此Java代码中的短路逻辑有什么问题? [英] What is wrong with the short circuit logic in this Java code?

问题描述

为什么func3无法在下面的程序中执行?在func1之后，无需对func2进行评估，但是对于func3，不是吗?

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;
}

推荐答案

您正在使用短路或.如果第一个参数为true，则整个表达式为true.

You're using a short-circuited or. If the first argument is true, the entire expression is true.

如果我添加编译器使用的隐式括号，这可能会有所帮助

It might help if I add the implicit parentheses that the compiler uses

编辑:正如Chris Jester-Young所说，这实际上是因为逻辑运算符必须从左到右的关联性:

Edit: As Chris Jester-Young noted, this is actually because logical operators have to left-to-right associativity:

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

func1返回后，它变为:

After func1 returns, it becomes this:

if (true || (func2() && func3()))

评估短路或后，它变为:

After evaluating the short-circuited or, it becomes:

if (true)

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