为什么1＆1 === 1返回true而2＆2 === 2返回false？

Question

In the console, when I type 1 & 1 it returns 1 , that's basically why 1 & 1 === 1 returns true .

For 2 & 2 it returns 2 but 2 & 2 === 2 returns false

Why ?

 console.log("1 & 1: ", 1 & 1); console.log("1 & 1 === 1: ", 1 & 1 === 1); console.log("2 & 2: ", 2 & 2); console.log("2 & 2 === 2: ", 2 & 2 === 2); console.log("typeof(2): ", typeof 2); console.log("typeof(2 & 2): ", typeof(2 & 2)); 

Answer 1

As @jonrsharpe said, it doesn't return false , it returns 0 . It's evaluated as 2 & (2 === 2) . (2 & 2) === 2 is true .

Thanks

Answer 2

在javascript中，当你写2和2时，这并不意味着2 && 2而是它将2转换为32位数字并执行AND操作，相同的AND你在二进制数中学习，其中数字不再是2因此（2和2）== = 2不会返回true，因为（1＆1）=== 1返回true，因为在二进制操作1和1上也返回1