What is the meaning of && in this JavaScript code?
function doIt(a) {
var b = a.parents(),
c = 1;
return b.each(function() {
jQuery(this).hasClass("free-content") && (c = 0)
}), c
}
normally I would think this would be a logical operator AND, but I can't figure out what it does in this line.
The logical AND operator in this case is used in place of an IF-statement. It will set c to 0 if jQuery(this).hasClass("free-content") returns a truthy value.
It's equivalent to:
if (jQuery(this).hasClass("free-content")) {
c = 0;
}
You wondering what it means is actually the reason I dislike this type of coding and consider it a bad practice. It's hard to read and can create confusion, which in turn can create bugs.
It's also worth noting what logical AND returns, if you want to use the returned value:
(Logical AND) 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.
Here's an example showing, in my opinion, bad code since it's hard to follow:
var foo = bar && baz || qux;
The above code is equivalent to:
var foo;
if (bar && baz) {
foo = baz;
} else {
foo = qux;
}
Summary: Do not use logical operators as a nifty way to replace IF-statements or to save keystrokes. It will most likely come back and bite you or someone else in the ass.
I know there will be people arguing against me on this (usually the same people who doesn't want to use semicolon because ASI), but it's just a really bad idea to write code that only an expert would understand. There's no argument against it.
return b.each(function () {
jQuery(this).hasClass("free-content") && (c = 0)
}), c
b.each loops over all entries in b, and checks whether the current element has the class free-content set. Only if that yields true, the second part of the expression is evaluated – because they are concatenated via &&, and JavaScript stops evaluating such an expression when the first part is false, because then the whole expression can’t become true any more.
So if there is an element with that class, the second part is evaluated – and thereby c is set to the value 0 (because that’s the assignment operator = there, and not a comparison).
And after that each loop is finished, the value of c is returned to the outside – because each() and c are connected via the comma operator , here, which evaluates its operands from left to right and then “returns” the second one.
JavaScript allows a lot of “fancy” coding like this, but as Marcus already said in his answer, that is hard to read, and there is no actual advantage here over, say
b.each(function() {
if(jQuery(this).hasClass("free-content")) {
c = 0;
return false;
}
});
return c;
I added the return false here inside the each loop, because once we found an element with that class, we don’t need to search the rest of them any more – that’s something that the person who came up with the original code forgot to do in all their “fancy-ness” … their loop will continue to iterate over all of b’s elements, not matter if it finds the element it is looking for in the first round already.
Related
I've been studying JavaScript for a couple months, and I have some questions.
Say we have some if statements like this:
if (something)
if (!something)
From what I have looked up, the first statement checks whether something evaluates to true, whereas the second one checks for whether the condition is not true(I recall reading the ! coerces it into boolean and then 'inverses' it).
However, I'd like to know what are these statements' direct equivalents when not "shortened". I'd like to understand code like this more in depth before really using it.
Is if (something) the same as if (something == true)? Or even if (something === true)?
And what would be the "non-shortened" (for my lack of a better term) direct equivalent of if (!something)?
Thank you for any help.
Part 1:
When we use if(something) javascript will check if it is a 'Truthy' or 'Falsy' value.
So if we say
if(1) {
alert("Yes, it's true...");
}
javscript will check if 1 is a "truthy" value and it is.
Everything that is not a falsy value is truthy by extension.
Falsy values are:
0, 0n, null, undefined, false, NaN, and the empty string “”.
So if we directly put any of the above into a parenthesis it will return false.
e.g. if (null) ... the if clause won't be executed because null is falsy.
Part 2:
When we use if(something == true), we check if one value is equal to another value. If the types are not the same, since we're using == coercion will happen.
What is this about 'coercion'?
If we say if('1' == 1) this will execute the if clause as it returns true. Javascript will convert 1(number) to '1'(string) in this case.
If we want to be strict and say "I only want true if they're both numbers, I don't care if 1 is a string" then we use
if(1 === '1') and this will not execute the if clause(because the condition returns false).
So if we say if(false == false) this will return true because false is equal to false, and the if clause will be executed.
Note the difference if we said if(false) or if('') the if statement will not be executed.
Next, if we want to use negation, we have to put the ! in front of the boolean we're negating, regardless of whether it's part of an if condition or... There is no way around it.
However, there is a shorter version of this:
var a = 3;
var b = 2;
if(a===b) {
alert('A is equal to B');
} else {
alert('B is not equal to A');
}
This is considered shorter(ternary operator):
Note that if we want to use more than one statement, we should use the if else
a = 3;
b = 2;
a===b ? alert('A is equal to B') : alert('B is not equal to A');
I had a bit of a weird result in the javascript console. I was trying to look for an alternative (more readable) version of the ternary operator, just for fun. Typing:
{ if(3===4) {5} else {6} }
Evaluates in my console to 6, but for some reason, I can not assign it to a variable, so running:
let a = { if(3===4) {5} else {6} }
Does not let me store it to the variable directly. So my main question is, if this block is returning something, why can't I assign it?
