英文:
Evaluating an expression with multiple boolean values
问题
这对于大多数程序员来说可能相当简单，但是看到以下代码打印出 true，我有点惊讶。
// 实际的实现有适当的逻辑，
// 为了简单起见，我在这里使用布尔值来代替那些逻辑
System.out.println(false && false || true && true || false && false);
因为我的理解是，如果第一个布尔值计算为 false，并且下一个操作数是 &&，那么 JVM 将不会尝试计算表达式的其余部分，而只会将整个表达式视为 false。
我期望像下面这样的代码是可以的，但上面的代码不行：
System.out.println((false && false) || (true && true) || (false && false));
这是不是编译器默认将 || 之间的 && 进行了分组呢？
英文:
This might seem quite straight forward to most of the programmers, but I was a little surprised to see the following code print true.
//The actual implementation has proper logic,
//just for simplicity I am using booleans in place of those logics
System.out.println(false && false || true && true || false && false);
Because my perception was that - if the 1st boolean value evaluates to false and the next operand is && then the jvm will not even try to evaluate the rest of the expression and just consider the whole as false
I was expecting code like below to work, but not the above one:
System.out.println((false && false) || (true && true) || (false && false));
Is it like the compiler by default groups the &&s between ||s
答案1
得分: 2
它是否类似于编译器默认将
&&放在||之间进行分组？
确实是这样。这被称为运算符优先级。语言明确地定义它，因为否则操作可能会产生歧义，而编译器不能容许歧义。
运算符优先级的一个更简单的示例是：
x = 1 + 2 * 3;
你会期望结果是 9 还是 7？运算符优先级明确地定义了结果必须是什么。
英文:
> Is it like the compiler by default groups the &&s between ||s
That's exactly what it does. It's called operator precedence. Languages explicitly define it because otherwise operations could be ambiguous, and ambiguity can't be allowed by a compiler.
A simpler example of operator precendence would be something like:
x = 1 + 2 * 3;
Would you expect the result to be 9 or 7? Operator precedence defines unambiguously what the result must be.
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