Having the following test case: (find fiddle here)
var a = new Date();
var b = null;
var c = {test: "test"};
if(a)
console.log(a); //--- prints the current date
if(b)
console.log('null'); //--- never reached
if(c)
console.log('test'); //--- prints 'test'
console.log(a && b); //--- prints null
Knowing that
console.log(typeof null); //--- prints "object"
console.log(typeof c); //--- prints "object"
I expect the result of
console.log(a && b);
to be false and not null as it shown in the example.
Any hint?
Having the following test case: (find fiddle here)
var a = new Date();
var b = null;
var c = {test: "test"};
if(a)
console.log(a); //--- prints the current date
if(b)
console.log('null'); //--- never reached
if(c)
console.log('test'); //--- prints 'test'
console.log(a && b); //--- prints null
Knowing that
console.log(typeof null); //--- prints "object"
console.log(typeof c); //--- prints "object"
I expect the result of
console.log(a && b);
to be false and not null as it shown in the example.
Any hint?
Share Improve this question asked Nov 9, 2015 at 11:53 BeNdErRBeNdErR 17.9k21 gold badges77 silver badges106 bronze badges 03 Answers
Reset to default 4From the MDN:
expr1 && expr2: Returns expr1 if it can be converted to false; otherwise, returns expr2
new Date can't be converted to false (it's not falsy), so b is returned.
I expect the result of
console.log(a && b);to be
falseand notnullas it shown in the example.
In many languages with && and || operators (or AND and OR), the result is always a boolean, yes.* But JavaScript's && and || are more useful than that: Their result is their left-hand operand's value or their right-hand operand's value, not coerced to boolean.
Here's how && works:
Evaluate the left-hand operand.
If the value is falsey (coerces to
falsewhen we make it a boolean), return the value (not the coerced value)If the value from #2 is truthy, evaluate and return the right-hand value (uncoerced)
The falsey values are null, undefined, 0, "", NaN, and of course, false. Everything else is truthy.
So for instance
console.log(0 && 'hi'); // 0
...shows 0 because 0 is falsey. Note it resulted in 0, not false. Similarly, the result here:
console.log(1 && 'hello'); // hello
...is 'hello' because the right-hand side isn't coerced at all by &&.
|| works similarly (I call it the curiously-powerful OR operator): It evaluates the left-hand side and, if that's truthy, takes that as its result; otherwise it evaluates the right-hand side and takes that as its result.
The only time you get true or false from && or || is if the selected operand is already a boolean.
* Of course, many (but not all) of those languages (Java, C#) also require that the operands to && and || already be booleans. C's an exception, it does some coercion, but still has the result of the operands as a boolean (or if you go back far enough, an int that will be 1 or 0).
From MDN:
Logical AND (&&) expr1 && expr2 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.
MDN Documentation