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I want to toggle a variable between 0 and 1. If it's 0 I want to set it to 1, else if it's 1 I want to set it to 0.
This is such a fundamental operation that I write so often I'd like to investigate the shortest, clearest possible way of doing it. Here's my best so far:
v = (v == 0 ? 1 : 0);
Can you improve on this?
Edit: the question is asking how to write the above statement in the fewest characters while retaining clarity - how is this 'not a real question'? This wasn't intended to be a code-golf exercise, though some interesting answers have come out of people approaching it as golf - it's nice to see golf being used in a constructive and thought-provoking manner.
You can simply use:
v = 1 - v;
This of course assumes that the variable is initialised properly, i.e. that it only has the value 0 or 1.
Another method that is shorter but uses a less common operator:
v ^= 1;
Edit:
To be clear; I never approached this question as code golf, just to find a short way of doing the task without using any obscuring tricks like side effects of operators.
Since 0 is a false value and 1 is a true value.
v = (v ? 0 : 1);
If you are happy to use true and false instead of numbers
v = !v;
or if they must be numbers:
v = +!v; /* Boolean invert v then cast back to a Number */
v = (v + 1) % 2 and if you need to cycle through more values just change 2 for (n + 1). Say you need to cycle 0,1,2 just do v = (v + 1) % 3.
You could write a function for it and use it like:
v = inv(v)
If you don't care about any possibility other than 1:
v = v ? 0 : 1;
In the above case, v will end up being 1 if v is 0, false, undefined or null. Take care using this kind of approach - v will be 0 even if v is "hello world".
Lines like v = 1 - v, or v ^= 1 or v= +!v will all get the job done, but they constitute what I would refer to as hacks. These are not beautiful lines of code, but cheap tricks to have the intended effect. 1 - v does not communicate "toggle the value between 0 and 1". This makes your code less expressive and introduces a place (albeit a small one) where another developer will have to parse your code.
Having instead a function like v = toggle(v) communicates the intent at the quickest glance.
(Honesty and mathematical integrity - given the number of votes on this "answer" - have led me to edit this answer. I held off as long as possible because it was intended as a short quip and not as anything "deep" so putting in any explanation seemed counter to the purpose. However, the comments are making it clear that I should be clear to avoid misunderstanding.)
My original answer:
The wording of this part of the specification:
If it's 0, I want to set it to 1, else set it to 0.
implies that the most accurate solution is:
v = dirac_delta(0,v)
First, the confession: I did get my delta functions confused. The Kronecker delta would have been slightly more appropriate, but not by much as I wanted something that was domain-independent (the Kronecker delta is mainly used just for integers). But I really shouldn't have used delta functions at all, I should have said:
v = characteristic_function({0},v)
Let me clarify. Recall that a function is a triple, (X,Y,f), where X and Y are sets (called the domain and codomain respectively) and f is a rule that assigns an element of Y to each element of X. We often write the triple (X,Y,f) as f: X → Y. Given a subset of X, say A, there is a characteristic function which is a function χA: X → {0,1} (it can also be thought of as a function to a larger codomain such as ℕ or ℝ). This function is defined by the rule:
χA(x) = 1 if x ∈ A and χA(x) = 0 if x ∉ A.
If you like truth tables, it's the truth table for the question "Is the element x of X an element of the subset A?".
So from this definition, it's clear that the characteristic function is what is needed here, with X some big set containing 0 and A = {0}. That's what I should have written.
And so to delta functions. For this, we need to know about integration. Either you already know it, or you don't. If you don't, nothing I can say here will tell you about the intricacies of the theory, but I can give a one sentence summary. A measure on a set X is in essence "that which is needed to make averages work". That is to say that if we have a set X and a measure μ on that set then there is a class of functions X → ℝ, called measurable functions for which the expression ∫X f dμ makes sense and is, in some vague sense, the "average" of f over X.
Given a measure on a set, one can define a "measure" for subsets of that set. This is done by assigning to a subset the integral of its characteristic function (assuming that this is a measurable function). This can be infinite, or undefined (the two are subtly different).
