Question

I have the following switch statement in JavaScript :

switch(scrollable.direction){
  case "top" :
    break;
  case "left" :
    break;
  case "right" :
    scrollable.select("."+lineCssClass).invoke("setStyle", {float: "right"});
    break;
  case "bottom" :
    alert("Bottom scrolling not implemented yet ! Sorry !");
    return;
}

(the “invoke” bit is prototype.js, but it’s no relevant to the question anyway)

It is inside a function. I want that if the value is "bottom" a message is displayed and the method execution stops.

The problem is that if the value is e.g. "top", the break is executed, but the execution jumps to the return; statement instead of exiting the switch statement.

I actually solved the problem by adding an additionnal break; after the return, which is actually dead code since it can never be executed.

But I would be curious to know why it executed the “return” in the first place ?

Edit: I am sorry, the “return” wasn’t actually executed. I was stepping through the code using Firebug and it actually stepped on and highlighted the “return” line, but it wasn’t executed.

There are other problems in my code that cause it not to work as expected, and I was wrongly blaming this.

Answer 1

As I said in the edit I made to the question, the problem was actually not that the return; was executed, but that Firebug gave the impression it was when I went step by step through the switch statement.

Alsot, after some more tests, I can reproduce this weird thing only when I enclose the switch/case statement in a try block.

As missingno suggested I will leave this question in case other people are confused by this behaviour of Firebug.