Develop Reference

How Javascript is evaluating expressions? [duplicate]

The 'Wat' talk for CodeMash 2012 basically points out a few bizarre quirks with Ruby and JavaScript.
I have made a JSFiddle of the results at http://jsfiddle.net/fe479/9/.
The behaviours specific to JavaScript (as I don't know Ruby) are listed below.
I found in the JSFiddle that some of my results didn't correspond with those in the video, and I am not sure why. I am, however, curious to know how JavaScript is handling working behind the scenes in each case.
Empty Array + Empty Array
[] + []
result:
<Empty String>
I am quite curious about the + operator when used with arrays in JavaScript.
This matches the video's result.
Empty Array + Object
[] + {}
result:
[Object]
This matches the video's result. What's going on here? Why is this an object. What does the + operator do?
Object + Empty Array
{} + []
result:
[Object]
This doesn't match the video. The video suggests that the result is 0, whereas I get [Object].
Object + Object
{} + {}
result:
[Object][Object]
This doesn't match the video either, and how does outputting a variable result in two objects? Maybe my JSFiddle is wrong.
Array(16).join("wat" - 1)
result:
NaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaNNaN
Doing wat + 1 results in wat1wat1wat1wat1...
I suspect this is just straightforward behaviour that trying to subtract a number from a string results in NaN.

Here's a list of explanations for the results you're seeing (and supposed to be seeing). The references I'm using are from the ECMA-262 standard.
[] + []
When using the addition operator, both the left and right operands are converted to primitives first (§11.6.1). As per §9.1, converting an object (in this case an array) to a primitive returns its default value, which for objects with a valid toString() method is the result of calling object.toString() (§8.12.8). For arrays this is the same as calling array.join() (§15.4.4.2). Joining an empty array results in an empty string, so step #7 of the addition operator returns the concatenation of two empty strings, which is the empty string.
[] + {}
Similar to [] + [], both operands are converted to primitives first. For "Object objects" (§15.2), this is again the result of calling object.toString(), which for non-null, non-undefined objects is "[object Object]" (§15.2.4.2).
{} + []
The {} here is not parsed as an object, but instead as an empty block (§12.1, at least as long as you're not forcing that statement to be an expression, but more about that later). The return value of empty blocks is empty, so the result of that statement is the same as +[]. The unary + operator (§11.4.6) returns ToNumber(ToPrimitive(operand)). As we already know, ToPrimitive([]) is the empty string, and according to §9.3.1, ToNumber("") is 0.
{} + {}
Similar to the previous case, the first {} is parsed as a block with empty return value. Again, +{} is the same as ToNumber(ToPrimitive({})), and ToPrimitive({}) is "[object Object]" (see [] + {}). So to get the result of +{}, we have to apply ToNumber on the string "[object Object]". When following the steps from §9.3.1, we get NaN as a result:
If the grammar cannot interpret the String as an expansion of StringNumericLiteral, then the result of ToNumber is NaN.
Array(16).join("wat" - 1)
As per §15.4.1.1 and §15.4.2.2, Array(16) creates a new array with length 16. To get the value of the argument to join, §11.6.2 steps #5 and #6 show that we have to convert both operands to a number using ToNumber. ToNumber(1) is simply 1 (§9.3), whereas ToNumber("wat") again is NaN as per §9.3.1. Following step 7 of §11.6.2, §11.6.3 dictates that
If either operand is NaN, the result is NaN.
So the argument to Array(16).join is NaN. Following §15.4.4.5 (Array.prototype.join), we have to call ToString on the argument, which is "NaN" (§9.8.1):
If m is NaN, return the String "NaN".
Following step 10 of §15.4.4.5, we get 15 repetitions of the concatenation of "NaN" and the empty string, which equals the result you're seeing.
When using "wat" + 1 instead of "wat" - 1 as argument, the addition operator converts 1 to a string instead of converting "wat" to a number, so it effectively calls Array(16).join("wat1").
As to why you're seeing different results for the {} + [] case: When using it as a function argument, you're forcing the statement to be an ExpressionStatement, which makes it impossible to parse {} as empty block, so it's instead parsed as an empty object literal.

