Mid*_*hun 6 java eclipse eclipse-jdt
我这里有一个代码:
public class TestOverride {
int foo() {
return -1;
}
}
class B extends TestOverride {
@Override
int foo() {
// error - quick fix to add "return super.foo();"
}
}
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如您所见,我已经提到了错误.我正试图在eclipse jdt ui中为此创建一个quickfix.但我无法获得类B的超类节点,即类TestOverride.
我尝试了以下代码
if(selectedNode instanceof MethodDeclaration) {
ASTNode type = selectedNode.getParent();
if(type instanceof TypeDeclaration) {
ASTNode parentClass = ((TypeDeclaration) type).getSuperclassType();
}
}
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在这里,我只将parentClass作为TestOverride.但是当我检查时,这不是TypeDeclaration类型,它也不是SimpleName类型.
我的查询是如何获得类TestOverride节点的?
编辑
for (IMethodBinding parentMethodBinding :superClassBinding.getDeclaredMethods()){
if (methodBinding.overrides(parentMethodBinding)){
ReturnStatement rs = ast.newReturnStatement();
SuperMethodInvocation smi = ast.newSuperMethodInvocation();
rs.setExpression(smi);
Block oldBody = methodDecl.getBody();
ListRewrite listRewrite = rewriter.getListRewrite(oldBody, Block.STATEMENTS_PROPERTY);
listRewrite.insertFirst(rs, null);
}
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您将不得不与bindings. 要绑定可用,这意味着resolveBinding()不返回 null,可能需要我发布的其他步骤。
要使用绑定,此访问者应该有助于朝着正确的方向前进:
class TypeHierarchyVisitor extends ASTVisitor {
public boolean visit(MethodDeclaration node) {
// e.g. foo()
IMethodBinding methodBinding = node.resolveBinding();
// e.g. class B
ITypeBinding classBinding = methodBinding.getDeclaringClass();
// e.g. class TestOverride
ITypeBinding superclassBinding = classBinding.getSuperclass();
if (superclassBinding != null) {
for (IMethodBinding parentBinding: superclassBinding.getDeclaredMethods()) {
if (methodBinding.overrides(parentBinding)) {
// now you know `node` overrides a method and
// you can add the `super` statement
}
}
}
return super.visit(node);
}
}
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