JS 0 === (0 || 6); 返回 false - 为什么？

Question

0 === (0 || 6) // returns false
false === (false || 6) // returns false
(false === false || false === 6) // returns true
(0 === 0 || 0 === 6) // returns true

 console.log( 0 === (0 || 6) ); console.log( false === (false || 6) ); console.log( (false === false || false === 6) ); console.log( (0 === 0 || 0 === 6) );

what gives? I ran across this scenario earlier and don't get it.

Answer 1

0 === (0 || 6) // returns false

It's all about order of operations. That returns false because the parenthesis are evaluated first. So (0 || 6) returns 6 and 0 === 6 is false

false === (false || 6) // returns false

(false === false || false === 6) // returns true

(0 === 0 || 0 === 6) // returns true

The rest are easy to follow after that explanation.

Answer 2

0 === (0 || 6) is equivalent to 0 === 6 which is of course false.

false === (false || 6) is equivalent to false === 6 which is also obviously false.

(false === false || false === 6) is equivalent to true || false true || false which is of course true.

(0 === 0 || 0 === 6) is also equivalent to true || false true || false .

Look up "operator precedence". That'll help you wrap your head around javascript in the future.

Answer 3

First, you should know the operators priority.

Second, you should know how the operators work and what they return.

Here is the documentation you can take it as reference .

 console.log( 0 === (0 || 6) ); // (0 || 6) returns 6 because 0 is falsy, // so it becomes 0 === 6. It is obviously return false. console.log( false === (false || 6) ); (false || 6) is equals (0 || 6), so we get (false === 6) = false. console.log( (false === false || false === 6) ); // this is three part. === is prior to ||, // so we should do (false === false) and (false === 6). // (false === false) is true and (false === 6) is false, // so we get (true || false) = true console.log( (0 === 0 || 0 === 6) ); // (0 === 0) is true and (0 === 6) is false, // so true || false is true.

By the way, most of the JavaScript operator do type conversion whether the two sides is not the type the operator want. Here's something you can take a look:

https://javascript.info/type-conversions

Answer 4

From your question, I am guessing you are not familiar with how AND & OR are used or works.

So,

| stands for bitwise OR

and, & stands for bitwise AND

It is a common method of writing AND as && and OR as ||.

For AND , following are the scenarios:

 1 && 1 => 1
 1 && 0 => 0
 0 && 1 => 0
 0 && 0 => 0

For OR , following are the scenarios:

 1 || 1 => 1
 1 || 0 => 1
 0 || 1 => 1
 0 || 0 => 0

In the above, 1 can be any non-zero integer or value (but we denote it by one, meaning high ).

So, if you write,

0 === ( 0 || 6 ) ,

then, ( 0 || 6 ) is already true , then, if you compare true with 0 , then it is always going to be false.

Whereas, if you you change the above condition as:

0 == ( 0 && 6 ) , ( EDITTED )

( 0 && 6 ) becomes false, and then 0 == false becomes true .

=== and == are different because the second one just checks if they are equal or not, but it does not check if they are exactly same. === checks if the values are identical, meaning exactly same.

I hope my answer was helpful.