The following works:
let x = 1 && console.log("true"); (-- logs true)
let y = 0 && console.log("true"); (-- logs nothing)
The above shows that the statement before && operator is acting like an expression.
Then I tried this:
console.log(let m = 5); // Error
What is going on here? If that's an expression then why it didn't work in this case and if it's not an expression then why it worked in the first two cases?
The following works -
let x = 1 && console.log("true"); (-- logs true)
let y = 0 && console.log("true"); (-- logs nothing)
The above shows that the statement before && operator is acting like an expression
This is where you are mistaken. The && operator is used to join expressions, not statements (even though it has control flow effects). This parses as follows:
let x = (1 && console.log("true")); (-- logs true)
let y = (0 && console.log("true")); (-- logs nothing)
console.log is acting as an expression here, not let .... Function calls always return a value (which may be undefined). If you logged x and y, you'd see that x === undefined (result of 1 && undefined) and y === 0 (result of 0 && <not evaluated>).
It is probably confusing you here that && short-circuits: In the first expression, the first operand of && is 1 (truthy), so the second operand - the expression which is a call to console.log - has to be evaluated; in the second expression, the first operand of && is the falsy 0, so && short-circuits to 0, not evaluating (not calling) console.log("true").
let statements are statements and not expressions, which is why you get a syntax error in your second example.
Related
Why do these logical operators return an object and not a boolean?
var _ = (obj.fn && obj.fn() ) || obj._ || ( obj._ = {} );
var _ = obj && obj._;
I want to understand why it returns result of obj.fn() (if it is defined) OR obj._ but not boolean result.
In JavaScript, both || and && are logical short-circuit operators that return the first fully-determined “logical value” when evaluated from left to right.
In expression X || Y, X is first evaluated, and interpreted as a boolean value. If this boolean value is “true”, then it is returned. And Y is not evaluated. (Because it doesn’t matter whether Y is true or Y is false, X || Y has been fully determined.) That is the short-circuit part.
If this boolean value is “false”, then we still don’t know if X || Y is true or false until we evaluate Y, and interpret it as a boolean value as well. So then Y gets returned.
And && does the same, except it stops evaluating if the first argument is false.
The first tricky part is that when an expression is evaluated as “true”, then the expression itself is returned. Which counts as "true" in logical expressions, but you can also use it. So this is why you are seeing actual values being returned.
The second tricky part is that when an expression is evaluated as “false”, then in JS 1.0 and 1.1 the system would return a boolean value of “false”; whereas in JS 1.2 on it returns the actual value of the expression.
In JS false, 0, -0, "", null, undefined, NaN and document.all all count as false.
Here I am of course quoting logical values for discussion’s sake. Of course, the literal string "false" is not the same as the value false, and is therefore true.
In the simplest terms:
The || operator returns the first truthy value, and if none are truthy, it returns the last value (which is a falsy value).
The && operator returns the first falsy value, and if none are falsy, it return the last value (which is a truthy value).
It's really that simple. Experiment in your console to see for yourself.
console.log("" && "Dog"); // ""
console.log("Cat" && "Dog"); // "Dog"
console.log("" || "Dog"); // "Dog"
console.log("Cat" || "Dog"); // "Cat"
var _ = ((obj.fn && obj.fn() ) || obj._ || ( obj._ == {/* something */}))? true: false
will return boolean.
UPDATE
Note that this is based on my testing. I am not to be fully relied upon.
It is an expression that does not assign true or false value. Rather it assigns the calculated value.
Let's have a look at this expression.
An example expression:
var a = 1 || 2;
// a = 1
// it's because a will take the value (which is not null) from left
var a = 0 || 2;
// so for this a=2; //its because the closest is 2 (which is not null)
var a = 0 || 2 || 1; //here also a = 2;
Your expression:
var _ = (obj.fn && obj.fn() ) || obj._ || ( obj._ = {} );
// _ = closest of the expression which is not null
// in your case it must be (obj.fn && obj.fn())
// so you are gettig this
Another expression:
var a = 1 && 2;
// a = 2
var a = 1 && 2 && 3;
// a = 3 //for && operator it will take the fartest value
// as long as every expression is true
var a = 0 && 2 && 3;
// a = 0
Another expression:
var _ = obj && obj._;
// _ = obj._
In most programming languages, the && and || operators returns boolean. In JavaScript it's different.
OR Operator:
It returns the value of the first operand that validates as true (if any), otherwise it returns the value of the last operand (even if it validates as false).
Example 1:
var a = 0 || 1 || 2 || 3;
^ ^ ^ ^
f t t t
^
first operand that validates as true
so, a = 1
Example 2:
var a = 0 || false || null || '';
^ ^ ^ ^
f f f f
^
no operand validates as true,
so, a = ''
AND Operator:
It returns the value of the last operand that validates as true (if all conditions validates as true), otherwise it returns the value of the first operand that validates as false.
