹配的位置。您需要记录的是遇到 next 时要跳回的位置end，但在那之前只需继续逐个解释。

这些位置与相应的计数器一起可以存储在堆栈结构中。

const script = `
DO
  A
  DO
    B
  LOOP 3
  DO
    C
    DO
      D
    LOOP 5
    E
  LOOP 4
  F
LOOP 2
`

const parse = (script) =>
  script
    .replace(/[\W_]+/g, " ")
    .split(" ")
    .filter(Boolean);

const interpret = (code) => {
  let loops = []; // Active loops: iteration count and jump target
  let ip = 0; // instruction pointer

  let result = "";
  while (ip < code.length) {
    const instruction = code[ip];
    switch (instruction) {
      case "DO": {
        ++ip;
        loops.push({count: 0, start: ip});
      } break;
      case "LOOP": {
        const limit = Number(code[++ip]);
        const {count, start} = loops.pop();
        if (count < limit) {
          loops.push({count: count+1, start});
          ip = start; // jump back
        } else {
          ++ip;
        }
      } break;
      default: {
        ++ip;
        result += instruction; // a print statement
      } break;
    }
  }
  return result;
};

console.log(interpret(parse(script)));

我已经稍微简化了结构以使用do-while循环，所以我永远不必跳过循环体。在解析器发出的真正字节码中，跳转目标（来回）将是指令本身的一部分，并且只有计数“变量”需要存储在堆栈中。跳转目标永远不会改变，因此您只需要在parse函数中生成一次。

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