If I know the variable will be object later, I use:
var obj;
but it doesn't really matter if I initialize it as null or undefined:
var obj = null;
or
var obj = undefined;
For strings I personally use:
var str = '';
as later on I can do
str += 'some text';
and if I use null for example I get "nullsome text".
null, undefined, empty string or 0 are all falsy values so to check if they are defined. which is the correct way to initialize variables if my variable will be used as object, dom node, etc.. ?
It's initialized to undefined by default, just use that, you don't have to write it out:
var obj;
If it you want to concatenate to a string, use '', for counters use 0... Just use common sense.
In an ideal scenario, you'd just declare your variables at the top of the scope like so:
var foo, bar, obj, domNode;
All of these variables will be undefined, as in their value is undefined. If you know for a fact one of these variables will be used as a string in a loop, for example, rather thant writing foo = (foo === undefined ? '' : foo) + 'add this';, just write:
var foo = '', bar, obj, domNode;
If you're going to use the obj var as an object literal, just assign it {my: 'objec', vals: 'here'} when the time comes, or initialize it to {}, and add properties obj.as = 'you'; obj[foo] = 'go';. If you're going to assign a reference to a DOM element, keep it undefined, as any assignment will just generate pointless overhead.
Same goes for any type of reference (function objects, arrays, objects...). Just postpone the assignment, because it'll just overwrite any previous assignment.
As for numbers, undefined + 123 will get evaluate to NaN, so in that case, initializing to 0 makes sense. However, in case of loop counters:
var foo, i, obj, arr = [1,2,3];
for (i=0;i<arr.length;i++)
{
console.log(arr[i]);
}
is equivalent to:
var foo, i= 0, arr = [1,2,3];
for (;i<arr.length;i++)
{
console.log(arr[i]);
}
Only the latter almost looks wantonly complicated. Just use your common sense and you'll be all right.
Just know that variable declarations are hoisted to the top of the scope, but assignments aren't:
console.log(foo);//undefined
var foo = 123;
console.log(foo);//123
because it's translated into:
var foo;
console.log(foo);
foo = 123;
console.log(foo);
So why not write your code as it'll be interpreted by the engine?
Related
So in my js code I have some global variable that changes its value several times, for example
var x = 0;
...
x = 10;
...
x = 5;
Is there any possibility to get "history" of x without saving its value in other variables? Like, is there some function to detect that at some point of time x was equal to 10?
No, once a value is assigned to a variable, that variable's previous value is overwritten. It isn't retained anywhere. (If it were, it would be a nightmare for memory management.)
You could make an object property that retained a history if you wanted, by using a setter function; rough example:
var obj = {
_fooValue: undefined,
fooHistory: [],
set foo(value) {
this.fooHistory.push(this._fooValue);
this._fooValue = value;
},
get foo() {
return this._fooValue;
}
};
obj.foo = 0;
obj.foo = 5;
obj.foo = 42;
console.log(obj.fooHistory);
In that example, the history doesn't contain the current value, just the previous ones, and it stores the current value in another object property which means code could bypass the setter. There are lots of tweaks you could do. If you thought it was important, you could lock it down more:
var obj = (function() {
// These two vars are entirely private to the object
var fooHistory = [];
var fooValue;
// The object we'll assign to `obj`
return {
set foo(value) {
fooHistory.push(fooValue);
fooValue = value;
},
get foo() {
return fooValue;
},
get fooHistory() {
// Being really defensive and returning
// a copy
return fooHistory.slice(0);
}
}
})();
obj.foo = 0;
obj.foo = 5;
obj.foo = 42;
console.log(obj.fooHistory);
You can use variable like array and unshift next value to this array. And to use it take first element:
var x = [];
...
x.unshift(10);
...
x.unshift(5);
var currentX = x[0];
var allValues = x;
Yes there is. Using the Time Travelling debugging in Microsoft Edge browser. Check this out.
I'm not JS specialist, but as common idea for any OOP language, I would suggest to create special class for x (inherited from Integer in your example), which has overriden setter and some history array list. So, when you set a new value it stored in your history.
