I have the following code:

if(!partialHits.get(req_nr).containsKey(z) || partialHits.get(req_nr).get(z) < tmpmap.get(z)){
    partialHits.get(z).put(z, tmpmap.get(z));

where partialHits is a HashMap.
What will happen if the first statement is true? Will Java still check the second statement? Because in order the first statement to be true, the HashMap should not contain the given key, so if the second statement is checked, I will get NullPointerException.
So in simple words, if we have the following code

if(a && b)
if(a || b)

would Java check b if a is false in the first case and if a is true in the second case?

Best Answer

No, it will not be evaluated. And this is very useful. For example if you need to test whether a string is not null or empty you can write

if (str != null && !str.isEmpty()) {

or, the other way around

if (str == null || str.isEmpty()) {
} else {

If we didn't have 'short-circuits' in Java, we'd receive a lot of NullPointerExceptions in the above lines of code.