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 functions are evaluated according to precedence rules

because "&&" is of higher precendence than "||", it is evaluated first because you did not have any brackets to set explicit precedence

so you expression of

(A || B && C) 

which is

(T || F && F)

is bracketed as

(T || (F && F)) 

because of the precedence rules.

Since the compiler understands that if 'A == true' it doesn't need to bother evaluating the rest of the expression, it stops after evaluating A.

If you had bracketed ((A || B) && C) Then it would evaluate to false.

EDIT

Another way, as mentioned by other posters is to use "|" and "&" instead of "||" and "&&" because that stops the expression from shortcutting. However, because of the precedence rules, the end result will still be the same.