title: Throwing exceptions in Java
body: <p>I have a question about throwing exceptions in Java, a kind of misunderstanding from my side, as it seems, which I would like to clarify for myself.</p>
  
  <p>I have been reading that the two basic ways of handling exception code are:</p>
  
  <p>1.) throwing an exception in a try-block with "throw new ...", and catching it immediately in a catch-block - the so called try-throw-catch mechanism.</p>
  
  <p>2.) throwing an exception in a method with "throw new ..." and then declaring in the header of the method that this method might throw an exception with "throws ..." - the so called pass-the-buck.</p>
  
  <p>I have recently read that "it doesn't make any sense to throw an exception and then catch it in the same method", which made me think whether I understand the thing in the wrong way, or the person who had written this had something else in mind. Doesn't the first way of handling exceptions does exactly this (the try-throw-catch mechanism) ? I mean, it throws an exception and catches it in the same method. I have read that it is a better practice to throw an exception in one method, and catch it in another method, but this is just one (probably better) way. The other way is also legal and correct, isn't it? </p>
  
  <p>Would you, please, give me a comment on this ? Thank you very much.</p>