You need change only the variable type, not the code, which works with that. I also don't think there is some standard solution for this in any language. Probably, some dynamic introspectors, but those would be even more complex than my idea.
This question already has answers here:
Is JavaScript a pass-by-reference or pass-by-value language?
(33 answers)
Closed 3 years ago.
Does JavaScript pass by references or pass by values?
Here is an example from JavaScript: The Good Parts. I am very confused about the my parameter for the rectangle function. It is actually undefined, and redefined inside the function. There are no original reference. If I remove it from the function parameter, the inside area function is not able to access it.
Is it a closure? But no function is returned.
var shape = function (config) {
var that = {};
that.name = config.name || "";
that.area = function () {
return 0;
};
return that;
};
var rectangle = function (config, my) {
my = my || {};
my.l = config.length || 1;
my.w = config.width || 1;
var that = shape(config);
that.area = function () {
return my.l * my.w;
};
return that;
};
myShape = shape({
name: "Unhnown"
});
myRec = rectangle({
name: "Rectangle",
length: 4,
width: 6
});
console.log(myShape.name + " area is " + myShape.area() + " " + myRec.name + " area is " + myRec.area());
Primitives are passed by value, and Objects are passed by "copy of a reference".
Specifically, when you pass an object (or array) you are (invisibly) passing a reference to that object, and it is possible to modify the contents of that object, but if you attempt to overwrite the reference it will not affect the copy of the reference held by the caller - i.e. the reference itself is passed by value:
function replace(ref) {
ref = {}; // this code does _not_ affect the object passed
}
function update(ref) {
ref.key = 'newvalue'; // this code _does_ affect the _contents_ of the object
}
var a = { key: 'value' };
replace(a); // a still has its original value - it's unmodfied
update(a); // the _contents_ of 'a' are changed
Think of it like this:
Whenever you create an object in ECMAscript, this object is formed in a mystique ECMAscript universal place where no man will ever be able to get. All you get back is a reference to that object in this mystique place.
var obj = { };
Even obj is only a reference to the object (which is located in that special wonderful place) and hence, you can only pass this reference around. Effectively, any piece of code which accesses obj will modify the object which is far, far away.
My two cents.... It's irrelevant whether JavaScript passes parameters by reference or value. What really matters is assignment vs. mutation.
I wrote a longer, more detailed explanation in this link.
When you pass anything (whether that be an object or a primitive), all JavaScript does is assign a new variable while inside the function... just like using the equal sign (=).
How that parameter behaves inside the function is exactly the same as it would behave if you just assigned a new variable using the equal sign... Take these simple examples.
var myString = 'Test string 1';
// Assignment - A link to the same place as myString
var sameString = myString;
// If I change sameString, it will not modify myString,
// it just re-assigns it to a whole new string
sameString = 'New string';
console.log(myString); // Logs 'Test string 1';
console.log(sameString); // Logs 'New string';
If I were to pass myString as a parameter to a function, it behaves as if I simply assigned it to a new variable. Now, let's do the same thing, but with a function instead of a simple assignment
function myFunc(sameString) {
// Reassignment... Again, it will not modify myString
sameString = 'New string';
}
var myString = 'Test string 1';
// This behaves the same as if we said sameString = myString
myFunc(myString);
console.log(myString); // Again, logs 'Test string 1';
The only reason that you can modify objects when you pass them to a function is because you are not reassigning... Instead, objects can be changed or mutated.... Again, it works the same way.
var myObject = { name: 'Joe'; }
// Assignment - We simply link to the same object
var sameObject = myObject;
// This time, we can mutate it. So a change to myObject affects sameObject and visa versa
myObject.name = 'Jack';
console.log(sameObject.name); // Logs 'Jack'
sameObject.name = 'Jill';
console.log(myObject.name); // Logs 'Jill'
// If we re-assign it, the link is lost
sameObject = { name: 'Howard' };
console.log(myObject.name); // Logs 'Jill'
If I were to pass myObject as a parameter to a function, it behaves as if I simply assigned it to a new variable. Again, the same thing with the exact same behavior but with a function.
function myFunc(sameObject) {
// We mutate the object, so the myObject gets the change too... just like before.
sameObject.name = 'Jill';
// But, if we re-assign it, the link is lost
sameObject = {
name: 'Howard'
};
}
var myObject = {
name: 'Joe'
};
// This behaves the same as if we said sameObject = myObject;
myFunc(myObject);
console.log(myObject.name); // Logs 'Jill'
Every time you pass a variable to a function, you are "assigning" to whatever the name of the parameter is, just like if you used the equal = sign.
