If you read the comments at the jQuery inArray page here, there's an interesting declaration:
!!~jQuery.inArray(elm, arr)
Now, I believe a double-exclamation point will convert the result to type boolean, with the value of true. What I don't understand is what is the use of the tilde (~) operator in all of this?
var arr = ["one", "two", "three"];
if (jQuery.inArray("one", arr) > -1) { alert("Found"); }
Refactoring the if statement:
if (!!~jQuery.inArray("one", arr)) { alert("Found"); }
Breakdown:
jQuery.inArray("one", arr) // 0
~jQuery.inArray("one", arr) // -1 (why?)
!~jQuery.inArray("one", arr) // false
!!~jQuery.inArray("one", arr) // true
I also noticed that if I put the tilde in front, the result is -2.
~!!~jQuery.inArray("one", arr) // -2
I don't understand the purpose of the tilde here. Can someone please explain it or point me towards a resource?
There's a specfic reason you'll sometimes see ~ applied in front of $.inArray.
Basically,
~$.inArray("foo", bar)
is a shorter way to do
$.inArray("foo", bar) !== -1
$.inArray returns the index of the item in the array if the first argument is found, and it returns -1 if its not found. This means that if you're looking for a boolean of "is this value in the array?", you can't do a boolean comparison, since -1 is a truthy value, and when $.inArray returns 0 (a falsy value), it means its actually found in the first element of the array.
Applying the ~ bitwise operator causes -1 to become 0, and causes 0 to become `-1. Thus, not finding the value in the array and applying the bitwise NOT results in a falsy value (0), and all other values will return non-0 numbers, and will represent a truthy result.
if (~$.inArray("foo", ["foo",2,3])) {
// Will run
}
And it'll work as intended.
!!~expr evaluates to false when expr is -1 otherwise true.
It is same as expr != -1, only broken*
It works because JavaScript bitwise operations convert the operands to 32-bit signed integers in two's complement format. Thus !!~-1 is evaluated as follows:
-1 = 1111 1111 1111 1111 1111 1111 1111 1111b // two's complement representation of -1
~-1 = 0000 0000 0000 0000 0000 0000 0000 0000b // ~ is bitwise not (invert all bits)
!0 = true // ! is logical not (true for falsy)
!true = false // duh
A value other than -1 will have at least one bit set to zero; inverting it will create a truthy value; applying ! operator twice to a truthy value returns boolean true.
When used with .indexOf() and we only want to check if result is -1 or not:
!!~"abc".indexOf("d") // indexOf() returns -1, the expression evaluates to false
!!~"abc".indexOf("a") // indexOf() returns 0, the expression evaluates to true
!!~"abc".indexOf("b") // indexOf() returns 1, the expression evaluates to true
* !!~8589934591 evaluates to false so this abomination cannot be reliably used to test for -1.
The tilde operator isn't actually part of jQuery at all - it's a bitwise NOT operator in JavaScript itself.
See The Great Mystery of the Tilde(~).
You are getting strange numbers in your experiments because you are performing a bitwise logical operation on an integer (which, for all I know, may be stored as two's complement or something like that...)
Two's complement explains how to represent a number in binary. I think I was right.
~foo.indexOf(bar) is a common shorthand to represent foo.contains(bar) because the contains function doesn't exist.
Typically the cast to boolean is unnecessary due to JavaScript's concept of "falsy" values. In this case it's used to force the output of the function to be true or false.
jQuery.inArray() returns -1 for "not found", whose complement (~) is 0. Thus, ~jQuery.inArray() returns a falsy value (0) for "not found", and a truthy value (a negative integer) for "found". !! will then formalise the falsy/truthy into real boolean false/true. So, !!~jQuery.inArray() will give true for "found" and false for "not found".
The ~ for all 4 bytes int is equal to this formula -(N+1)
SO
~0 = -(0+1) // -1
~35 = -(35+1) // -36
~-35 = -(-35+1) //34
Tilde is bitwise NOT - it inverts each bit of the value. As a general rule of thumb, if you use ~ on a number, its sign will be inverted, then 1 will be subtracted.
Thus, when you do ~0, you get -1 (0 inverted is -0, subtract 1 is -1).
It's essentially an elaborate, super-micro-optimised way of getting a value that's always Boolean.
The ~ operator is the bitwise complement operator. The integer result from inArray() is either -1, when the element is not found, or some non-negative integer. The bitwise complement of -1 (represented in binary as all 1 bits) is zero. The bitwise-complement of any non-negative integer is always non-zero.
Thus, !!~i will be true when integer "i" is a non-negative integer, and false when "i" is exactly -1.
Note that ~ always coerces its operand to integer; that is, it forces non-integer floating point values to integer, as well as non-numeric values.
You're right: This code will return false when the indexOf call returns -1; otherwise true.
As you say, it would be much more sensible to use something like
return this.modifiedPaths.indexOf(path) !== -1;
The ~ operator is the bitwise NOT operator. What this means is that it takes a number in binary form and turns all zeroes into ones and ones into zeroes.
For instance, the number 0 in binary is 0000000, while -1 is 11111111. Likewise, 1 is 00000001 in binary, while -2 is 11111110.
My guess is that it is there because it's a few characters shorter (which library authors are always after). It also uses operations that only take a few machine cycles when compiled into the native code (as opposed to the comparison to a number.)
