Why is !true ? 'false' : 'true' returning 'true'?

Question

Why is it that when I do

(!true) ? 'false' : 'true'

it returns 'true' ?

Answer 1

It simply means

if (!true) {
  return 'false';
} else {
  return 'true';
}

!true (not true) means false , so the else is returned.

Answer 2

The syntax of A? B: C A? B: C means that if A is TRUE, then return the value B. Else return value C. Since A is FALSE , it returns the value C which happens to be true .

Answer 3

Because (!true) is false, and then the right side of the : is chosen.

Answer 4

Because the above is equivalent to:

if (false) {
    return 'false';
} else {
    return 'true';
}

Though perhaps the confusion is coming from the difference between:

if (false) // which is false

And

if (false == false) // which is true

Answer 5

This can be expanded to:

if(!true){
   return 'false';
} else {
   return 'true';
}

Answer 6

if(!true) is equivalent to if(!true= true) which is equivalent to if(false=true) which is false . Therefore return (true) which is on the false side of the if statement.

Answer 7

The confusion lies here because of the use of string literals to represent boolean values. If you reverse the 'false' and 'true' , it makes more sense:

(?true): 'true' : 'false'

Would return the string literal false , which is much different than a boolean value.

Your original statement (?true): 'false' : 'true' reads as

"If not true, then return the string literal true".

The statement I posted first reads as

"If not true, then return the string literal false".

Which, if you know the opposite ( not ) value of true is false , then it explains the logic illustrated.

Answer 8

 const test = true; // change this value to see the result change if (.test) { // if test is NOT truthy console.log('false') } else { // if test === true console.log('true') }

// Since in the original question a ternary expression was used, the above code is equivalent to this:

!test ? console.log('false') : console.log('true');