Example 1:
var a = 1 && 2 && 3 && 4;
^ ^ ^ ^
t t t t
^
last operand that validates as true
so, a = 4
Example 2:
var a = 2 && '' && 3 && null;
^ ^ ^ ^
t f t f
^
return first operand that validates as false,
so, a = ''
Conclusion:
If you want JavaScript to act the same way how other programming languages work, use Boolean() function, like this:
var a = Boolean(1 || 2 || 3);// a = true
You should think of the short-circuit operators as conditionals rather than logical operators.
x || y roughly corresponds to:
if ( x ) { return x; } else { return y; }
and x && y roughly corresponds to:
if ( x ) { return y; } else { return x; }
Given this, the result is perfectly understandable.
From MDN documentation:
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 will return a non-Boolean value.
And here's the table with the returned values of all logical operators.
I think you have basic JavaScript methodology question here.
Now, JavaScript is a loosely typed language. As such, the way and manner in which it treats logical operations differs from that of other standard languages like Java and C++. JavaScript uses a concept known as "type coercion" to determine the value of a logical operation and always returns the value of the first true type. For instance, take a look at the code below:
var x = mystuff || document;
// after execution of the line above, x = document
This is because mystuff is an a priori undefined entity which will always evaluate to false when tested and as such, JavaScript skips this and tests the next entity for a true value. Since the document object is known to JavaScript, it returns a true value and JavaScript returns this object.
If you wanted a boolean value returned to you, you would have to pass your logical condition statement to a function like so:
var condition1 = mystuff || document;
function returnBool(cond){
if(typeof(cond) != 'boolean'){ //the condition type will return 'object' in this case
return new Boolean(cond).valueOf();
}else{ return; }
}
// Then we test...
var condition2 = returnBool(condition1);
window.console.log(typeof(condition2)); // outputs 'boolean'
We can refer to the spec(11.11) of JS here of:
Semantics
The production LogicalANDExpression :LogicalANDExpression &&BitwiseORExpression is evaluated as follows:
Evaluate LogicalANDExpression.
2.Call GetValue(Result(1)).
3.Call ToBoolean(Result(2)).
4.If Result(3) is false, return Result(2).
5.Evaluate BitwiseORExpression.
6.Call GetValue(Result(5)).
7.Return Result(6).
see here for the spec
First, it has to be true to return, so if you are testing for truthfulness then it makes no difference
Second, it lets you do assignments along the lines of:
function bar(foo) {
foo = foo || "default value";
Compare:
var prop;
if (obj.value) {prop=obj.value;}
else prop=0;
with:
var prop=obj.value||0;
Returning a truthy expression - rather than just true or false - usually makes your code shorter and still readable. This is very common for ||, not so much for &&.
I am having trouble understanding the return line in this function.
I'm creating a parser in C# and found this tutorial (http://lisperator.net/pltut/parser/the-parser) which was written in Javascript. I can't understand the return line of the function
function is_punc(ch) {
var tok = input.peek();
return tok && tok.type == "punc" && (!ch || tok.value == ch) && tok;
}
I believe it will return true if: tok is not null AND its type is punc AND (the char ch is not null OR the value of tok equals ch) AND tok is not null.
If I am reading this right why is he evaluating tok is not null twice.
It's not meaningful in this context, but the logical operators in JavaScript work like those in Python and Lisp (the only connection to Lisp in this question).
That is, a && b returns a if it is "false-y", otherwise it returns b.
Similarly, a || b returns a if it is "truth-y", otherwise it returns b.
Examples (from my Firefox console):
> "" && 23
< ""
> 0 && 23
< 0
> 1 && 23
< 23
> 1 && "hello"
< "hello"
> "hello" || "world"
< "hello"
> 0 || "world"
< "world"
This means that the final tok makes the function return the token itself if all the other conditions are truth-y.
You can express boolean logic with more than true/false, and some languages exploit this to return more information from boolean operators. For example, in a JS console, you can write the following:
> x = 10
10
> (x < 100) && x
10
An integer like 10 belongs to the set of True values when used in a boolean context. The result is thus 10.
The same happens in Lisp, where (or (try-first-this) (maybe-that)) evaluates to the first expression that holds true.
In your case the last term is necessary for the function to return the token.
Why do these logical operators return an object and not a boolean?
var _ = (obj.fn && obj.fn() ) || obj._ || ( obj._ = {} );
var _ = obj && obj._;
I want to understand why it returns result of obj.fn() (if it is defined) OR obj._ but not boolean result.
In JavaScript, both || and && are logical short-circuit operators that return the first fully-determined “logical value” when evaluated from left to right.
