CODEWITHSUNDEEP
javascript
qiuyuntao
qiuyuntao

Reputation: 2394

Why 0 && true echo 0 in js?

As title

As I know

(if this part is true) && (this part will execute)

if(condition){
   (this part will execute)
}

0 is false, so why not echo false but 0?

enter image description here

Upvotes: 6

Views: 3490

Answers (7)

shashank
shashank

Reputation: 89

For && operator comparing with false is always FALSE.

0 && false => 0
false && 0 => false

for better understanding :- In case of && operator (always start from left-right), when you get the value 0 (false) it will print 0(false); if start with false it will directly print false. it won't check the second operand when get false. but in case

true && 1 => 1
1 && true => true

as it has to check till end and ultimately give the end operand as result if won't get false.

For || operator comparing with true is always TRUE.

1 || true => 1
true || 1 => true

for better understanding :- In case of || operator (always start from left-right), when you get the value 1 (true) it will print 1(true). Starting with true it will directly print true. it won't check the second operand when get true. but in case

false || 1 => 1
0 || true => true

Upvotes: 0

Rakesh Makluri
Rakesh Makluri

Reputation: 647

From MDN:

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.

expr1 || expr2 -- 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.

!expr -- Returns false if its single operand can be converted to true; otherwise, returns true.

Some expressions that can be converted to false are:

  • null
  • NaN
  • 0
  • empty string("" or '' or ``)
  • undefined

Short-circuit evaluation

As logical expressions are evaluated left to right, they are tested for possible "short-circuit" evaluation using the following rules:

  • false && (anything) is short-circuit evaluated to false.
  • true || (anything) is short-circuit evaluated to true.

Upvotes: 5

Josu Goñi
Josu Goñi

Reputation: 1274

a && b is equivalent to

if (a) {
  return b;
} else {
  return a;
}

So in your case a equals to 0, which is falsy, therefore you get a.

Unlike other languages, JavaScript does not return true or false on && and ||, it returns the first truthy operand for a || operator, or the last one, and the first falsy operand for the && operator, or the last one.

You can find more info here.

Upvotes: 0

Sreekanth Reddy Balne
Sreekanth Reddy Balne

Reputation: 3424

In javascript all except for null, undefined, false, 0, and NaN are Truthy.

In your case, why not echo false but 0?.

Javascript's ToBoolean function evaluates it to the first falsey value. i.e,

0 && true
=> 0

true && undefined
=> undefined

null && undefined
=> null

And if you need either strictly true or false, then go for not-not i.e, !!.

!!0 && true
=> false

!!true && undefined
=> false

!!null && undefined
=> false

Upvotes: 1

Vadim Hulevich
Vadim Hulevich

Reputation: 1833

Because operator && return first falsey element otherwise they return last element

1 && 0 && false // 0
1 && 2 && 3     // 3

Upvotes: 9

Karan
Karan

Reputation: 74

(something falsy) && (anything) will always return false.


Also 0 is falsy and '0' (a non empty string) is truthy

Upvotes: 0

axiac
axiac

Reputation: 72186

The JavaScript documentation of logical operators explains:

Logical operators are typically used with Boolean (logical) values. When they are, they return a Boolean value. However, the && and || operators actually return the value of one of the specified operands, so if these operators are used with non-Boolean values, they may return a non-Boolean value.

Upvotes: 1

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