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. */

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.

Answer 3

You can do other things too like:

var myVar = Math.random() > 0.5;

myVar && doFunc();

which is the same as

if(myVar) {
    doFunc();
}

The || basically means "If the first thing is false, go to second"

The && basically means "If the first thing is true, go to the second"

This is why you see things like this at the top of functions:

function myFunction(options) {
    options = options || {};
}

Which means: If options is falsey, make it {}

Answer 4

This has nothing to do with type conversion.

|| returns the first truthy operand. 1 || true will give you 1

&& returns the first falsy operand, or the second operand if both are truthy.

See page 58 of the ECMAScript specification

Answer 5

true && 1 => this will always return the last value of && if all conditions are true or false otherwise

true || 1 => this will always return the first 'truthy' value

Resources:

http://www.ecma-international.org/publications/files/ECMA-ST-ARCH/ECMA-262,%203rd%20edition,%20December%201999.pdf

Answer 6

You're confusing casting (the * / + operators will do this) with logical evaluation (which &&, ||) do.