Question

我不明白为什么只有num1才会打印输出。我在这里错过了什么吗？

var num1 = 0x200127;
var num2 = 0x200124;
    
if(num1 & 0x100 == 0x100){
    console.log("num1: " + (num1 & 0x100 ) );
}
if(num2 & 0x100 == 0x100){
    console.log("num2: " + (num2 & 0x100 ) );
}

Answer 1

这是操作顺序的问题。如需参考，请查看table of JavaScript operator precedence。

==的预测值为10，而&的预测值为9，因此首先评估==。

所以你的代码基本上是在评估：

num & (0x100 == 0x100)

相当于：

num & true

输出

num1而num2不是因为：

0x200127 & true == 1 (true)
0x200124 & true == 0 (false)

尝试将您的按位操作放在括号中，因为grouping operator具有最高优先级。

if((num1 & 0x100) == 0x100){
    console.log("num1: " + (num1 & 0x100 ) );
}
if((num2 & 0x100) == 0x100){
    console.log("num2: " + (num2 & 0x100 ) );
}

测试如下：

var num1 = 0x200127,
  num2 = 0x200124,
  output = document.getElementById('output');

if ((num1 & 0x100) == 0x100) {
  output.innerHTML += "<p>num1: " + (num1 & 0x100) + "</p>";
}
if ((num2 & 0x100) == 0x100) {
  output.innerHTML += "<p>num2: " + (num2 & 0x100) + "</p>";
}

<div id="output"></div>

Answer 2

==运算符的higher precedence大于&。所以

x & y == z

评估为

x & (y == z)

在第二种情况下，这会使条件评估为0，从而false：

num2 & 0x100 == 0x100
0x200124 & 0x100 == 0x100
0x200124 & true
0x200124 & 1
0

您想使用分组运算符更改优先级：

if((num1 & 0x100) == 0x100){
    console.log("num1: " + (num1 & 0x100 ) );
}
if((num2 & 0x100) == 0x100){
    console.log("num2: " + (num2 & 0x100 ) );
}

Answer 3

我找到了答案，我猜= =的优先级高于＆amp;运营商。如果我将代码更改为此，则可以正常工作。

＆＃13;

var num1 = 0x200127;
var num2 = 0x200124;
    
if(num1 & 0x100 == 0x100){
    console.log("num1: " + (num1 & 0x100 ) );
}
if((num2 & 0x100) == 0x100){
    console.log("num2: " + (num2 & 0x100 ) );
}

＆＃13;

Answer 4

在javascript ==具有更高的优先顺序，然后按位AND运算符因此比较0x100 == 0x100 always result in 1是真的，

if we do Bitwise AND of 1 with even number, result will always be 0. 
And if we do Bitwise AND of 1 with odd number, result will always be 1

按位＆amp; Javascript中的运算符

4 个答案: