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Odd behaviour of comparison of object literals

I have searched but could not find logic behind following in JavaScript.
When I type in the Chrome console:
{} == null
it returns
Uncaught SyntaxError: Unexpected token ==
But
{} == {}
and
{} == function(){}
returns false
Why?

I assume you understand why {} == null throws SyntaxError. Long story short it is because { in the begining starting a block statement not an object literal. You could check the answer here
As of why {} == {} this works.
If you check chromium code that evaluates expressions in console. You could find the following (code)
if (/^\s*\{/.test(text) && /\}\s*$/.test(text))
text = '(' + text + ')';
executionContext.evaluate(text, "console", !!useCommandLineAPI, false, false, true, printResult);
This code wraps {} == {} code with parentheses making it a valid expression ({} == {}) comparing two empty object literals. Which evaluates to false because objects are compared by reference.
Node repl has the same behaviour src
if (/^\s*\{/.test(code) && /\}\s*$/.test(code)) {
// It's confusing for `{ a : 1 }` to be interpreted as a block
// statement rather than an object literal. So, we first try
// to wrap it in parentheses, so that it will be interpreted as
// an expression.
code = `(${code.trim()})\n`;
wrappedCmd = true;
}

You can find this in the specs under Statement (art. 12)
12 - Statement
Statement :
Block.
VariableStatement
EmptyStatement
ExpressionStatement
.
.
.
The first applicable rules are either Block or Expression Statement. So we need to look at 12.4.
In 12.4 the specs clearly state that an expression statement cannot start with a {.
though i haven’t yet found what makes example 2 an expression, maybe it’s implementation specific
12.4 Expression Statement
Syntax
ExpressionStatement :
[lookahead ∉ {{, function}] Expression ;
NOTE An ExpressionStatement cannot start with an opening curly brace because that might make it ambiguous with a Block. Also, an ExpressionStatement cannot start with the function keyword because that might make it ambiguous with a FunctionDeclaration.
Semantics
The production ExpressionStatement : [lookahead ∉ {{, function}]Expression; is evaluated as follows:
Let exprRef be the result of evaluating Expression.
Return (normal, GetValue(exprRef), empty).

I would say this is a parsing issue, rather than a logic issue per se.
In Chrome I get the observed behavior.
In IE I get syntax errors whenever I put {} (or, it seems, any object literal) on the LHS of the ==.
In both browsers, putting () around the expression fixed things. Or first assigning the object to a variable
var x = {}
x == null
I would say it seems to me like a bug in the parsing. Whether that is true in an academic sense would take digging through specs and grammars; the practical answer is, there are simple enough work-arounds that the best bet is to not do that.

It’s because JavaScript sucks
The reason it doesn’t work is because
JavaScript takes the type the comparison is taking from the first object.
For instance
3+”1” = 4
But
“3”+1 = 31
The first example does the operation as a number because the first object is a number
The second example sees a string as the first object and treats the operation as concatenation of a string.
For your example
{} is an object but null can’t be converted to an object
It works the other way because
{} can be represented as a null object.

JSHint -Expected a conditional expression and instead saw an assignment

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.

meaning of the AND operator in this line

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.

What do these characters do in Javascript [duplicate]

This question already has answers here:
JavaScript plus sign in front of function expression
(4 answers)
Closed 8 years ago.
I was going through a piece of code here http://cssdeck.com/labs/bjiau4dy and I saw this in the Javascript box -
!+-+-+!+-+-+!+-+-+!+-+-+!+-+-+!+-+-+!+-+-+!+-+-+!
What does that do ? and why does it not throw errors in the console ?
Thanks!

Any of those symbols turn the function that follows it into a function expression, instead of a function declaration. Putting them all in there together is just for fun.
If you try to call a regular function by putting () right after the declaration:
function () {
// this is a syntax error...
}();
you'll get a syntax error:
SyntaxError: Unexpected token (
since you cannot call a function declaration.
So people usually wrap an anonymous function in parentheses to turn it into a function expression:
(function () {
// this will execute immediately
}());
You can achieve the same thing be prepending any of those symbols:
!function () {
// this will also execute immediately
}();
For more information, see here: http://kangax.github.com/nfe/#expr-vs-decl

There is zero point to doing the long set of symbols except as a stylistic watermark.
That set of symbols acts as an operation on the function that follows, but does nothing.
In JS you can write 1 or +1 (one) or !1 (false) or !+1 (still false). This is just a long chain of those that are acting on the following function.
So !+-+-+!+-+-+!+-+-+!+-+-+!+-+-+!+-+-+!+-+-+!+-+-+!1 doesn't throw an error. Its value is false though.
And !+-+-+!+-+-+!+-+-+!+-+-+!+-+-+!+-+-+!+-+-+!+-+-+!function(){ ... } just executes the function and evaluates to false (but nothing is done with the value anyway)
However, since something is there, it turns the function definition into an operation and evaluates the function immediately. But just +function(){ ... } would do the same thing.

To elaborate a bit more, each of these operators will actually perform type coercion on the operand proceeding it. + will convert the operand following it into a number, as will the - operator. ! is the not operator, and will convert the operand into a boolean (true/false).
Another thing to bear in mind is that in Javascript, everything can have evaluate to some kind of value. That could be a "truthy" or "falsey" value, and can also have a "number" value (even if that value is not a number, AKA NaN).
So, if you open up JSFiddle, or Firebug, and mess around with what these values do to functions, you can see that they'll also yield some kind of new return value.
For example:
!function(){} will evaluate to a value of false (since coercing a function object to a boolean yields a value of true).
+function(){} will evaluate to a value of NaN (since coercing a function object to a number yields NaN). The same result can be seen by using -.
!+function(){} yields true (coercing a number of value NaN will yield false and not false yields true.
!+-+-+!function(){} yields true (because !function(){} yields false, +false yields 0, and will continue to throughout all those + and - operators until finally !0 is evaluated to true).
Applying the operators like they're listed in your example will go back and forth between false, -1, 0, 1, true until all of the operators have been evaluated.
Note that I checked these using Firebug. There could possibly be differences between browsers, and perhaps what Firebug shows us on evaluation. The TL;DR is that Javascript does lots of type coercion, and will evaluate expressions differently than declarations.

What does !!~ do? [duplicate]

This question already exists:
What does `!!~` mean in javascript? [duplicate]
Closed 8 years ago.
I was reading this blog post which mentioned using:
!!~
I have no idea what this does? at first I thought it would give an error, but the code below does run:
var _sessions = [
"_SID_1",
"_SID_2",
"_SID_3",
"_SID_4"
];
if(!!~_sessions.indexOf("_SID_5")) {
console.log('found');
} else {
console.log('!found');
}
output:
node test.js
!found

~ is the bitwise not operator. It inverts the bits of its operand. ! is the logical not operator. The bitwise not operator will return 0 when applied to -1, which is what indexOf returns when the value is not found in the array. Since 0 evaluates to false, doubly negating it will simply return false (a boolean value, rather than a numeric one):
var index = _sessions.indexOf("_SID_5");
console.log(~index); // 0
console.log(!~index); // true
console.log(!!~index); //false
The bitwise not operator will return a value less than 0 for any other possible value returned by indexOf. Since any other value will evaluate to true, it's just a shorthand method (kind of... they are both the same number of characters!) of checking whether an element exists in an array, rather than explicitly comparing with -1:
if (_sessions.indexOf("_SID_5") > -1) {
// This would work the same way
}
Update
With regards to the performance of this, it appears (in Chrome at least) to be marginally slower than the more common comparison with -1 (which itself is marginally slower than a comparison with 0).
Here's a test case and here's the results:
Update 2
In fact, the code in your question can be shortened, which may have been what the author was attempting to do. You can simply remove the !!, since the ~ will always result in 0 or below (and 0 is the only value that will evaluate to false):
if (~_sessions.indexOf("_SID_5")) {
// This works too
}
However, in a slightly different situation it could make sense to add in the ! operators. If you were to store the result of the bitwise operator in a variable, it would be a numeric value. By applying the logical not operator, you get a boolean value (and applying it again ensures you get the correct boolean value). If for some reason you require a boolean value over a numeric one, it makes a little bit more sense (but you can still just use the normal comparison with -1 or 0):
var inArray = !!~_sessions.indexOf("_SID_5");
console.log(typeof inArray); // boolean

Donald Knuth: "[...] premature optimization is the root of all evil"
For the sake of readability: please use
.indexOf !== -1

This explains it well:
The tilde operator in Javascript
Mixing the two NOT operators together can produce some interesting results:
!~(-2) = false
!~(-1) = true
!~(0) = false
!~(1) = false
!~(2) = false
So this just checks if the value equals -1 or not, and indexOf returns -1 if it does not find a match