Always remember that the equals sign = means assignment.
And passing a parameter to a function also means assignment.
They are the same and the two variables are connected in exactly the same way.
The only time that modifying a variable affects a different variable is when the underlying object is mutated.
There is no point in making a distinction between objects and primitives, because it works the same exact way as if you didn't have a function and just used the equal sign to assign to a new variable.
Function arguments are passed either by-value or by-sharing, but never ever by reference in JavaScript!
Call-by-Value
Primitive types are passed by-value:
var num = 123, str = "foo";
function f(num, str) {
num += 1;
str += "bar";
console.log("inside of f:", num, str);
}
f(num, str);
console.log("outside of f:", num, str);
Reassignments inside a function scope are not visible in the surrounding scope.
This also applies to Strings, which are a composite data type and yet immutable:
var str = "foo";
function f(str) {
str[0] = "b"; // doesn't work, because strings are immutable
console.log("inside of f:", str);
}
f(str);
console.log("outside of f:", str);
Call-by-Sharing
Objects, that is to say all types that are not primitives, are passed by-sharing. A variable that holds a reference to an object actually holds merely a copy of this reference. If JavaScript would pursue a call-by-reference evaluation strategy, the variable would hold the original reference. This is the crucial difference between by-sharing and by-reference.
What are the practical consequences of this distinction?
var o = {x: "foo"}, p = {y: 123};
function f(o, p) {
o.x = "bar"; // Mutation
p = {x: 456}; // Reassignment
console.log("o inside of f:", o);
console.log("p inside of f:", p);
}
f(o, p);
console.log("o outside of f:", o);
console.log("p outside of f:", p);
Mutating means to modify certain properties of an existing Object. The reference copy that a variable is bound to and that refers to this object remains the same. Mutations are thus visible in the caller's scope.
Reassigning means to replace the reference copy bound to a variable. Since it is only a copy, other variables holding a copy of the same reference remain unaffected. Reassignments are thus not visible in the caller's scope like they would be with a call-by-reference evaluation strategy.
Further information on evaluation strategies in ECMAScript.
As with C, ultimately, everything is passed by value. Unlike C, you can't actually back up and pass the location of a variable, because it doesn't have pointers, just references.
And the references it has are all to objects, not variables. There are several ways of achieving the same result, but they have to be done by hand, not just adding a keyword at either the call or declaration site.
JavaScript is pass by value.
For primitives, primitive's value is passed. For Objects, Object's reference "value" is passed.
Example with Object:
var f1 = function(inputObject){
inputObject.a = 2;
}
var f2 = function(){
var inputObject = {"a": 1};
f1(inputObject);
console.log(inputObject.a);
}
Calling f2 results in printing out "a" value as 2 instead of 1, as the reference is passed and the "a" value in reference is updated.
Example with primitive:
var f1 = function(a){
a = 2;
}
var f2 = function(){
var a = 1;
f1(a);
console.log(a);
}
Calling f2 results in printing out "a" value as 1.
In the interest of creating a simple example that uses const...
const myRef = { foo: 'bar' };
const myVal = true;
function passes(r, v) {
r.foo = 'baz';
v = false;
}
passes(myRef, myVal);
console.log(myRef, myVal); // Object {foo: "baz"} true
In practical terms, Alnitak is correct and makes it easy to understand, but ultimately in JavaScript, everything is passed by value.
What is the "value" of an object? It is the object reference.
When you pass in an object, you get a copy of this value (hence the 'copy of a reference' that Alnitak described). If you change this value, you do not change the original object; you are changing your copy of that reference.