There are lots of measures around, but there are two that are important here. One is the standard measure on the real line, ℝ. For this measure, then ∫ℝ f dμ is pretty much what you get taught in school (is calculus still taught in schools?): sum up little rectangles and take smaller and smaller widths. In this measure, the measure of an interval is its width. The measure of a point is 0.
Another important measure, which works on any set, is called the point measure. It is defined so that the integral of a function is the sum of its values:
∫X f dμ = ∑x ∈X f(x)
This measure assigns to each singleton set the measure 1. This means that a subset has finite measure if and only if it is itself finite. And very few functions have finite integral. If a function has a finite integral, it must be non-zero only on a countable number of points. So the vast majority of functions that you probably know do not have finite integral under this measure.
And now to delta functions. Let's take a very broad definition. We have a measurable space (X,μ) (so that's a set with a measure on it) and an element a ∈ X. We "define" the delta function (depending on a) to be the "function" δa: X → ℝ with the property that δa(x) = 0 if x ≠ a and ∫X δa dμ = 1.
The most important fact about this to get a-hold of is this: The delta function need not be a function. It is not properly defined: I have not said what δa(a) is.
What you do at this point depends on who you are. The world here divides in to two categories. If you are a mathematician, you say the following:
Okay, so the delta function might not be defined. Let's look at its hypothetical properties and see if we can find a proper home for it where it is defined. We can do that, and we end up with distributions. These are not (necessarily) functions, but are things that behave a little like functions, and often we can work with them as if they were functions; but there are certain things that they don't have (such as "values") so we need to be careful.
If you are not a mathematician, you say the following:
Okay, so the delta function might not be properly defined. Who says so? A bunch of mathematicians? Ignore them! What do they know?
Having now offended my audience, I shall continue.
The dirac delta is usually taken to be the delta function of a point (often 0) in the real line with its standard measure. So those who are complaining in the comments about me not knowing my deltas are doing so because they are using this definition. To them, I apologise: although I can wriggle out of that by using the Mathematician's defence (as popularised by Humpty Dumpty: simply redefine everything so that it is correct), it is bad form to use a standard term to mean something different.
But there is a delta function which does do what I want it to do and it is that which I need here. If I take a point measure on a set X then there is a genuine function δa : X → ℝ which satisfies the criteria for a delta function. This is because we are looking for a function X → ℝ which is zero except at a and such that the sum of all of its values is 1. Such a function is simple: the only missing piece of information is its value at a, and to get the sum to be 1 we just assign it the value 1. This is none other than the characteristic function on {a}. Then:
∫X δa dμ = ∑x ∈ X δa(x) = δa(a) = 1.
So in this case, for a singleton set, the characteristic function and the delta function agree.
In conclusion, there are three families of "functions" here:
The characteristic functions of singleton sets,
The delta functions,
The Kronecker delta functions.
The second of these is the most general as any of the others is an example of it when using the point measure. But the first and third have the advantage that they are always genuine functions. The third is actually a special case of the first, for a particular family of domains (integers, or some subset thereof).
So, finally, when I originally wrote the answer I wasn't thinking properly (I wouldn't go so far as to say that I was confused, as I hope I've just demonstrated I do know what I'm talking about when I actually think first, I just didn't think very much). The usual meaning of the dirac delta is not what is wanted here, but one of the points of my answer was that the input domain was not defined so the Kronecker delta would also not have been right. Thus the best mathematical answer (which I was aiming for) would have been the characteristic function.
I hope that that is all clear; and I also hope that I never have to write a mathematical piece again using HTML entities instead of TeX macros!
in general whenever you need to toggle between two values , you can just subtract the current value from the sum of the two toggle values :
0,1 -> v = 1 - v
1,2 -> v = 3 - v
4,5 -> v = 9 - v
You could do
v = Math.abs(--v);
The decrement sets the value to 0 or -1, and then the Math.abs converts -1 to +1.
If it must be the integer 1 or 0, then the way you're doing it is fine, though parentheses aren't needed. If these a are to be used as booleans, then you can just do:
v = !v;
v = v == 0 ? 1 : 0;
Is enough !
List of solutions
There are three solutions I would like to propose. All of them convert any value to 0 (if 1, true etc.) or 1 (if 0, false, null etc.):
v = 1*!v
v = +!v
v = ~~!v
and one additional, already mentioned, but clever and fast (although works only for 0s and 1s):
v = 1-v
Solution 1
You can use the following solution:
v = 1*!v
This will first convert the integer to the opposite boolean (0 to True and any other value to False), then will treat it as integer when multiplying by 1. As a result 0 will be converted to 1 and any other value to 0.
As a proof see this jsfiddle and provide any values you wish to test: jsfiddle.net/rH3g5/
The results are as follows:
-123 will convert to integer 0,
-10 will convert to integer 0,
-1 will convert to integer 0,
0 will convert to integer 1,
1 will convert to integer 0,
2 will convert to integer 0,
60 will convert to integer 0,
Solution 2
As mblase75 noted, jAndy had some other solution that works as mine:
v = +!v
It also first makes boolean from the original value, but uses + instead of 1* to convert it into integer. The result is exactly the same, but the notation is shorter.
Solution 3
The another approach is to use ~~ operator:
v = ~~!v
It is pretty uncommon and always converts to integer from boolean.
To sum up another answer, a comment and my own opinion, I suggest combining two things:
Use a function for the toggle
Inside this function use a more readable implementation
Here is the function which you could place in a library or maybe wrap it in a Plugin for another Javascript Framework.
function inv(i) {
if (i == 0) {
return 1
} else {
return 0;
}
}
And the usage is simply:
v = inv(v);
The advantages are:
No code Duplication
If you or anybody read this again in the future, you will understand your code in a minimum of time.
This is missing:
v = [1, 0][v];
It works as round robin as well:
v = [2, 0, 1][v]; // 0 2 1 0 ...
v = [1, 2, 0][v]; // 0 1 2 0 ...
v = [1, 2, 3, 4, 5, 0][v]; // 0 1 2 3 4 5 ...
v = [5, 0, 1, 2, 3, 4][v]; // 0 5 4 3 2 1 0 ...
Or
v = {0: 1, 1: 0}[v];
The charme of the last solution, it works with all other values as well.
v = {777: 'seven', 'seven': 777}[v];
For a very special case, like to get a (changing) value and undefined, this pattern may be helpful:
v = { undefined: someValue }[v]; // undefined someValue undefined someValue undefined ...
I don't know why you want to build your own booleans? I like the funky syntaxes, but why not write understandable code?
This is not the shortest/fastest, but the most clearest (and readable for everyone) is using the well-known if/else state:
if (v === 0)
{
v = 1;
}
else
{
v = 0;
}
If you want to be really clear, you should use booleans instead of numbers for this. They are fast enough for most cases. With booleans, you could just use this syntax, which will win in shortness:
v = !v;
Another form of your original solution:
v = Number(v == 0);
EDIT: Thanks TehShrike and Guffa for pointing out the error in my original solution.
I would make it more explicit.
What does v mean?
For example when v is some state. Create an object Status. In DDD an value object.
Implement the logic in this value object. Then you can write your code in a more functional way which is more readable. Switching status can be done by creating a new Status based on the current status. Your if statement / logic is then encapsulated in your object, which you can unittest. An valueObject is always immutable, so it has no identity. So for changing it's value you have to create a new one.
Example:
public class Status
{
private readonly int _actualValue;
public Status(int value)
{
_actualValue = value;
}
public Status(Status status)
{
_actualValue = status._actualValue == 0 ? 1 : 0;
}
//some equals method to compare two Status objects
}
var status = new Status(0);
Status = new Status(status);
Since this is JavaScript, we can use the unary + to convert to int:
v = +!v;
This will logical NOT the value of v (giving true if v == 0 or false if v == 1). Then we convert the returned boolean value into its corresponding integer representation.
Another way to do it:
v = ~(v|-v) >>> 31;
One more:
v=++v%2
(in C it would be simple ++v%=2)
ps. Yeah, I know it's double assignment, but this is just raw rewrite of C's method (which doesn't work as is, cause JS pre-increment operator dosen't return lvalue.
