Question

我正在使用javaparser-1.0.8而我正在尝试生成以下通用方法。

public <T extends SomeInterface> T get(int param) {
  return (T) doSomeMagic(param);
}

我有以下代码来构建方法：

public static void main(String[] args) throws Exception {

    // creates an input stream for the file to be parsed
    File mainActivity = new File("<path>/Main.java");
    FileInputStream in = new FileInputStream(mainActivity);

    CompilationUnit cu;
    try {
        // parse the file
        cu = JavaParser.parse(in);
        addMethod(cu);
    } finally {
        in.close();
    }

    // prints the resulting compilation unit to default system output
    System.out.println(cu.toString());
}

private static void addMethod(CompilationUnit cu) {
        WildcardType wildcardType = new japa.parser.ast.type.WildcardType(ASTHelper.createReferenceType("SomeInterface", 0));
        MethodDeclaration method = new MethodDeclaration(ModifierSet.PUBLIC, wildcardType, "get");
        method.setModifiers(ModifierSet.addModifier(method.getModifiers(), ModifierSet.PUBLIC));
        Parameter param = ASTHelper.createParameter(ASTHelper.INT_TYPE, "id");
        ASTHelper.addParameter(method, param);
        BlockStmt block = new BlockStmt();
        method.setBody(block);
        ASTHelper.addMember(cu.getTypes().get(0), method);
}

输出：

public ? extends SomeInterface get(int id) {
}

Answer 1

我有类似的情况（但并不完全）。我想生成

public Class<? extends SomeInterface> getSomeClass() {
   return SomeImplementation.class;
}

并且这样做：

Type w = new WildcardType(ASTHelper.createReferenceType("SomeInterface", 0));
ClassOrInterfaceType returnType = new ClassOrInterfaceType("Class");
ret.setTypeArgs(Arrays.asList(w));
MethodDeclaration mesthod = new MethodDeclaration(
    ModifierSet.PUBLIC, returnType, "getSomeClass");

JavaParser将Generic Type添加为方法返回类型

1 个答案: