What is wrong with the short circuit logic in this Java code?
-
21-08-2019 - |
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() && func3()) {
System.out.println("true");
} else {
System.out.println("false");
}
}
public static boolean func1() {
System.out.println("func1");
return true;
}
public static boolean func2() {
System.out.println("func2");
return false;
}
public static boolean func3() {
System.out.println("func3");
return false;
}
Solution
You're using a short-circuited or. If the first argument is true, the entire expression is true.
It might help if I add the implicit parentheses that the compiler uses
Edit: As Chris Jester-Young noted, this is actually because logical operators have to left-to-right associativity:
if (func1() || (func2() && func3()))
After func1 returns, it becomes this:
if (true || (func2() && func3()))
After evaluating the short-circuited or, it becomes:
if (true)
OTHER TIPS
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.
Java short-circuits boolean expressions. That means that, once func1()
is executed and returns true
, the rest of that boolean doesn't matter since you are using an or
operator. No matter what func2() && func3()
evaluates to, the whole expression will evaluate to true
. Thus, Java doesn't even bother evaluating the func2()
or func3()
.
You're using the shortcut-operators || and &&. These operators don't execute the rest of the expression, if the result is already defined. For || that means if the first expression is true and for && if the first expression is false.
If you want to execute all parts of the expression use | and & instead, that is not shortcut.
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
short answer: short-circuit evaluation
since func1() yelds true there is not need to continue evaluation since it is always true
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.
If you want all functions to be executed you can drop the short-cut variants
if (func1() | func2() & func3()) {