This is more of a comment than an answer, but for some reason I can't comment on your question. I wanted to correct your JSFiddle code. However, I posted this on Hacker News and someone suggested that I repost it here.
The problem in the JSFiddle code is that ({}) (opening braces inside of parentheses) is not the same as {} (opening braces as the start of a line of code). So when you type out({} + []) you are forcing the {} to be something which it is not when you type {} + []. This is part of the overall 'wat'-ness of Javascript.
The basic idea was simple JavaScript wanted to allow both of these forms:
if (u)
v;
if (x) {
y;
z;
}
To do so, two interpretations were made of the opening brace: 1. it is not required and 2. it can appear anywhere.
This was a wrong move. Real code doesn't have an opening brace appearing in the middle of nowhere, and real code also tends to be more fragile when it uses the first form rather than the second. (About once every other month at my last job, I'd get called to a coworker's desk when their modifications to my code weren't working, and the problem was that they'd added a line to the "if" without adding curly braces. I eventually just adopted the habit that the curly braces are always required, even when you're only writing one line.)
Fortunately in many cases eval() will replicate the full wat-ness of JavaScript. The JSFiddle code should read:
function out(code) {
function format(x) {
return typeof x === "string" ?
JSON.stringify(x) : x;
}
document.writeln('>>> ' + code);
document.writeln(format(eval(code)));
}
document.writeln("<pre>");
out('[] + []');
out('[] + {}');
out('{} + []');
out('{} + {}');
out('Array(16).join("wat" + 1)');
out('Array(16).join("wat - 1")');
out('Array(16).join("wat" - 1) + " Batman!"');
document.writeln("</pre>");
[Also that is the first time I have written document.writeln in many many many years, and I feel a little dirty writing anything involving both document.writeln() and eval().]

I second #Ventero’s solution. If you want to, you can go into more detail as to how + converts its operands.
First step (§9.1): convert both operands to primitives (primitive values are undefined, null, booleans, numbers, strings; all other values are objects, including arrays and functions). If an operand is already primitive, you are done. If not, it is an object obj and the following steps are performed:
Call obj.valueOf(). If it returns a primitive, you are done. Direct instances of Object and arrays return themselves, so you are not done yet.
Call obj.toString(). If it returns a primitive, you are done. {} and [] both return a string, so you are done.
Otherwise, throw a TypeError.
For dates, step 1 and 2 are swapped. You can observe the conversion behavior as follows:
var obj = {
valueOf: function () {
console.log("valueOf");
return {}; // not a primitive
},
toString: function () {
console.log("toString");
return {}; // not a primitive
}
}
Interaction (Number() first converts to primitive then to number):
> Number(obj)
valueOf
toString
TypeError: Cannot convert object to primitive value
Second step (§11.6.1): If one of the operands is a string, the other operand is also converted to string and the result is produced by concatenating two strings. Otherwise, both operands are converted to numbers and the result is produced by adding them.
More detailed explanation of the conversion process: “What is {} + {} in JavaScript?”

We may refer to the specification and that's great and most accurate, but most of the cases can also be explained in a more comprehensible way with the following statements:
+ and - operators work only with primitive values. More specifically +(addition) works with either strings or numbers, and +(unary) and -(subtraction and unary) works only with numbers.
All native functions or operators that expect primitive value as argument, will first convert that argument to desired primitive type. It is done with valueOf or toString, which are available on any object. That's the reason why such functions or operators don't throw errors when invoked on objects.
So we may say that:
[] + [] is same as String([]) + String([]) which is same as '' + ''. I mentioned above that +(addition) is also valid for numbers, but there is no valid number representation of an array in JavaScript, so addition of strings is used instead.
[] + {} is same as String([]) + String({}) which is same as '' + '[object Object]'
{} + []. This one deserves more explanation (see Ventero answer). In that case, curly braces are treated not as an object but as an empty block, so it turns out to be same as +[]. Unary + works only with numbers, so the implementation tries to get a number out of []. First it tries valueOf which in the case of arrays returns the same object, so then it tries the last resort: conversion of a toString result to a number. We may write it as +Number(String([])) which is same as +Number('') which is same as +0.
Array(16).join("wat" - 1) subtraction - works only with numbers, so it's the same as: Array(16).join(Number("wat") - 1), as "wat" can't be converted to a valid number. We receive NaN, and any arithmetic operation on NaN results with NaN, so we have: Array(16).join(NaN).

