Using the logical operator || will yield a numeric result rather than a boolean value

Here's an example with two if conditions. The first if condition functions correctly, but the second if condition returns 11 unexpectedly. It's clear that the second if condition is incorrect, but it's worth exploring why JavaScript outputs 11 in this case.

function exception(number) {
// if(number === 10 || number === 11) { // Working as expected
   if(number === 10 || 11) { // Why 11?
        console.log(number);
   }
}

function loop(f) {
    for (i = 0; i <= 100; i++) {
        f(i);
    }
}

loop(exception);

Answer №1

Referenced from this particular discussion.

(expr1 || expr2)

"If expr1 can be converted to true, it is returned; otherwise, expr2 is returned."

source

This means that when expr1 results in values like 0, "", false, null, undefined, or NaN, then expr2 is returned. Otherwise, expr1 is returned.

Answer №2

Here is some insight into your intended goal:

  • number === 10 || number === 11 can be written as
    (number === 10) || (number === 11)
  • number === 10 || 11 can be written as (number === 10) || (11), it does not compare 11 to number in this context

Now, let's delve deeper into the expressionnumber === 10 || 11:

  • number === 10 will return true if the variable number is a number and equal to 10
  • If the first condition is false, it will then evaluate the boolean value of the next statement: 11 (which is considered true, as it is a number not equal to 0)

Answer №3

due to the fact that Boolean(11) evaluates to true (you can test this in your console)

thus, in the event that the initial condition is false (if the number is not 10), the subsequent condition will always be true

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