Java 7精确的重新抛出和遗留代码

Question

更精确的重新抛出允许编写抛出异常抛出的异常的代码:

   public void foo(String bar) throws FirstException, SecondException {
     try{
       // Code that may throw both FirstException and SecondException
     }
     catch (Exception e){
       throw e;
     }
   }

在Java 7之前,你必须写:

   public void foo(String bar) throws Exception {
     try{
       // Code that may throw both FirstException and SecondException
     }
     catch (Exception e){
       throw e;
     }
   }

我的问题:是否有一个工具允许检测不精确的投掷,以便用" Exception" 替换" FirstException, SecondException"？

到目前为止,我已经检查过Eclipse中没有编译器警告.FindBugs或CodePro中没有规则.

Answer 1

尝试在抛出之前强制转换异常，也许可以解决问题？

public void foo(String bar) throws FirstException, SecondException {
  try{
    // Code that may throw both FirstException and SecondException
  }
  catch (Exception e){
    throw ((e instanceof FirstException) ? (FirstException)e : (SecondException)e);
  }
}

顺便说一句，重新抛出相同的异常对我来说似乎相当笨拙......