java - 围绕 Spring-AOP 切入点和继承的澄清

Question

在 my.package 中给出以下示例类...

public class Foo {
    public void logicNotInBar()     {/*code*/}
    public void logicBarOverrides() {/*code*/}
}

public class Bar extends Foo {
    public void logicBarOverrides() {/*code*/}
}

以及以下 Spring-AOP 切入点...

<aop:pointcut id="myPointcutAll" expression="execution(* my.package.*.*(..))"   />
<aop:pointcut id="myPointcutFoo" expression="execution(* my.package.Foo.*(..))" />
<aop:pointcut id="myPointcutBar" expression="execution(* my.package.Bar.*(..))" />

将建议应用于 Bar 实例上的上述切入点的结果是什么？特别是……

Bar bar = new Bar();
bar.logicNotInBar();      // will myPointcutBar advice trigger?
bar.logicBarOverrides();  // is myPointcutFoo ignored here?

我想我遗漏了一些关于切入点如何与继承交互的基本事实，因此底层解释/文档可能会有很长的路要走。

Answer 1

来自 aspectj documentation :

When matching method-execution join points, if the execution pointcut method signature specifies a declaring type, the pointcut will only match methods declared in that type, or methods that override methods declared in or inherited by that type. So the pointcut

execution(public void Middle.*())

picks out all method executions for public methods returning void and having no arguments that are either declared in, or inherited by, Middle, even if those methods are overridden in a subclass of Middle. So the pointcut would pick out the method-execution join point for Sub.m() in this code:

  class Super {
    protected void m() { ... }
  }
  class Middle extends Super {
  }
  class Sub extends Middle {
    public void m() { ... }
  }