A faster way to build model in inria Spoon

Question

I am using inria Spoon to parse Java projects, and then extract information about classes, Interfaces, fields, and methods and all their references.

I am using the below code to build model of an input project.

SpoonAPI spoonAPI = new Launcher();
spoonAPI.addInputResource(projectDirectory);
spoonAPI.buildModel();
CtModel ctModel = spoonAPI.getModel();

However, buildModel() is very time consuming in big projects.

PS: I used JavaParser using the below code and it is more faster than Spoon.

Path pathToSource = Paths.get(projectDirectory);
SourceRoot sourceRoot = new SourceRoot(pathToSource);
sourceRoot.tryToParse();
List<CompilationUnit> compilations = sourceRoot.getCompilationUnits();

I was wondering if there is any faster way to create the CtModel in Spoon.

Yes, JavaParser is fast to build a model but may not provide the API you need. — Martin Monperrus, Sep 02 '20 at 04:49
It is correct as I am facing some problems in converting my current application (currently based on spoon) to JavaParser. — Iman, Sep 02 '20 at 06:26

score 0 · Answer 1 · answered Sep 02 '20 at 04:50

0

Instead of parsing the full source directory, you can build the model only for the files you need to analyze or transform:

spoonAPI.addInputResource("path/Foo.java"); 
spoonAPI.addInputResource("path/Bar.java");

answered Sep 02 '20 at 04:50

Martin Monperrus

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Yes, but I need to parse the whole project. – Iman Sep 02 '20 at 06:24

A faster way to build model in inria Spoon

1 Answers1