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?
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?
Because operator && return first falsey element otherwise they return last element
1 && 0 && false // 0
1 && 2 && 3 // 3
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:
Short-circuit evaluation
As logical expressions are evaluated left to right, they are tested for possible "short-circuit" evaluation using the following rules:
The JavaScript documentation of logical operators explains:
Logical operators are typically used with
Boolean(logical) values. When they are, they return aBooleanvalue. However, the&&and||operators actually return the value of one of the specified operands, so if these operators are used with non-Booleanvalues, they may return a non-Booleanvalue.
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
(something falsy) && (anything) will always return false.
Also 0 is falsy and '0' (a non empty string) is truthy
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.
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