如何为小语言建立翻译？ [英] How to build an interpreter for a small language?

问题描述

我正在试图弄清楚从这个项目开始的位置。也许有人可以引导我朝着正确的方向前进。

I'm trying to figure out where to start with this project. Perhaps someone can steer me in the right direction.

我得到一种小语言，我必须为其编写翻译。该语言由括号中的表达式组成：

I am given a small language that I must write an interpreter for. The language consists of either an expression in parentheses:

(integer integer operator)

或由表达式组成的算术IF语句：

or an arithmetic IF statement made up of expressions in the form:

IF exp1 exp2 exp3 exp4

如果exp1为负数，则返回exp2，$ b如果exp1为零，则返回$ b exp3，如果exp1为正，则返回exp4。

where exp2 is returned if exp1 is negative, exp3 is returned if exp1 is zero, and exp4 is returned if exp1 is positive.

运算符为+或x（用于加法和乘法）分别）。

The operator is either + or x (for addition and multiplication respectively).

我必须一起实现一个扫描器/解析器，然后是将输出结果的解释器。解释器部分并不困难，但我无法弄清楚如何开始扫描/解析过程。

I have to implement a scanner/parser together, then the interpreter that will output the result. The interpreter part will not be difficult, but I'm having trouble figuring out how to start the scanning/parsing process.

我已经开始使用Java了，并且有一个Scanner对象收集输入并将其存储在字符串中。然后我将String拆分为一个String数组，不使用任何作为分隔符（这样每个单个字符，符号，空格等都存储在它自己的字符串索引中）。这可能不是最好的方法，因为我无法弄清楚从哪里开始。我不能理解的部分是如果没有遵循这种语法，如何返回错误，或者如何检测括号和/或IF等等。

I have started by using Java, and have a Scanner object collect the input and store it in a string. Then I split the String into a String array using nothing as the delimiter (so that each single character, symbol, space, etc. is stored in its own index of a string). This may not be the best way to do this, as I cannot figure out where to go from here. The part I cannot grasp is how to return errors if this syntax isn't followed, or how to detect the parenthesis and/or IF and etc.

这是以下的片段我在上一段描述的代码：

Here's the snippet of code that I described in the last paragraph:

public void run() {
    Scanner sc = new Scanner(System.in);

    while (sc.hasNext()) {
        String sLine = sc.nextLine();
        String[] scanned = sLine.split("");

输入示例：

(7 2 +)

Output: 9

IF (2 -2 +) (5 2 +) (5 -2 x) (5 2 x)

Output: -10

如果有人有好的方向对我来说，请分享。 ：）

If anybody has a good direction for me to take, please share. :)

推荐答案

我认为使用ANTLR，JavaCC，SampleCC或其他解析器生成器工具将会使用大锤来解决问题。如果语法定义中没有递归，只需要几个方法就足够了。下面的代码给出了一个基本的想法（它可能无法编译或工作，我从头开始编写如何开始的插图）：

I think using ANTLR, JavaCC, SampleCC or other parser generator tools would be using a sledgehammer to crack a nut. if there is no recursion in grammar definition just a few methods would be sufficient. the following code gives a basic idea (it may not compile or work, I wrote it from scratch as an illustration how to start):

public int parse(String input) {
Scanner scanner = new Scanner(input);

    return consumeLine(scanner);
}

public int consumeLine(Scanner scanner) {
    if( scanner.hasNext("(") ) {
        return consumeExpression(scanner);

    } else if( scanner.hasNext("IF") ) {
        return consumeIf(scanner);
    }
}


public int consumeExpression(Scanner scanner) {
    scanner.next("(");
    int a = scanner.nextInt();
    int b = scanner.nextInt();
    String op = scanner.next("[+-/*]");
    scanner.next(")");

    if( "+".equals(op) ) {
        return a + b;

    } else if( "-".equals(op) ) {
        return a - b;
    } ...

    throw new RuntimeException("parsing error");
}

public int consumeIf(Scanner scanner) {
    scanner.next("IF");
    int exp1 = consumeExpression(scanner);
    int exp2 = consumeExpression(scanner);
    int exp3 = consumeExpression(scanner);
    int exp4 = consumeExpression(scanner);

    if( exp1 < 0 ) {
        return exp2;
    } else if( exp1 == 0 ) {
        return exp3;
    } ...

    throw new RuntimeException("should not be here (TM)");
}

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