Javascript - 为什么这个 if 语句会在控制台中打印两次“Hello”？

Question

For this sample code, it prints "Hello" twice into the console:

 var x = 10; if ( (null) || (console.log("Hello")) || x > 5 ) { console.log("Hello"); }

I assume when the "if" statement is read, it reads the console.log, then prints it, and since x > 5, the return statement is true, then consoles the second "Hello".

Can someone tell me if this is correct? Thanks!

Answer 1

Your function:

1. Sets variable and it's value var x = 10

2. Checks the conditions, until one of them "is" true (returns true )

3. (null) returns itself, so the functions checks other condition

4. (console.log("Hello")) returns nothing ( null ), so the function checks other condition

5. x > 5 returns true , so the execution of code inside if{ < code > } happens

6. (console.log("Hello")) inside if{ } executes.

As you can see, the (console.log("Hello")) is executed both, while checking conditions inside if statement and while executing "the insides" of that if . That's why you see "Hello" logged 2 times.

Answer 2

console.log has a return type of undefined.

You're seeing the console.log output twice, because you've called it twice:

Once in your if condition, and the next in your if statement.

Your if statement will run because you're using OR, and x = 10 > 5

Answer 3

The if statement first encounters null , which is false, then does a console.log() , which returns undefined so is also false, and finally x > 5 , which is true. Thus the if statement is true, and it goes into the body and does another print.

Answer 4

I assume when the "if" statement is read, it reads the console.log, then prints it, and since x > 5, the return statement is true, then consoles the second "Hello".

More accurately, the "if" statement is executed . Specifically, the condition is evaluated. Since null is "falsey", the right of the ||operator is evaluated. Which executes the first console.log() . This in turn returns undefined which is also "falsey" and the next operand of the next ||is evaluated. x>5 evaluates to true , so the second console.log() is executed.

Note that this is not a very good way to write code. console.log() shouldn't be used inside an if condition like this. It's return value is meaningless since it always returns undefined .

Answer 5

The first console.log you wrote is in the if statement, that function will be called when the if statement is evaluated. The second statement will be called if the condition in the if statement is met. The console.log itself returns undefined and has no real purpose to be in your if statement. I would recommend removing it as I don't believe it is achieving what you intended Good luck :)