Regarding Java switch statements - using return and omitting breaks in each case

Question

Given this method, does this represent some egregious stylistic or semantic faux pas:

private double translateSlider(int sliderVal) {     switch (sliderVal) {         case 0:             return 1.0;         case 1:             return .9;         case 2:             return .8;         case 3:             return .7;         case 4:             return .6;         default:             return 1.0;     } }   

It's clearly not in line with the Java tutorials here.

However, It's clear, concise and so far has yielded exactly what I need. Is there a compelling, pragmatic reason to create a local variable, assign a value to it within each case, add a break to each case and return the value at the end of the method?

Answer 1

Assigning a value to a local variable and then returning that at the end is considered a good practice. Methods having multiple exits are harder to debug and can be difficult to read.

That said, thats the only plus point left to this paradigm. It was originated when only low-level procedural languages were around. And it made much more sense at that time.

While we are on the topic you must check this out. Its an interesting read.

Answer 2

From human intelligence view your code is fine. From static code analysis tools view there are multiple returns, which makes it harder to debug. e.g you cannot set one and only breakpoint immediately before return.

Further you would not hard code the 4 slider steps in an professional app. Either calculate the values by using max - min, etc., or look them up in an array:

public static final double[] SLIDER_VALUES = {1.0, 0.9, 0.8, 0.7, 0.6}; public static final double SLIDER_DEFAULT = 1.0;   private double translateSlider(int sliderValue) {   double result = SLIDER_DEFAULT;   if (sliderValue >= 0 && sliderValue < SLIDER_VALUES.length) {       ret = SLIDER_VALUES[sliderValue];   }    return result; }