In expression X || Y, X is first evaluated, and interpreted as a boolean value. If this boolean value is “true”, then it is returned. And Y is not evaluated. (Because it doesn’t matter whether Y is true or Y is false, X || Y has been fully determined.) That is the short-circuit part.
If this boolean value is “false”, then we still don’t know if X || Y is true or false until we evaluate Y, and interpret it as a boolean value as well. So then Y gets returned.
And && does the same, except it stops evaluating if the first argument is false.
The first tricky part is that when an expression is evaluated as “true”, then the expression itself is returned. Which counts as "true" in logical expressions, but you can also use it. So this is why you are seeing actual values being returned.
The second tricky part is that when an expression is evaluated as “false”, then in JS 1.0 and 1.1 the system would return a boolean value of “false”; whereas in JS 1.2 on it returns the actual value of the expression.
In JS false, 0, -0, "", null, undefined, NaN and document.all all count as false.
Here I am of course quoting logical values for discussion’s sake. Of course, the literal string "false" is not the same as the value false, and is therefore true.
In the simplest terms:
The || operator returns the first truthy value, and if none are truthy, it returns the last value (which is a falsy value).
The && operator returns the first falsy value, and if none are falsy, it return the last value (which is a truthy value).
It's really that simple. Experiment in your console to see for yourself.
console.log("" && "Dog"); // ""
console.log("Cat" && "Dog"); // "Dog"
console.log("" || "Dog"); // "Dog"
console.log("Cat" || "Dog"); // "Cat"
var _ = ((obj.fn && obj.fn() ) || obj._ || ( obj._ == {/* something */}))? true: false
will return boolean.
UPDATE
Note that this is based on my testing. I am not to be fully relied upon.
It is an expression that does not assign true or false value. Rather it assigns the calculated value.
Let's have a look at this expression.
An example expression:
var a = 1 || 2;
// a = 1
// it's because a will take the value (which is not null) from left
var a = 0 || 2;
// so for this a=2; //its because the closest is 2 (which is not null)
var a = 0 || 2 || 1; //here also a = 2;
Your expression:
var _ = (obj.fn && obj.fn() ) || obj._ || ( obj._ = {} );
// _ = closest of the expression which is not null
// in your case it must be (obj.fn && obj.fn())
// so you are gettig this
Another expression:
var a = 1 && 2;
// a = 2
var a = 1 && 2 && 3;
// a = 3 //for && operator it will take the fartest value
// as long as every expression is true
var a = 0 && 2 && 3;
// a = 0
Another expression:
var _ = obj && obj._;
// _ = obj._
In most programming languages, the && and || operators returns boolean. In JavaScript it's different.
OR Operator:
It returns the value of the first operand that validates as true (if any), otherwise it returns the value of the last operand (even if it validates as false).
Example 1:
var a = 0 || 1 || 2 || 3;
^ ^ ^ ^
f t t t
^
first operand that validates as true
so, a = 1
Example 2:
var a = 0 || false || null || '';
^ ^ ^ ^
f f f f
^
no operand validates as true,
so, a = ''
AND Operator:
It returns the value of the last operand that validates as true (if all conditions validates as true), otherwise it returns the value of the first operand that validates as false.
Example 1:
var a = 1 && 2 && 3 && 4;
^ ^ ^ ^
t t t t
^
last operand that validates as true
so, a = 4
Example 2:
var a = 2 && '' && 3 && null;
^ ^ ^ ^
t f t f
^
return first operand that validates as false,
so, a = ''
Conclusion:
If you want JavaScript to act the same way how other programming languages work, use Boolean() function, like this:
var a = Boolean(1 || 2 || 3);// a = true
You should think of the short-circuit operators as conditionals rather than logical operators.
x || y roughly corresponds to:
if ( x ) { return x; } else { return y; }
and x && y roughly corresponds to:
if ( x ) { return y; } else { return x; }
Given this, the result is perfectly understandable.
From MDN documentation:
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 will return a non-Boolean value.
And here's the table with the returned values of all logical operators.
I think you have basic JavaScript methodology question here.
Now, JavaScript is a loosely typed language. As such, the way and manner in which it treats logical operations differs from that of other standard languages like Java and C++. JavaScript uses a concept known as "type coercion" to determine the value of a logical operation and always returns the value of the first true type. For instance, take a look at the code below:
var x = mystuff || document;
// after execution of the line above, x = document
This is because mystuff is an a priori undefined entity which will always evaluate to false when tested and as such, JavaScript skips this and tests the next entity for a true value. Since the document object is known to JavaScript, it returns a true value and JavaScript returns this object.
If you wanted a boolean value returned to you, you would have to pass your logical condition statement to a function like so:
var condition1 = mystuff || document;
function returnBool(cond){
if(typeof(cond) != 'boolean'){ //the condition type will return 'object' in this case
return new Boolean(cond).valueOf();
}else{ return; }
}
// Then we test...
var condition2 = returnBool(condition1);
window.console.log(typeof(condition2)); // outputs 'boolean'
We can refer to the spec(11.11) of JS here of:
Semantics
The production LogicalANDExpression :LogicalANDExpression &&BitwiseORExpression is evaluated as follows:
Evaluate LogicalANDExpression.