"Global" JavaScript variables are members of the window object. You could access the reference as a member of the window object.
var v = "initialized";
function byref(ref) {
window[ref] = "changed by ref";
}
byref((function(){for(r in window){if(window[r]===v){return(r);}}})());
// It could also be called like... byref('v');
console.log(v); // outputs changed by ref
Note, the above example will not work for variables declared within a function.
Without purisms, I think that the best way to emulate scalar argument by reference in JavaScript is using object, like previous an answer tells.
However, I do a little bit different:
I've made the object assignment inside function call, so one can see the reference parameters near the function call. It increases the source readability.
In function declaration, I put the properties like a comment, for the very same reason: readability.
var r;
funcWithRefScalars(r = {amount:200, message:null} );
console.log(r.amount + " - " + r.message);
function funcWithRefScalars(o) { // o(amount, message)
o.amount *= 1.2;
o.message = "20% increase";
}
In the above example, null indicates clearly an output reference parameter.
The exit:
240 - 20% Increase
On the client-side, console.log should be replaced by alert.
★ ★ ★
Another method that can be even more readable:
var amount, message;
funcWithRefScalars(amount = [200], message = [null] );
console.log(amount[0] + " - " + message[0]);
function funcWithRefScalars(amount, message) { // o(amount, message)
amount[0] *= 1.2;
message[0] = "20% increase";
}
Here you don't even need to create new dummy names, like r above.
I can't see pass-by-reference in the examples where people try to demonstrate such. I only see pass-by-value.
In the case of variables that hold a reference to an object, the reference is the value of those variables, and therefore the reference is passed, which is then pass-by-value.
In a statement like this,
var a = {
b: "foo",
c: "bar"
};
the value of the 'a' is not the Object, but the (so far only) reference to it. In other words, the object is not in the variable a - a reference to it is. I think this is something that seems difficult for programmers who are mainly only familiar with JavaScript. But it is easy for people who know also e.g. Java, C#, and C.
Objects are always pass by reference and primitives by value. Just keep that parameter at the same address for objects.
Here's some code to illustrate what I mean (try it in a JavaScript sandbox such as https://js.do/).
Unfortunately you can't only retain the address of the parameter; you retain all the original member values as well.
a = { key: 'bevmo' };
testRetain(a);
document.write(' after function ');
document.write(a.key);
function testRetain (b)
{
document.write(' arg0 is ');
document.write(arguments[0].key);
b.key = 'passed by reference';
var retain = b; // Retaining the original address of the parameter
// Address of left set to address of right, changes address of parameter
b = {key: 'vons'}; // Right is a new object with a new address
document.write(' arg0 is ');
document.write(arguments[0].key);
// Now retrieve the original address of the parameter for pass by reference
b = retain;
document.write(' arg0 is ');
document.write(arguments[0].key);
}
Result:
arg0 is bevmo arg0 is vons arg0 is passed by reference after function passed by reference
Primitives are passed by value. But in case you only need to read the value of a primitve (and value is not known at the time when function is called) you can pass function which retrieves the value at the moment you need it.
function test(value) {
console.log('retrieve value');
console.log(value());
}
// call the function like this
var value = 1;
test(() => value);
I understand that the assignment operator is right associative.
So for example x = y = z = 2 is equivalent to (x = (y = (z = 2)))
That being the case, I tried the following:
foo.x = foo = {a:1}
I expected that the object foo would be created with value {a:1} and then the property x will be created on foo which will just be a reference to the foo object.
(This is actually what happens if I was to separate the multiple assignment statement into two separate statements foo = {a:1};foo.x = foo; )
The outcome was actually:
ReferenceError: foo is not defined(…)
So then I tried the following:
var foo = {};
foo.x = foo = {a:1};
Now I don't get the exception anymore but foo.x is undefined!
Why is the assignment not working as I expected?
Disclaimer: The 'duplicate' question seems to be very different to the one that I'm asking, as the issue there is that the variables that were created in the assignment were global, as apposed to variables created with the var keyword. That's not the issue here.
