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2012-06-27T10:04:21.933
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Since when is RTFM an answer? Anyways. What you are trying to do is not at all easy, since Java is context free (type-2). Try, for starters writing a syntax-analyser for a language of type-3 (<a href="http://en.wikipedia.org/wiki/Chomsky_hierarchy" rel="nofollow">Chomsky Hierarchy</a>). But I'll try to explain to you what you'd need to do anyways. You would have to define rules for Java which look like that (in my example I'll define a function within a java class, where lowercase letters are terminals and uppercase letters are non-terminals). Terminals cannot be derived any further while non-terminals can. X -> Y means X derives to Y. X -> Y | Z means X derives either to Y or Z. f is any name. t is a Type, this would not be a terminal, if I'd try to go all the way, but since I define types to be non-declarable to make my life less of a pain, it's a terminal. '(', ')', '{', '}', ',' and ' ' are terminals. Eps is Epsilon and means nothing. <pre><code>S -> K t f(T) { N } T -> t f | t f , T F -> F, f | f K -> k K | k N -> L N | L L -> f(F); </code></pre> With this I could parse <pre><code>final boolean equals(Object obj) { compare(this, obj); compare(obj, this); } </code></pre> Which would result in: <pre><code>S -> K t f(T) { N } with K -> k -> k t f(T) { N } with T -> t f -> k t f(t f) { N } with N -> L N -> k t f(t f) { L N } with L -> f(F); -> k t f(t f) { f(F); N } with F -> f, F -> k t f(t f) { f(f, F); N } with F -> f -> k t f(t f) { f(f, f); N } with N -> L -> k t f(t f) { f(f, f); L } with L -> f(F) -> k t f(t f) { f(f, f); f(F) } ... -> k t f(t f) { f(f, f); f(f, f); } -> k (=final) t(=boolean) f(=equals) (t(=Object) f(=obj)) { ... } </code></pre> Which proves that S defines my simplyfied java (Well it doesn't, but at least I gave an example). So the next thing we have to do is figure out how to get a syntax-tree out of these rules. Thankfully, this is the easy part, because all you have to do is change the lines to be a tree. So S has children K t f(T) { N }. K has children K and k... Uppercase means a Node has children, Lowercase says it doesn't. Last problem, you do not start with S but you start with the code already being written. Which leaves you with <pre><code>K t f(T) { N } -> S t f -> T t f , T -> T F, f -> F f -> F k K -> K k -> K L N -> N N -> L f(F); -> L </code></pre> Parsing in reverse would look like this: <pre><code>final boolean equals(Object obj) { compare(this, obj); compare(obj, this); } final -> k boolean -> t equals -> f Object -> t obj -> f compare -> f this -> f k t f(t f) { f(f, f); f(f,f); } with k -> K K t f(t f) ... with t f -> T K t f(T) ... ... </code></pre> Which would build the tree from down below.
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1. USSebastian van Wickern
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