What is wrong with the short circuit logic in this Java code?

Question

Why doesn\'t func3 get executed in the program below? After func1, func2 doesn\'t need to get evaluated but for func3, shouldn\'t it?

if (func1() || func2()          


        

Answer 1

Java uses Lazy evaluation.

Since Func1 always returns true, the entire expression MUST be true, so it shortcuts the rest of the expression.

true || (???)

and

false && (???)

will always shortcut.

To turn off shortcut evaluation, use | and & instead of || and &&

We can use this to good effect:

String s;
if (s != null && !s.equals("")) ...

Meaning that if s is null, we won't even have to try to call s.equals, and we won't end up throwing an NullPointerException

Answer 2

short answer: short-circuit evaluation

since func1() yelds true there is not need to continue evaluation since it is always true

Answer 3

If function 1 always returns true, then Java doesn't need to evaluate the rest of the expression to determine that the whole expression will be true.