methods – Libera #java

Someone on ##java asked about how to return a void value. This is a terrible question, but let’s look at it anyway, just because.
Here’s some example code they asked about, translated:

void foo(boolean bar) {
  if(bar) {
    return baz();
  }
}
void baz() {
  // do stuff here
}

The intent was to call baz() as a replacement for foo()‘s execution; the one with the question didn’t include code after the if() block, but further questions indicated that the remaining part of foo() was to be avoided.
There’s a lot of monstrosity here. The worst is “return baz();“, which won’t compile, and shouldn’t compile; void‘s whole purpose is to avoid being put on the call stack, so “return void” makes no sense.
One way to rewrite this code is obvious:

void foo(boolean bar) {
  if(bar) {
    baz();
    return;
  }
  // extra code goes here,
  // not to be executed if bar is true
}
void baz() {
  // do stuff here
}

This calls baz() and terminates the method execution immediately after. You could also do something else that’s obvious:

void foo(boolean bar) {
  if(bar) {
    baz();
  } else {
    // extra code goes here
  }
}

This has the advantage of a single termination point for the method. From “Code Complete:”

17.1 return

Minimize the number of returns in each routine. It’s harder to understand a routine if, reading it at the bottom, you’re unaware of the possibility that it returned somewhere above.
Use a return when it enhances readability. In certain routines, once you know the answer, you want to return it to the calling routine immediately. If the routine is defined in such a way that it doesn’t require any cleanup, not returning immediately means that you have to write more code.
(Content copied shamelessly from an excellent answer on StackOverflow.)

Tag: methods

Returning void type from a method call

17.1 return