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\" r

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. */