isBooleanTooLongAndComplex

This is another post in our Code Health series. A version of this post originally appeared in Google bathrooms worldwide as a Google Testing on the Toilet episode. You can download a printer-friendly version to display in your office.

By Yiming Sun

You may have come across some complex, hard-to-read Boolean expressions in your codebase and wished they were easier to understand. For example, let's say we want to decide whether a pizza is fantastic:

// Decide whether this pizza is fantastic.

if ((!pepperoniService.empty() || sausages.size() > 0)

    && (useOnionFlag.get() || hasMushroom(ENOKI, PORTOBELLO)) && hasCheese()) {

  ...

}

A first step toward improving this is to extract the condition into a well-named variable:

boolean isPizzaFantastic

    (!pepperoniService.empty() || sausages.size() > 0)

    && (useOnionFlag.get() || hasMushroom(ENOKI, PORTOBELLO)) && hasCheese();

if (isPizzaFantastic) {

  ... 

}

However, the Boolean expression is still too complex. It's potentially confusing to calculate the value of isPizzaFantastic from a given set of inputs. You might need to grab a pen and paper, or start a server locally and set breakpoints. 

Instead, try to group the details into intermediate Booleans that provide meaningful abstractions. Each Boolean below represents a single well-defined quality, and you no longer need to mix && and || within an expression. Without changing the business logic, you’ve made it easier to see how the Booleans relate to each other:

boolean hasGoodMeat = !pepperoniService.empty() || sausages.size() > 0;

boolean hasGoodVeggies = useOnionFlag.get() || hasMushroom(ENOKI, PORTOBELLO);

boolean isPizzaFantastic = hasGoodMeat && hasGoodVeggies && hasCheese();

Another option is to hide the logic in a separate method. This also offers the possibility of early returns using guard clauses, further reducing the need to keep track of intermediate states:

boolean isPizzaFantastic() {

  if (!hasCheese()) {

    return false;

  }

  if (pepperoniService.empty() && sausages.size() == 0) {

    return false;

  }

  return useOnionFlag.get() || hasMushroom(ENOKI, PORTOBELLO);
}