If you're guaranteed your input is either a 1 or a 0, then you could use:
v = 2+~v;
Just for kicks: v = Math.pow(v-1,v) also toggles between 1 and 0.
define an array{1,0}, set v to v[v], therefore v with a value of 0 becomes 1, and vica versa.
Another creative way of doing it, with v being equal to any value, will always return 0 or 1
v = !!v^1;
If possible values for v are only 0 and 1, then for any integer x, the expression:
v = Math.pow((Math.pow(x, v) - x), v);
will toggle the value.
I know this is an ugly solution and the OP was not looking for this...but I was thinking about just another solution when I was in the loo :P
Untested, but if you're after a boolean I think var v = !v will work.
Reference: http://www.jackfranklin.co.uk/blog/2011/05/a-better-way-to-reverse-variables
v=!v;
will work for v=0 and v=1; and toggle the state;
If there are just two values, as in this case(0, 1), i believe it's wasteful to use int. Rather go for boolean and work in bits. I know I'm assuming but in case of toggle between two states boolean seems to be ideal choice.
v = Number(!v)
It will type cast the Inverted Boolean value to Number, which is the desired output.
Well, As we know that in javascript only that Boolean comparison will also give you expected result.
i.e. v = v == 0 is enough for that.
Below is the code for that:
var v = 0;
alert("if v is 0 output: " + (v == 0));
setTimeout(function() {
v = 1;
alert("if v is 1 Output: " + (v == 0));
}, 1000);
JSFiddle: https://jsfiddle.net/vikash2402/83zf2zz0/
Hoping this will help you :)
I'm studying for an exam on JavaScript at the moment. I've also got a little knowledge of C and Perl so I'm familiar with prefix and postfix operators in all three languages.
I did an online practice exam for it and one mistake I made was in evaluating the following code:
var x = 10;
x += x--;
Now, I thought it would evaluate to 19 because it would be 10 + 10, then subtract 1 to make 9. But the feedback I got was that it was wrong and it actually evaluates to 20. I thought that sounded a bit suspicious so I tested it out in an HTML document, and it came out with 20 again. I then tried the equivalents in C and Perl and both evaluated to 19.
Can anyone explain to me why JavaScript evaluates the answer as 20 when other languages evaluate it to 19? The answer I got from the test wasn't too clear to me:
The increment ++
and decrement -- operators can be placed
either before or after an operand. If the increment or decrement
operator is placed before the operand, the operation occurs immediately.
If the increment or decrement operator is placed after the operand, the
change in the operand's value is not apparent until the next time the
operand is accessed in the program. Thus the expression x += x-- is equivalent to x = x +
10 which evaluates to 20.
Expanding the statement
x += x--;
to the more verbose JS code
x = x + (function(){ var tmp = x; x = x - 1; return tmp; })();
the result makes perfect sense, as it will evaluate to
x = 10 + (function(){ var tmp = 10; x = 10 - 1; return tmp; })();
which is 20. Keep in mind that JS evaluates expressions left-to-right, including compound assignments, ie the value of x is cached before executing x--.
You could also think of it this way: Assuming left-to-right evaluation order, JS parses the assignment as
x := x + x--
whereas Perl will use
x := x-- + x
I don't see any convincing arguments for or against either choice, so it's just bad luck that different languages behave differently.
In C/C++, every variable can only be changed once in every statement (I think the exact terminology is: only once between two code points, but I'm not sure).
If you write
x += x--;
you are changing the value of x twice:
you are decrementing x using the postfix -- operator
you are setting the value of x using the assignment
Although you can write this and the compiler won't complain about it (not sure, you may want to check the different warning levels), the outcome is undefined and can be different in every compiler.
Basically, the value of x is decemented after assignment. This example might make it clearer (run in Firebug console)
var x = y =10;
x += y--;
console.log(x , y); // outputs 20 9
In C, the line
x += x--;
is undefined behaviour. It seems like your particular compiler is treating it like:
oldx = x--;
x = x + oldx
However, the ECMAScript specification does specify op= - and it gets the value of the left-hand-side before evaluating the right-hand-side.
So it would be equivalent to:
oldx = x--;
x = oldx + oldx