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

问题

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;
}

回答1:

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)

回答2:

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.

回答3:

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().

回答4:

http://en.wikipedia.org/wiki/Short-circuit_evaluation

回答5:

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.

回答6:

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

回答7:

short answer: short-circuit evaluation

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

回答8:

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.

回答9:

If you want all functions to be executed you can drop the short-cut variants

if (func1() | func2() & func3()) {

来源：https://stackoverflow.com/questions/852734/what-is-wrong-with-the-short-circuit-logic-in-this-java-code