To buttress what has been shared earlier.
The underlying cause of this behaviour is partly due to the weakly-typed nature of JavaScript. For example, the expression 1 + “2” is ambiguous since there are two possible interpretations based on the operand types (int, string) and (int int):
User intends to concatenate two strings, result: “12”
User intends to add two numbers, result: 3
Thus with varying input types,the output possibilities increase.
The addition algorithm
Coerce operands to primitive values
The JavaScript primitives are string, number, null, undefined and boolean (Symbol is coming soon in ES6). Any other value is an object (e.g. arrays, functions and objects). The coercion process for converting objects into primitive values is described thus:
If a primitive value is returned when object.valueOf() is invoked, then return this value, otherwise continue
If a primitive value is returned when object.toString() is invoked, then return this value, otherwise continue
Throw a TypeError
Note: For date values, the order is to invoke toString before valueOf.
If any operand value is a string, then do a string concatenation
Otherwise, convert both operands to their numeric value and then add these values
Knowing the various coercion values of types in JavaScript does help to make the confusing outputs clearer. See the coercion table below
+-----------------+-------------------+---------------+
| Primitive Value | String value | Numeric value |
+-----------------+-------------------+---------------+
| null | “null” | 0 |
| undefined | “undefined” | NaN |
| true | “true” | 1 |
| false | “false” | 0 |
| 123 | “123” | 123 |
| [] | “” | 0 |
| {} | “[object Object]” | NaN |
+-----------------+-------------------+---------------+
It is also good to know that JavaScript's + operator is left-associative as this determines what the output will be cases involving more than one + operation.
Leveraging the
Thus 1 + "2" will give "12" because any addition involving a string will always default to string concatenation.
You can read more examples in this blog post (disclaimer I wrote it).

javascript +function ($) { ... }(jQuery) [duplicate]

I’ve been looking for information about immediately invoked functions, and somewhere I stumbled on this notation:
+function(){console.log("Something.")}()
Can someone explain to me what the + sign in front of the function means/does?

It forces the parser to treat the part following the + as an expression. This is usually used for functions that are invoked immediately, e.g.:
+function() { console.log("Foo!"); }();
Without the + there, if the parser is in a state where it's expecting a statement (which can be an expression or several non-expression statements), the word function looks like the beginning of a function declaration rather than a function expression and so the () following it (the ones at the end of the line above) would be a syntax error (as would the absense of a name, in that example). With the +, it makes it a function expression, which means the name is optional and which results in a reference to the function, which can be invoked, so the parentheses are valid.
+ is just one of the options. It can also be -, !, ~, or just about any other unary operator. Alternately, you can use parentheses (this is more common, but neither more nor less correct syntactically):
(function() { console.log("Foo!"); })();
// or
(function() { console.log("Foo!"); }());

Subsidiary to #TJCrowder's answer, + is usually used to force numerical casting of a value as this SO answer explains. In this instance it is called the 'unary plus operator' (for ease of googling).
var num = +variant;
So in front of a function it can be a way to force the function's result to be interpreted as a number. I doubt it happens yet, but theoretically the JIT could use that to compile the function as a numerical-only function etc. However, to prevent the unary plus being a concatenation when used in a larger expression, you would need parentheses:
blah + (+(function(){ var scope; return "4"; })());

So the short answer is that it prevents a syntax error, by using the function results in one way or another.
You can also instruct the engine that you're not even interested in the return value by using the void operator:
void function() { console.log("Foo!"); }();
Of course, putting braces around the whole thing also serves that purpose.

TL;DR (Quick Answer)
Plus Sign Casts (Converts) the proceeding operand to a Number if it isn't already.
Solution & Origins
The + sign before the function, actually called Unary plus and is part of a group called a Unary Operators and (the Unary Plus) is used to convert string and other representations to numbers (integers or floats).
A unary operation is an operation with only one operand, i.e. a single
input. This is in contrast to binary operations, which use two
operands
Basic Uses:
const x = "1";
const y = "-1";
const n = "7.77";
console.log(+x);
// expected output: 1
console.log(+n);
// expected output: 7.77
console.log(+y);
// expected output: -1
console.log(+'');
// expected output: 0
console.log(+true);
// expected output: 1
console.log(+false);
// expected output: 0
console.log(+'hello');
// expected output: NaN
When the + sign is positioned before a variable, function or any returned string representations the output will be converted to integer or float; the unary operator (+) converts as well the non-string values true, false, and null.
Advanced Uses
The right way to use the function you mentioned above will be:
+function(){return "3.141"}()
// expected output: 3.141
I love to use + to turn a new Date() object to a timestamp, like this:
+new Date()
// expected output: 1641387991035
Other Unary Operators
- The unary negation operator converts its operand to Number type
and then negates it.
~ Bitwise NOT operator.
! Logical NOT operator.
delete The delete operator deletes a property from an object.
void The void operator discards an expression's return value.
typeof The typeof operator determines the type of a given object.