I agree with another answer that it's an overkill but perhaps might make sense in a tight loop (requires performance gain estimation, though, otherwise may turn out to be premature optimization.)
I assume, since it is a bitwise operation, it is the fastest (computationally cheap) way to check whether path appears in modifiedPaths.
As (~(-1)) === 0, so:
!!(~(-1)) === Boolean(~(-1)) === Boolean(0) === false
My question is how it became -6 when do the negation ?
Edit : Let's say like this, if we need to represent 6 on 2's complement it should be 110.But on the 2nd row of above is having '4294967290' (decimal) value when It has been converted by using cal Here
So how can it be a -6 then ?
The negation as you call it is a strict bit inversion, but decimal values in JavaScript are handled as twos-complement.
So you'd basically need '~5 + 1' to get to the equivalent representation as '-5'.
In two's-complement representation, positive numbers are simply represented as themselves, and negative numbers are represented by the two's complement of their absolute value
See http://en.wikipedia.org/wiki/Two's_complement for more details.
I'm reading this javascript function:
if (~['bacon', 'burger'].indexOf(type)) {
this.res.writeHead(200, { 'Content-Type': 'text/plain' });
this.res.end('Serving ' + type + ' sandwich!\n');
}
But I'm not sure what means ~ some one know when I use it or what meaning?
~ is the bitwise NOT operator. It toggles every bit of a number.
0 becomes -1.
-1 becomes 0.
No other numbers become zero.
That means that
if (~['bacon', 'burger'].indexOf(type)) {
is a confusing way of writing
if (['bacon', 'burger'].indexOf(type) == -1) {
indexOf returns -1 when it doesn't find the string.
~ is a Bitwise NOT operator...
Read more
In this instance, the ~ allows that code to turn the return value of .indexOf() — which is a number indicating the position of the searched-for value in the array — into a boolean. In other words, it takes the "where is the value" result and turns it into a "is the value in the list" result.
How? Well, .indexOf() returns -1 when the value is not found, and a number greater than or equal to zero if it is. The ~ operator converts its numeric argument to a 32-bit integer and then inverts every bit. That process happens to turn -1 to 0, and any positive integer to some negative non-zero value, and 0 to -1. When such results are subsequently examined as boolean values, the original -1 will be false (because 0 is "falsy") while the integers greater than or equal to zero will be true (because they're all converted to some non-zero value).
Bitwise NOT (~ a) Inverts the bits of its operand.
EXAMPLE
9 (base 10) = 00000000000000000000000000001001 (base 2)
--------------------------------
~9 (base 10) = 11111111111111111111111111110110 (base 2) = -10 (base
10)
https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Operators/Bitwise_Operators
A questions that stumped me on this JavaScript test was that ~null evaluates to -1.
Why does ~null evaluate to -1?
That's because ~ is a numeric operator, so it casts null to 0 first:
> ~0
-1
It would be equivalent to this expression:
~(+null)
Likewise:
> ~[]
-1
> ~{}
-1
First of all, ~ is a bitwise NOT operator. That means it flips all the bits in the number representation. 0010 1010 becomes 1101 0101.
As a consequence of computers using 2's complement for storing numbers, this equality holds:
~number == -number - 1
As can be shown from my previous example:
0010 1010 (this represents number 42)
1101 0101 (this represents number -43)
Now, because ~ is an operator that operates on numbers, its argument gets cast to a number first. Since null gets cast to a 0, you get -1 as a result (according the above equation).
This question already has answers here:
Using bitwise OR 0 to floor a number
(7 answers)
Closed 8 years ago.
Anyone able to explain what "|" and the value after does? I know the output for 0 creates sets of 13, the numbers, 3, 2, 1, 0. But what about | 1, or | 2.
var i = 52;
while(i--) {
alert(i/13 | 0);
}
It is the bitwise OR operator. There is both an explanation and an example over at MDC. Since doing bitwise OR with one operand being 0 produces the value of the other operand, in this case it does exactly nothing rounds the result of the division down.
If it were written | 1 what it would do is always print odd numbers (because it would set the 1-bit to on); specifically, it would cause even numbers to be incremented by 1 while leaving odd numbers untouched.
Update: As the commenters correctly state, the bitwise operator causes both operands to be treated as integers, therefore removing any fraction of the division result. I stand corrected.
This is a clever way of accomplishing the same effect as:
Math.floor(i/13);
JavaScript developers seem to be good at these kinds of things :)
In JavaScript, all numbers are floating point. There is no integer type. So even when you do:
var i = 1;
i is really the floating point number 1.0. So if you just did i/13, you'd end up with a fractional portion of it, and the output would be 3.846... for example.
When using the bitwise or operator in JavaScript, the runtime has to convert the operands to 32 bit integers before it can proceed. Doing this chops away the fractional part, leaving you with just an integer left behind. Bitwise or of zero is a no op (well, a no op in a language that has true integers) but has the side effect of flooring in JavaScript.
It's a bitwise operator. Specifically the OR Bitwise Operator.
What it basically does is use your var as an array of bits and each corresponding bit is with eachother. The result is 1 if any of them is 1. And 0 if both are 0.
Example:
24 = 11000
10 = 1010
The two aren't of equal length so we pad with 0's
24 = 11000
10 = 01010
26 = 11010
24 | 10 = 26
Best way to learn this is to readup on it.
That is the bitwise OR. In evaluating the expression, the LHS is truncated to an integer and returned, so | is effecively the same as Math.floor().