There's an important difference between associativity and order of evaluation.
In JavaScript, even though the assignment operator groups right to left, the operands are evaluated left to right before the actual assignments are performed (which do occur right to left). Consider this example:
var a = {};
var b = {};
var c = a;
c.x = (function() { c = b; return 1; })();
The variable c initially references a, but the right-hand side of the assignment sets c to b. Which property gets assigned, a.x or b.x? The answer is a.x because the left-hand side is evaluated first, when c still references a.
In general, the expression x = y is evaluated as follows:
Evaluate x and remember the result.
Evaluate y and remember the result.
Assign the result from step 2 to the result of step 1 (and return the former as the result of the expression x = y).
What happens with multiple assignments, as in x = (y = z)? Recurse!
Evaluate x and remember the result.
Evaluate y = z and remember the result. To do this:
Evaluate y and remember the result.
Evaluate z and remember the result.
Assign the result from step 2.2 to the result of step 2.1 (and return the former as the result of the expression y = z).
Assign the result from step 2 to the result of step 1 (and return the former as the result of the expression x = (y = z)).
Now let's look at your example, slightly edited:
var foo = {};
var bar = foo; // save a reference to foo
foo.x = (foo = {a:1}); // add parentheses for clarity
foo.x is evaluated before foo gets assigned to {a:1}, so the x property gets added to the original {} object (which you can verify by examining bar).
Edited the answer to make it simple
First of all you have to understand the differnce between Reference- and Value- Type.
var foo = {};
foo variable holds a Reference to an object in memory, lets say A
Now, there are two arts of accessors: Variable Accessor and Property Accessor.
So foo.x = foo = {a:1} can be understood as
[foo_VARIABLE_ACCESSOR][x_PROPERTY_ACCESSOR] = [foo_VARIABLE_ACCESSOR] = {a:1}
!!! Accessor chain is evaluated first to get the last accessor, which is then evaluated associative.
A['x'] = foo = {a:1}
Property Accessor are seperated into setters and getters
var foo = { bar: {} };
foo.bar.x = foo = {a:1}
Here where have decared two nested objects foo and bar. In memory we have then two object A and B.
[foo_VAR_ACCESSOR][bar_PROP_GETTER][x_PROP_ACCESSOR] = [foo_VAR_ACCESSOR] = {a:1}
> A[bar_PROP_GETTER][x_PROP_ACCESSOR] = [foo_VAR_ACCESSOR] = {a:1}
> B[x_PROP_ACCESSOR] = [foo_VAR_ACCESSOR] = {a:1}
> B['x'] = foo = {a: 1}
Here you have little example
var A = {};
var B = {}
Object.defineProperty(A, 'bar', {
get () {
console.log('A.bar::getter')
return B;
}
})
Object.defineProperty(B, 'x', {
set () {
console.log('B.x::getter')
}
});
var foo = A;
foo.bar.x = foo = (console.log('test'), 'hello');
// > A.bar.getter
// > test
// > B.x.setter
Great question. The thing to remember here is that JavaScript uses pointers for everything. It's easy to forget this since it is impossible to access the values representing memory addresses in JavaScript (see this SO question). But realizing this is very important in order to understand many things in JavaScript.
So the statement
var foo = {};
creates an object in memory and assigns a pointer to that object to foo. Now when this statement runs:
foo.x = foo = {a: 1};
the property x is actually getting added to the original object in memory, while foo is getting assigned a pointer to a new object, {a: 1}. For example,
var foo, bar = foo = {};
foo.x = foo = {a: 1};
shows that if foo and bar are pointing to the same object initially, bar (which will still point to that original object) will look like {x: {a: 1}}, while foo is simply {a: 1}.
So why doesn't foo look like {a: 1, x: foo}?
While you are right in that assignments are right associative, you must also realize that the interpreter still reads from left to right. Let's take an in-depth example (with some bits abstracted out):
var foo = {};
Okay, create an object in memory location 47328 (or whatever), assign foo to a pointer that points to 47328.
foo.x = ....