When to use parseInt

Which rule do I have to follow when extracting numbers out of DOM and calcluation with them? How does javascript knows that a value is a number or not? Should I always use parseInt?
Given following Code:
HTML
<div id="myvalue">5</div>
<div id="withParseInt"></div>
<div id="withoutParseInt"></div>
<div id="withoutParseIntButIncrement"></div>
JS & jQuery:
var value = $('#myvalue').text();
$('#withParseInt').text(parseInt(value) + 1);
$('#withoutParseInt').text(value + 1);
$('#withoutParseIntButIncrement').text(value++);
Gives following output:
5
6
51
5
Fiddle: http://jsfiddle.net/ytxKU/3/

The .text() method will always return a string. Some operators, like the + operator, are overloaded to perform both arithmetic and string operations. In the case of strings, it performs concatenation, hence the "51" result.
If you have a string and need to use a non-coercing operator, you will have to use parseInt (or some other method of converting to a number).
However, the * operator for example implicity performs this coercion, so you wouldn't need the parseInt call in that situation (see an updated fiddle for example).
Note that the increment ++ operator does coerce its operand, but you've used the postfix operator so it won't have any effect. Use the prefix operator and you can see it working:
$('#withoutParseIntButIncrement').text(++value);
So, to summarise:
// Parses string to number and adds 1
$('#withParseInt').text(parseInt(value) + 1);
// Coerces number 1 to string "1" and concatenates
$('#withoutParseInt').text(value + 1);
// Implicity coerces string to number, but after it's been inserted into the DOM
$('#withoutParseIntButIncrement').text(value++);
// Implicity coerces string to number, before it's been inserted into the DOM
$('#withoutParseIntButIncrement').text(++value);
// Implicity coerces to number
$('#withoutParseIntButMultiply').text(value * 2);
Side note: it's considered good practice to always pass the second argument (the radix) to parseInt. This ensures the number is parsed in the correct base:
parseInt(value, 10); // For base 10

One and only rule:
Every value that you retrieve from the DOM is a string.

Yes, you should always use parseInt() or Number() to be on the safe side. Otherwise Javascript will decide what to do with it
The value itself is a string
Using operator + will concatenate two strings
Using operator - will calculate the numerical difference
...

It's always good to use parseInt just to be on the safe side, especially as you can supply a second parameter for the numerical system to use.
By the way, in your final example it should be ++value if you want it to equal 6.

JavaScript plus sign in front of function expression

I’ve been looking for information about immediately invoked functions, and somewhere I stumbled on this notation:
+function(){console.log("Something.")}()
Can someone explain to me what the + sign in front of the function means/does?

It forces the parser to treat the part following the + as an expression. This is usually used for functions that are invoked immediately, e.g.:
+function() { console.log("Foo!"); }();
Without the + there, if the parser is in a state where it's expecting a statement (which can be an expression or several non-expression statements), the word function looks like the beginning of a function declaration rather than a function expression and so the () following it (the ones at the end of the line above) would be a syntax error (as would the absense of a name, in that example). With the +, it makes it a function expression, which means the name is optional and which results in a reference to the function, which can be invoked, so the parentheses are valid.
+ is just one of the options. It can also be -, !, ~, or just about any other unary operator. Alternately, you can use parentheses (this is more common, but neither more nor less correct syntactically):
(function() { console.log("Foo!"); })();
// or
(function() { console.log("Foo!"); }());

Subsidiary to #TJCrowder's answer, + is usually used to force numerical casting of a value as this SO answer explains. In this instance it is called the 'unary plus operator' (for ease of googling).
var num = +variant;
So in front of a function it can be a way to force the function's result to be interpreted as a number. I doubt it happens yet, but theoretically the JIT could use that to compile the function as a numerical-only function etc. However, to prevent the unary plus being a concatenation when used in a larger expression, you would need parentheses:
blah + (+(function(){ var scope; return "4"; })());

So the short answer is that it prevents a syntax error, by using the function results in one way or another.
You can also instruct the engine that you're not even interested in the return value by using the void operator:
void function() { console.log("Foo!"); }();
Of course, putting braces around the whole thing also serves that purpose.