2.Call GetValue(Result(1)).
3.Call ToBoolean(Result(2)).
4.If Result(3) is false, return Result(2).
5.Evaluate BitwiseORExpression.
6.Call GetValue(Result(5)).
7.Return Result(6).
see here for the spec
First, it has to be true to return, so if you are testing for truthfulness then it makes no difference
Second, it lets you do assignments along the lines of:
function bar(foo) {
foo = foo || "default value";
Compare:
var prop;
if (obj.value) {prop=obj.value;}
else prop=0;
with:
var prop=obj.value||0;
Returning a truthy expression - rather than just true or false - usually makes your code shorter and still readable. This is very common for ||, not so much for &&.
a = 1;
b = "1";
if (a == b && a = 1) {
console.log("a==b");
}
The Javascript code above will result in an error in the if statement in Google Chrome 26.0.1410.43:
Uncaught ReferenceError: Invalid left-hand side in assignment
I think this is because the variable a in the second part of the statement &&, a=1 cannot be assigned. However, when I try the code below, I'm totally confused!
a = 1;
b = "1";
if (a = 1 && a == b) {
console.log("a==b");
}
Why is the one statement right but the other statement wrong?
= has lower operator precendence than both && and ==, which means that your first assignment turns into
if ((a == b && a) = 1) {
Since you can't assign to an expression in this way, this will give you an error.
The second version is parsed as a = (1 && a == b); that is, the result of the expression 1 && a == b is assigned to a.
The first version does not work because the lefthand side of the assignment is not parsed as you expected. It parses the expression as if you're trying to assign a value to everything on the righthand side--(a == b && a) = 1.
This is all based on the precedence of the various operators. The problem here stems from the fact that = has a lower precedence than the other operators.
Because the order of operations is not what you expect. a == b && a = 1 is equivalent to (a == b && a) = 1 which is equivalent to false = 1.
If you really want to do the assignment, you need to use parentheses around it: a == b && (a = 1).
In if (a = 1 && a == b),
The operations to be first performed is 1 && a == b. 1 && the result of a == b is performed. The result of this && operation is assigned to a.
I'm working with this JS plugin, and I've encountered some syntax I've never seen before. I understand what it's doing, but I'm not sure why it works.
Here's an example of one instance of it:
settings.maxId != null && (params.max_id = settings.maxId);
Is this just taking advantage of conditionals and the single = ? Is this common syntax for JS?
In JavaScript the = operator is an expression and evaluates the assigned value. Because it is an expression it can be used anywhere an expression is allowed even though it causes a side-effect.
Thus:
settings.maxId != null && (params.max_id = settings.maxId)
Means: If settings.maxId is not null then (and only then, since && is short circuiting) evaluate the right-expression (params.max_id = settings.maxId) which in turn causes the value of settings.maxId to be assigned to params.max_id.
This is much more clearly written as:
if (settings.maxId != null) {
params.max_id = settings.maxId
}
Happy coding.
The && operator is known as "boolean AND". Typically, you'd see it in an if statement:
if (x == true && y == false) {
but that's not a restriction. You may use it in any valid expression to "combine" the boolean values of its operands into a single boolean result, according to the logical "AND" operation:
var z = (x == true && y == false);
// z is now true or false, accordingly
One of the lovely things about && is that it "short circuits". In false && true, because the first operand is false the entire expression may only evaluate to false, so the second operand is not even evaluated.
Let's check that again:
var z = (false && foo());
// z is now false
In this statement, the function foo is never even called! It doesn't have to be, for the program to know that z will be false.
This is more than an optimisation — you can rely on it.
Some silly people use this technique to rewrite conditional statements:
if (x == 0) {
foo();
}
into hard-to-read single expressions:
(x == 0) && foo();
Now, consider that assignment can be an expression just like a function call:
var a = (b = c);
Or:
var a = (b = foo());
And add in a conditional via the above technique:
var a = ((x == 0) && (b = foo()));
Now the entire expression b = foo() won't be evaluated at all if x is not 0, because of short circuiting.
We don't even need to do anything with the result of the && operation, and if we don't store it to a you're left with just:
(x == 0) && (b = foo());
which is a statement that'll assign b to the value of foo() only if x is 0.
Avoid it. It's hard to read. Just use an if statement.
this statement will assign params.max_id = settings.maxId only if settings.maxId != null due to the fact that && is a short-circuit logic operator
this behaviour is due to the fact that javascript will evaluate the condition until it's necessary. thus, if first condition is false and the second is in AND there's no need to check further