Okay, grab the object that foo currently points to at memory location 47328, add a property x to it, and get ready to assign x to the memory location of whatever's coming next.
foo = ....
Okay, grab the pointer foo and get ready to assign it to the memory location of whatever's coming next.
{a: 1};
Okay, create a new object in memory at location 47452. Now go back up the chain: Assign foo to point to memory location 47452. Assign property x of the object at memory location 47328 to also point to what foo now points to--memory location 47452.
In short, there is no shorthand way to do
var foo = {a: 1};
foo.x = foo;
I'm trying to use a string as a reference to a variable to pass into a function. For example:
var names = ['Peter', 'John'],
var hasName = function(name){
var params = ['names'];
return $.inArray(name, eval( params[0] )) === -1;
};
How to avoid eval()?
EDIT:
The string from params[0] is comming from a data-qval of an input in my html. The array that contains the actual data can be declared anywhere, params[0] is just a reference to that array passed in as string in data-qval, it's a parameter. I pasted my plugin's code here .
http://pastebin.mozilla.org/1598528 Line 101.
Full example: http://jsfiddle.net/elclanrs/ZsS2D/29/
It currently works, I'm just looking for a way get rid of eval()...
In that particular case, just use names:
var names = ['Peter', 'John'],
var hasName = function(name){
var params = ['names'];
return $.inArray(name, names ) === -1;
};
(See also the note below.) (Your edit makes the above not applicable.)
If you're trying to look up the names array in some container using the string "names", you'd have to have a reference to the container, e.g.:
var obj = {
names: ['Peter', 'John'
};
var hasName = function(name){
var params = ['names'];
return $.inArray(name, obj[params[0]] ) === -1;
};
If there is no container other than the variable scope in which you're doing this, you'll have to use eval. But you can (and usually should) adjust things so you have a container (as above) so you can avoid it. Note that if names is declared at global scope, you do have a container (window).
So to summarize:
If names is a var at global scope (or an implicit global), window[params[0]] will give you a reference to it.
If names is already in some container object, you can use container[params[0]] to get a reference to it.
If names is a var within a function, you cannot get at it using a runtime string without eval; ideally, rather than var names = [...];, use var container = {names: [...]}; and then you can use container[params[0]].
Note that your function is called hasName, but it returns true when the array doesn't have the name and false when it does. You probably want !== -1, not === -1.
Isn't this enough?
var hasName = function(name){
return $.inArray(name, names) > -1;
};
Also, notice the comparison
Because JavaScript treats 0 as loosely equal to false (i.e. 0 ==
false, but 0 !== false), if we're checking for the presence of value
within array, we need to check if it's not equal to (or greater than)
-1.
Strings as references to variables? If it's a global object in a browser, it will be in the window object, so you can just do window[variableName] to get its value. Same for objects, i.e. instead of object.foo, you can do object['foo'] or bar = 'foo', object[bar]. For locally scoped variables, you cannot do it without using an object or eval.
if names is global, you can use the global namespace, i.e. window
var hasName = function(name,namespace){
namespace = namespace || window;
return $.inArray(name, namespace.names) > -1;
};
hasName('Peter'); //=> true;
This may also be an idea:
var MYNS = { names:['Peter','John']
,hasName: function(name){
return $.inArray(name, this.names) > -1;
}
};
MYNS.hasName('Peter'); //=>true
I've got a feeling this might not be possible, but I would like to determine the original variable name of a variable which has been passed to a function in javascript. I don't know how to explain it any better than that, so see if this example makes sense.
function getVariableName(unknownVariable){
return unknownVariable.originalName;
}
getVariableName(foo); //returns string "foo";
getVariableName(bar); //returns string "bar";
This is for a jquery plugin i'm working on, and i would like to be able to display the name of the variable which is passed to a "debug" function.
You're right, this is very much impossible in any sane way, since only the value gets passed into the function.