The fact that blocks (and other statements) return values isn't accessible to your code. Consoles can see the result, and it exists at a language specification level, but not in your code.
Your options are the conditional operator¹ (which is quite readable when you're used to it, but you've said you're looking for alternatives to it) or the if/else assigning to a in both parts:
let a;
if (3 === 4) {
a = 5;
} else {
a = 6;
}
Or you could use an inline function (IIFE):
let a = (() => { if (3 === 4} return 5 else return 6; })();
There is also a proposal being floated for "do expressions", which would look like this:
// Currently a proposal, not implemented in engines yet
let a = do { if (3 === 4) 5; else 6; };
That proposal is at Stage 1 of the process, so it may or may not progress, and if it progresses it could change markedly before doing so.
¹ Although you see "the ternary operator" a lot, the proper name is the conditional operator. It is a ternary operator (an operator accepting three operands), and currently JavaScript's only ternary operator, but that could change someday. :-)
Use ternary operator, because you can't assign if statement:
let a = 3 === 4 ? 5 : 6;
The reason this doesn't work is because if is, as you pointed out, a statement. An assignment requires an expression holding a value to be assigned. A statement doesn't (per se) have a value--it simply performs some side-effect instead.
What you can use if you want to assign a value conditionally is the ternary conditional operator, which is an expression:
let a = (3 === 4 ? 5 : 6)
At this point you're probably wondering, why does the console print a value if a statement merely performs side-effects? This is because statements have a completion value, which for an if statement is the completion value of the corresponding branch, and for an expression statement (which is what the 5 and 6 inside the blocks are), is whatever the expression evaluates to. This is mainly useful for when you want to evaluate an expression in the console and see what it produces, without having to issue a console.log() each time you do.
It's not returning anything. if is not an expression, it's a statement; it has no value.
The point of the ternary operator was to provide a more readable version of this:
let a;
if (condition) {
a = 1;
}
else {
a = 2;
}
In the book 'Functional javascript' by Michael Fogus, I faced with one expression that I still can't undrestand.
Here is the function:
function defaults(d){
return function(o, k){
var val = fnull(_.identity, d[k]);
return o && val(o[k]);
}
}
where
function fnull(fun /*, defaults*/){
var defaults = _.rest(arguments);
return function(/* args */){
var args = _.map(arguments, function(e, i){
return existy(e)?e : defaults[i];
});
return fun.apply(null, args);
};
};
function existy(x){return x != null}
(underscore is the object of Underscore.js library)
and the example of use:
function doSomething(config){
var lookup = defaults({critical:108});
return lookup(config, 'critical');
}
doSomething({critical: 9});
//=> 9
doSomething({});
//=> 108
I've recreated exapmle in node.js and it works fine, but I wonder why is the logical 'and' in the return line of 'default' function?
return o && val(o[k]);
What is the point of doing that? I checked the exapmle with
return val(o[k]);
and it also worked well.
It's hard to believe that this is just a mistake...
The logical and will make sure the second part is only evaluated if the first part is true. If o does not evaluate to true, the expression returns false. Otherwise, it returns val(o[k]).
This is used as a quick check to see if o is not false / null / undefined.
return o && val(o[k])
mean that if "o" is TRUE it will return "val(o[k])"
given "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.
That is smart usage of something called short "short-circuiting" in logical expressions.
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.
The rules of logic guarantee that these evaluations are always correct. Note that the (anything) part of the above expressions is not evaluated (meaning not ran at all), so any side effects of doing so do not take effect. There are also benefits to it.
Usage:
Make your expressions evaluate faster by putting something easily calculable or likely to be true/fail at leftmost position in expression.
For example parOfExpressionLikelyToBeTrue && (rest of expression) will in most cases not even calculate the other part of expression. Same goes for parOfExpressionLikelyToBeTrue || (rest of espression).
Same can be used if something is very time consuming to calculate, you push it as far back to the right in expression. For example (rest of the expression) && ThisCostsALotOfTime or (rest of the expression) || ThisCostsALotOfTime. Here when first parts of expression short-circuit you save time on your time consuming part.
Short circuit existence evaluation. Lets say you need to check if your object's property pr is 3? What would you do? obj.pr === 3? Yes and no. What if property is missing? It's fine you will get undefined and that is not === 3. But what if object is not there. You will get trying to read pr of undefined error. You can use short-circuit logic here to you benefit by being defensive and writing the expression as if (obj || obj.pr === 3). This ensures there are no errors, only true and false.
Short circuit initialization. Let's say you wanna say variable a is b. But b might be undefined. And you wanna your variable to have a default. You could write a=b and then check if a is undefined and set it to default or you can be clever and write it as a = b || 3. This way a is b or if be is undefined it's 3. Ofc, you can use this for late initialization as well a = a || 3.
