I would like to extend Number with a plusPlus2 function that will increment the number by 2.
The problem is that I don't know how to assign the result back to the number in the extension function. Something like:
Number.prototype.plusPlus2 = function() {
this = this + 2;
}
And the usage would be:
x = 1;
x.plusPlus2(); // expect x to be 3
Primitives (numbers, strings, booleans) are immutable in javascript.
So anytime you change a primitive, you need to assign it to another variable (or even reassign it to itself).
That being said, you cannot do what you propose, you need to return a new value, containing the value you want, let's say:
Number.prototype.plusplus2 = function() {
return this + 2;
}
And then reassign it:
let x = 5;
x = x.plusplus2();
Then, you may be wondering: how x++ works?
And the answer is, x++ is a syntax sugar for x = x + 1, meaning that, in fact, you are not changing x, but instead, adding 1 to x and reassigning it to itself.
You cannot. That has various reasons:
1) this is read only, as you don't expect it to change during the execution of a method.
2) What you access with this is a Number object that wraps the primitive number. It gets thrown away after the call. So even if you could change the internal value property containing the number, the value of x won't change.
Number.prototype.example = function() { this.stuff = "you see" };
let x = 1; // primitive
x.example(); // wrapped object
// wrapped object gets thrown away
console.log(x.stuff); // undefined, another wrapped object
3) Numbers are immutable and primitive. You can write a new number into x, but you can't turn all 1s into 3s.
You could create a new number, but then you have the problem that you have to write that number into x.
Number.prototype.plus2 = function() { return this + 2 };
let x = 1;
x = x.plus2();
Number.prototype.plus2 = function() {
return this + 2
}
let a = 2
console.log(a.plus2())
I was watching a JavaScript talk, and the tutor said that if we pass a property of an object in a function it will actually change the real value, because we will be passing the variable by reference. Here is the slide:
but when I tried to practice the concept, that wasn't the case. Here is my code:
var obj = {val: 5};
function changeVal(x) {
x = x+5;
return x;
}
console.log(obj.val) // 5
console.log(changeVal(obj.val)) // 10
console.log(obj.val) // 5
I was expecting obj.val to change to 10.
Please tell me what's wrong here, and correct me if I am wrong. Thanks
You are passing not the object, but the primitive type. So when you pass the val of the obj, it is a number and is a primitive type.It copies the val and passes the copy to the object.
If you pass like this, it will work
var obj = {val: 5};
function changeVal( param ) {
param.val = param.val + 5;
return param.val ;
}
console.log(obj.val) // 5
console.log(changeVal(obj)) // 10
console.log(obj.val) // 10
You are not actually passing an object, just passing the value of property(val).
If you will pass obj in changeVal(), then it will actually change the value of the property of passed object.
For that you need to do like:
var obj = {val: 5};
function changeVal(x)
{
x = x+5;
return x;
}
console.log(obj.val); // 5
changeVal(obj); // Need to pass object instead of value of the property's value
console.log(obj.val); // 10
Primitive types (string, integer, boolean, etc...) are immutable, which means if you change one of the values inside a function, the callee (scope which calls your function) will not see the change.
function doSomething(a) {
a = a + 1;
}
var value = 2;
console.log(value); // result: 2
doSomething(value);
console.log(value); // result: 2
Pass-by-reference only works for objects. Like this:
function doSomething(obj) {
obj.attribute = obj.attribute + 1;
}
var myObject = {attribute: 2};
console.log(myObject.attribute); // result: 2
doSomething(myObject);
console.log(myObject.attribute); // result: 3
More reading about Javascript types:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Data_structures
Say for instance you have an Iphone . Now lets say a manufacturing company calls you and asks to borrow your Iphone for a reference just so they can design an Iphone that is similar and sell it to customers . Your original Iphone still exists and is never gone , but every now and then the factory needs to use it for a reference , think of your function as the factory that just make a copy of obj.
//Original data
var obj = {val: 5};
Once your function returns something , it technically becomes a value
Example :
return 3; is a value of 3
so
function changeVal(x) {
x = x+5;
return x;
}
is a new value of x which in this case would be x + 5;
x is a copy of whatever you pass into the function .
Hope this helps.
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);
This question already has answers here:
Is JavaScript a pass-by-reference or pass-by-value language?
(33 answers)
Closed 8 years ago.
I can't figure out what passing by reference and passing by value looks like (I know what they mean in theory). Can you tell me whether the below function is passing arguments by value ore reference? BTW this is my first post :)
var number_1 = 32;
var number_2 = 43;
function addition(num1,num2) {
alert (num1*num2) ;
}
addition(number_1,number_2); /*is this passing by reference?*/
Primitive types
In javascript number, strings, boolean fall under the category of primitives.
And whenever such types are passed as arguments to a function, a separate copy is created inside the function scope which has no impact on the outer scope
for e.g
var number_1 = 32;
var number_2 = 43;
addition(32,43);
//number_1 is 32 here
//number_2 is 43 here
function addition(number_1,number_2) {
number_1 += 1; // number_1 is 33
number_2 += 1; // number_2 is 44
}
Reference types
Reference types are slightly different
Take the following example
var obj = new Object();
fn(obj);
function fn(object){
object.property = "test";
//obj.property is "test"
object = new Object();
object.property = "test 2";
//obj.property is still "test"
//obj.property should have changed to "test 2",
//if it had been passed by reference
}
if it had been passed by reference , obj.property should have got changed to "test 2" after the last statement inside fn, but it didnt.
so when passing reference values to functions, a separate copy of the pointer to the object is passed.
In javascript every variable is passed by value not reference but there is an exception for objects. objects are references (like it was answered here).
which means a number, a string, a boolean or a function don't change at all after you put'em as argument in a function. but an the propertues and the elements inside an object or an array change if they are changed inside a function.
Here is an example
Simple variable
function nullify(a)
{
a = null;
}
var myVariable = 123;
nullify(myVariable); // myVariable still equals to 123
Object
function changeProperty (obj)
{
obj.key = null;
}
var myObject = new Object();
myObject.key = "value";
changeProperty(myObject); // the property key is null after calling the function
Array
function changeArray (Arr)
{
Arr[0] = null;
}
var myArray = [1,2,3];
changeArray(myArray); // now the value of myArray is [null,2,3]
I have recently encountered a nasty issue in JS.
Let say we pass a map, an array of objects to a function f.
var o=[{a:0}];
function f(a){
for(var i in a){
if (a.hasOwnProperty(i)){
a[i]=null;
}
}
return a;
};
var outp=f(o);
alert(outp[0]+" === "+o[0]+" : "+(outp[0]===o[0]));
// here we expect loose equality, and equality in type,
//furthermore it should identically equal as well, and we got right!
But, we can not pass total responsibility of an object to a function as argument, same like in functional paradigm o=(function(o){return o})(), because any kind of modification to o is not referenced!
var o=[];
function ff(a){
return (a=undefined);
};
var outp=ff(o);
alert(outp+" === "+o.constructor+" : "+(outp===o));
// here we expect true, but we got false!
Why is the above described reference loss and
presumably different referencce handling in the second use case,
though in both case, functions got the array argument in the 0. position?
Javascript always passes arguments by value, so this won't work:
function foo(x) {
x = 100;
}
y = 5
foo(y)
y == 100 // nope
However this does work:
function foo(x) {
x.bar = 100;
}
y = {}
foo(y)
y.bar == 100 // yes
In the second snippet x is still passed by value, but this very value is a reference (pointer) to an object. So it's possible in a function to dereference it and access what's "inside" the object.