This is now somehow possible thanks to ES6:
function getVariableName(unknownVariableInAHash){
return Object.keys(unknownVariableInAHash)[0]
}
const foo = 42
const bar = 'baz'
console.log(getVariableName({foo})) //returns string "foo"
console.log(getVariableName({bar})) //returns string "bar"
The only (small) catch is that you have to wrap your unknown variable between {}, which is no big deal.
As you want debugging (show name of var and value of var),
I've been looking for it too, and just want to share my finding.
It is not by retrieving the name of the var from the var but the other way around : retrieve the value of the var from the name (as string) of the var.
It is possible to do it without eval, and with very simple code, at the condition you pass your var into the function with quotes around it, and you declare the variable globally :
foo = 'bar';
debug('foo');
function debug(Variable) {
var Value = this[Variable]; // in that occurrence, it is equivalent to
// this['foo'] which is the syntax to call the global variable foo
console.log(Variable + " is " + Value); // print "foo is bar"
}
Well, all the global variables are properties of global object (this or window), aren't they?
So when I wanted to find out the name of my variables, I made following function:
var getName = function(variable) {
for (var prop in window) {
if (variable === window[prop]) {
return prop;
}
}
}
var helloWorld = "Hello World!";
console.log(getName(helloWorld)); // "helloWorld"
Sometimes doesn't work, for example, if 2 strings are created without new operator and have the same value.
Global w/string method
Here is a technique that you can use to keep the name and the value of the variable.
// Set up a global variable called g
var g = {};
// All other variables should be defined as properties of this global object
g.foo = 'hello';
g.bar = 'world';
// Setup function
function doStuff(str) {
if (str in g) {
var name = str;
var value = g[str];
// Do stuff with the variable name and the variable value here
// For this example, simply print to console
console.log(name, value);
} else {
console.error('Oh snap! That variable does not exist!');
}
}
// Call the function
doStuff('foo'); // log: foo hello
doStuff('bar'); // log: bar world
doStuff('fakeVariable'); // error: Oh snap! That variable does not exist!
This is effectively creating a dictionary that maps variable names to their value. This probably won't work for your existing code without refactoring every variable. But using this style, you can achieve a solution for this type of problem.
ES6 object method
In ES6/ES2015, you are able to initialize an object with name and value which can almost achieve what you are trying to do.
function getVariableName(unknownVariable) {
return Object.keys(unknownVariable)[0];
}
var foo = 'hello';
var output = getVariableName({ foo }); // Note the curly brackets
console.log(output);
This works because you created a new object with key foo and value the same as the variable foo, in this case hello. Then our helper method gets the first key as a string.
Credit goes to this tweet.
Converting a set of unique variable into one JSON object for which I wrote this function
function makeJSON(){ //Pass the variable names as string parameters [not by reference]
ret={};
for(i=0; i<arguments.length; i++){
eval("ret."+arguments[i]+"="+arguments[i]);
}
return ret;
}
Example:
a=b=c=3;
console.log(makeJSON('a','b','c'));
Perhaps this is the reason for this query
I think you can use
getVariableName({foo});
Use a 2D reference array with .filter()
Note: I now feel that #Offermo's answer above is the best one to use. Leaving up my answer for reference, though I mostly wouldn't recommend using it.
Here is what I came up with independently, which requires explicit declaration of variable names and only works with unique values. (But will work if those two conditions are met.)
// Initialize some variables
let var1 = "stick"
let var2 = "goo"
let var3 = "hello"
let var4 = "asdf"
// Create a 2D array of variable names
const varNames = [
[var1, "var1"],
[var2, "var2"],
[var3, "var3"]
]
// Return either name of variable or `undefined` if no match
const getName = v => varNames.filter(name => name[0] === v).length
? varNames.filter(name => name[0] === v)[0][1]
: undefined
// Use `getName` with OP's original function
function getVariableName(unknownVariable){
return getName(unknownVariable)
}
This is my take for logging the name of an input and its value at the same time:
function logVariableAndName(unknownVariable) {
const variableName = Object.keys(unknownVariable)[0];
const value = unknownVariable[variableName];
console.log(variableName);
console.log(value);
}
Then you can use it like logVariableAndName({ someVariable })