Making sure object for function exists before trying to run function. Same as before mentioned with properties you might wanna test if object containing the function exists before running the function. Let's say you got object obj and function fn as it's property. You might call that function like obj.fn(). It's fine, but if obj is undefined you will get an error. Being defensive you might wanna write: obj && obj.fn().
Running the function only if there is one. Since functions in JS can be passed to other functions you can not be sure at run time it's there. Being defensive you might wanna run your function as (typeof passedFunction === "function" && passedFunction() instead just passedFunction() which my produce an error.
other smart things like guardian expressions etc which are complicated and too many to for me to remember all and you should avoid them anyway for better code readability.
I am well aware of similar questions, but I still can't solve mine basing on those.
So, my code looks like this:
for (var e, i = 0; e = formErrors[i]; i += 1)
JSHint returns said error in char 39, so with ; after formErrors[i]. What can I do?
JSHint is warning you of a potential bug. It is expected that the 2nd part of a for statement will be a Boolean expression. Normally, you'd use one of the comparison operators for this (==, ===, !=, > etc..). Since the expression is e = formErrors[i] it looks like it might be a mistake, possibly due to a missing equal sign. This is a common typo that causes a lot of bugs.
Apparently, in this case it is not a bug, but rather an intentional usage of the fact that any expression evaluates to something, and an assignment expression evaluates to the assigned value:
var x;
alert(x = 1);
So the for statement actually assigns a new value to e, but also evaluates that value as a condition, casting it to Boolean if required.
You could refactor your code such that it both assigns the value and uses a casting operation that would satisfy JSHint and make the code a bit more obvious to the reader:
for (var e, i = 0; !!(e = formErrors[i]); i += 1)
The 2nd ! (the one directly in front of (e...) causes a casting to Boolean, but also negates it, the 1st ! reverts this negation.
That's just a very weird way of writing a loop. JsHint expects a boolean expression ("conditional") there, and judges that your assignment is a mistake and you actually wanted a comparison (== instead of =).
What you should do is switch to the following common array iteration idiom:
for (var i = 0; i < formErrors.length; i += 1) {
var e = formErrors[i];
…
(Which works the same as your original code for non-sparse formErrors arrays that contain no falsy values.)
Or alternatively, if you want to write non-idiomatic code, dump jshint :-)
e = formErrors[i] is an assignment, don't you want a condition there? use a comma after i=0 in that case, else put a condition after the semicolon.
Typically, the middle element in a for loop is the condition that is used to decide whether to continue the loop or not. You have an assignment there, if you wanted a condition you should use e === formErrors[i] (or with a double =, but that is usually not recommended).
An assignment can technically work, because e can be, for example, some object (true) or null (false). But that is considered bad coding style, usually, and won't make for very readable code.
I am relatively new to Javascript and am working through ch. 5 of Eloquent Javascript. I came across some code that I don't quite understand. I know HOW it works (the general method and steps), but I don't understand WHY it works.
The code is here:
function filter(array, test) {
var passed = [];
for (var i = 0; i < array.length; i++) {
if (test(array[i]))
passed.push(array[i]);
}
return passed;
}
Basically the function takes the element the 'for loop is iterating over' from the array, and compares it to the test parameter.
I am wondering how/why this works:
if (test(array[i]))
There is no || && or other 'comparison operators'. How does it compare values with only using parenthesis?
How is test compared to the array[i] value with no operators?
Link to file: http://eloquentjavascript.net/05_higher_order.html
go to 'Filtering an Array' exercise
Thanks!
Whatever is inside the parentheses of an if statement will be evaluated. If the result is falsy (false, 0, "", NaN, null, undefined) it fails and if the result is truthy (everything else) it passes. So if your if statement contains a function call if (test(something)) {}, then the if statement just says if the result of the function call is truthy then pass.
Also, && and || are not comparison operators, they are boolean operators. They just connect the result of two statements, like true || false which evaluates to true.
I am not quite sure, but I think this is a custom function. Most likely there is some comparison there and the result of the function is True/False. If you give us the whole code we could explain it better to you.
This code is accepting a test parameter that is what is called a "predicate function" i.e. a function that given an element will return true or false.
It's going to be used for example with
var big_numbers = filter(numbers, function(x){ return x > 100; });
i.e. the expected parameter test is actually code.
In Javascript passing code is very common and idiomatic. It's something that is more annoying in other languages that don't support the concept of "closure" and of "nested function", forcing all code to live at the top level, being given a name and to have no context (e.g. the C language).
'test' here is a function, not a value. In Javascript, each function is an object and can be passed as parameter. In this case, test is a function that take one parameter and return true or false base on the parameter value.
So in the for loop, test function is called with each array element and if the result is true, it will be store in another array. Eventually, passed elements would be return to the function caller.