JavaScript combining && statements in a variable to be true or false
javascript, variables
Solution
In Javascript the `&&` and `||` operators are slightly strange. It depends on if the value is "falsy" (zero, `undefined`, `null`, empty string, `NaN`) or truthy (anything else, including empty arrays).
With `&&` if the first value is "falsy", then the result of the operation will be the first value, otherwise it will be the second value. With `||` if the first value is "falsy" then the result of the operation will be the second value, otherwise it will be the first value.
Example:
var a = 5 && 3; // a will be 3
var a = 0 && 7; // a will be 0
var a = 1 || 2; // a will be 1
var a = 0 || 2; // a will be 2
This is very useful if you want to replace this:
if (x == null){
x = 5;
}
With:
x = x || 5;
So in short, if `isEnabled()` is truthy then `is_enabled` will be set to whatever `isSupported()` returns. If `isEnabled()` is falsy, then `is_enabled` will be set to whatever that falsy value is.
Also as Robert pointed out, there is short-circuiting:
var x = 5 || infinite_loop();
var x = false && infinite_loop();
In both cases, the `infinite_loop()` call doesn't happen, since the two operations are short-circuited - `||` doesn't evaluate the second value when the first value is truthy, and `&&` doesn't evaluate the second value when the first value is falsy.
Problem
just a quick question. I cannot find anything relating to this since I don't really see how to explain it... but, if I combine two bool values using an && to make another variable, what will happen? ``` var is_enabled = isEnabled() && isSupported(); ``` If isEnabled() is false and isSupported() is true, will it equal false?