JavaScript type conversion: (true && 1) vs (true | | 1)

Question

JavaScript is non-strictly typed language as Java,for example.

As we know, it converts value of result dependently upon context:

"2" + "3" results "23"

"2" * "3" results 6

This is quite clear and OK for understanding.

I just tried following expressions and got confused:

true && 1 results 1
true || 1 results true

Why the first gives Number and the second gives boolean?

Considering JavaScript conversion rules,I expect to get boolean values in both cases,due to boolean context of expression.

Answer 1

Check Douglas Crockford's site, it says:

The && operator is commonly called logical and. It can also be called guard. If the first operand is false, null, undefined, "" (the empty string), or the number 0 then it returns the first operand. Otherwise, it returns the second operand. This provides a convenient way to write a null-check:

var value = p && p.name; /* The name value will only be retrieved from
p if p has a value, avoiding an error. */

The || operator is commonly called logical or. It can also be called default. If the first operand is false, null, undefined, "" (the empty string), or the number 0, then it returns the second operand. Otherwise, it returns the first operand. This provides a convenient way to specify default values:

value = v || 10; /* Use the value of v, but if v doesn't have a value,
use 10 instead. */

Answer 2

To quote MDC;

&&; Returns expr1 if it can be converted to false; otherwise, returns expr2. Thus, when used with Boolean values, && returns true if both operands are true; otherwise, returns false.

||; Returns expr1 if it can be converted to true; otherwise, returns expr2. Thus, when used with Boolean values, || returns true if either operand is true; if both are false, returns false.

So in the first example, 1 is being returned because expr1 cannot be converted to false.

In the second example, true can be converted to true, so it's returned.