Javascript: OR and AND

Javascript has an odd and interesting way of processing the || and && operators. Contrary to what might be expected, their return value is not a boolean(!). Instead, both will always return one of the two parameters passed.

OR (||)

a = b || c;

Javascript will return b if b is truthy, and c if b is falsy. Instead of directly returning true or false, it will simply pass one of the parameters that has a truthy or falsy value! Note how this perfectly executes the logic of the OR operator. This allows you to use it in an extended context. For example, it allows you to easily define a default value for a parameter:

opacity = this.options.opacity || 0.5;

If this.options.opacity was undefined (falsy), the || operator will simply return the second argument (0.5).

AND (&&)

a = b && c;

Javascript will return b if b is falsy, and c ifb is truthy. Again this makes sense, since it perfectly executes the logic of the AND operator.

Falsy / Truthy

For completeness, I will list the Javascript falsy values here - quite a list compared to most other languages! 

• 0 (Number)
• NaN (Number)
• '' (empty String)
• false (Boolean)
• null (Object)
• undefined (undefined)

I will not discuss the miraculous behaviour (and type) of NaN or undefined here. Nor will I go into detail about null being an Object. JavaScript has its design errors... For completeness, I'll state that everything else is truthy (!).

Ternary Operator (exp ? value1 : value2)

As a consequence the ternary operator has a pattern that can allways be rewritten into a shorter || form:

depends = a ? a : b;

always equals the simpler

depends = a || b;

which is something, most people will overlook. Personally, I think the first example is much clearer to people unfamiliar with Javascript's odd || operator. However, once you are, the second example is much easier to read (the ternary operator usually is not).