TL;DR (Quick Answer)
Plus Sign Casts (Converts) the proceeding operand to a Number if it isn't already.
Solution & Origins
The + sign before the function, actually called Unary plus and is part of a group called a Unary Operators and (the Unary Plus) is used to convert string and other representations to numbers (integers or floats).
A unary operation is an operation with only one operand, i.e. a single
input. This is in contrast to binary operations, which use two
operands
Basic Uses:
const x = "1";
const y = "-1";
const n = "7.77";
console.log(+x);
// expected output: 1
console.log(+n);
// expected output: 7.77
console.log(+y);
// expected output: -1
console.log(+'');
// expected output: 0
console.log(+true);
// expected output: 1
console.log(+false);
// expected output: 0
console.log(+'hello');
// expected output: NaN
When the + sign is positioned before a variable, function or any returned string representations the output will be converted to integer or float; the unary operator (+) converts as well the non-string values true, false, and null.
Advanced Uses
The right way to use the function you mentioned above will be:
+function(){return "3.141"}()
// expected output: 3.141
I love to use + to turn a new Date() object to a timestamp, like this:
+new Date()
// expected output: 1641387991035
Other Unary Operators
- The unary negation operator converts its operand to Number type
and then negates it.
~ Bitwise NOT operator.
! Logical NOT operator.
delete The delete operator deletes a property from an object.
void The void operator discards an expression's return value.
typeof The typeof operator determines the type of a given object.

What is the difference between Number(...) and parseFloat(...)

What is the difference between parseInt(string) and Number(string) in JavaScript has been asked previously.
But the answers basically focused on the radix and the ability of parseInt to take a string like "123htg" and turn it into 123.
What I am asking here is if there is any big difference between the returns of Number(...) and parseFloat(...) when you pass it an actual number string with no radix at all.

The internal workings are not that different, as #James Allardic already answered. There is a difference though. Using parseFloat, a (trimmed) string starting with one or more numeric characters followed by alphanumeric characters can convert to a Number, with Number that will not succeed. As in:
parseFloat('3.23abc'); //=> 3.23
Number('3.23abc'); //=> NaN
In both conversions, the input string is trimmed, by the way:
parseFloat(' 3.23abc '); //=> 3.23
Number(' 3.23 '); //=> 3.23

No. Both will result in the internal ToNumber(string) function being called.
From ES5 section 15.7.1 (The Number Constructor Called as a Function):
When Number is called as a function rather than as a constructor, it performs a type conversion...
Returns a Number value (not a Number object) computed by ToNumber(value) if value was supplied, else returns +0.
From ES5 section 15.1.2.3 (parseFloat (string)):
...
If neither trimmedString nor any prefix of trimmedString satisfies the syntax of a StrDecimalLiteral (see 9.3.1)
...
And 9.3.1 is the section titled "ToNumber Applied to the String Type", which is what the first quote is referring to when it says ToNumber(value).
Update (see comments)
By calling the Number constructor with the new operator, you will get an instance of the Number object, rather than a numeric literal. For example:
typeof new Number(10); //object
typeof Number(10); //number
This is defined in section 15.7.2 (The Number Constructor):
When Number is called as part of a new expression it is a constructor: it initialises the newly created object.

Not a whole lot of difference, as long as you're sure there's nothing but digits in your string. If there are, Number will return NaN. Another problem that you might get using the Number constructor is that co-workers might think you forgot the new keyword, and add it later on, causing strict comparisons to fail new Number(123) === 123 --> false whereas Number(123) === 123 --> true.
In general, I prefer to leave the Number constructor for what it is, and just use the shortest syntax there is to cast to an int/float: +numString, or use parse*.

When not using new to create a wrapper object for a numerical value, Number is relegated to simply doing type conversion from string to number.
'parseFloat' on the other hand, as you mentioned, can parse a floating point number from any string that starts with a digit, a decimal, or +/-
So, if you're only working with strings that contain only numerical values, Number(x) and parseFloat(x) will result in the same values

Please excuse me posting yet another answer, but I just got here via a Google search and did not find all of the details that I wanted. Running the following code in Node.js:
var vals = ["1", "1.1", "0", "1.1abc", "", " ", null];
for(var i = 0; i < vals.length; i++){
var ifTest = false;
if(vals[i])
{
ifTest = true;
}
console.log("val=" + vals[i] + ", Number()=" + Number(vals[i])+ ", parseFloat()=" + parseFloat(vals[i]) + ", if()=" + ifTest);
}
gives the following output:
val=1, Number()=1, parseFloat()=1, if()=true
val=1.1, Number()=1.1, parseFloat()=1.1, if()=true
val=0, Number()=0, parseFloat()=0, if()=true
val=1.1abc, Number()=NaN, parseFloat()=1.1, if()=true
val=, Number()=0, parseFloat()=NaN, if()=false
val= , Number()=0, parseFloat()=NaN, if()=true
val=null, Number()=0, parseFloat()=NaN, if()=false
Some noteworthy takeaways:
If protecting with an if(val) before trying to convert to number, then parseFloat() will return a number except in the whitespace case.
Number returns a number in all cases except for non-numeric characters aside from white-space.
Please feel free to add any test cases that I may be missing.