Parameter Passing: In and Out - javascript

Consider:
Primitive data types are passed by value in JavaScript. This means
that a copy is effectively made of a variable when it is passed to a
function, so any manipulation local to the function leaves the
original variables untouched.
function fiddle(arg1) {
arg1 = "Fiddled with";
console.log("In function fiddle str = "+arg1+"<br>");
}
var str = "Original Value";
console.log("Before function call str = "+str+"<br>");
fiddle(str);
console.log("After function call str ="+str+"<br>");
EASY
composite types such as arrays and objects if used, they are passed
by reference rather than value
Consider the following modification
of the previous fiddle() function:
function fiddle(arg1) {
arg1[0] = "Fiddled with";
console.log("In function fiddle arg1 = "+arg1+"<br>");
}
var arr = ["Original", " Original ", " Original "];
console.log("Before function call arr = "+arr+"<br>");
fiddle(arr);
console.log("After function call arr ="+arr+"<br>");
FINE till here
From here I find myself really confused,
function fiddle(arg1) {
arg1 = ["Blasted!","Blasted!"];
console.log("In function fiddle arg1 = "+arg1+"<br>");
}
var arr = ["Original", " Original ", " Original "];
console.log("Before function call arr = "+arr+"<br>");
fiddle(arr);
console.log("After function call arr ="+arr+"<br>"); // Why this hasn't changed?
Any suggestion? And will it be OK to say
composite types such as arrays and objects if used, they are passed
by reference rather than value
in this case too?

Seems pretty clear, inside fiddle you're overwriting the reference to arr with an array you created on the spot. You never touched arr itself through its reference, just the reference, so once you're outside again, arr is still the original array.

That is because the bracket notation (for arrays) or the dot notation (for objects) allows you to mutate objects within an array/object, which will propagate up and outwards the scope of the function as they are passed by reference, even though the "parent" object is passed by value. This feature is known as call-by-sharing, as addressed in a previous question. In short:
Everything is passed by value, but...
Objects within the "everything" (if exists, i.e. for arrays and objects) is passed by reference
If you want to modify arr outside of the scope of the fiddle() function, you will have to return it, i.e.: return arr1; and then assign arr = fiddle(arg1[,arg2,...]), like as follow:
function fiddle(arg1) {
arg1 = ["Blasted!","Blasted!"];
$('body').append("In function fiddle arg1 = "+arg1+"<br>");
return arg1;
}
var arr = ["Original", " Original ", " Original "];
$('body').append("Before function call arr = "+arr+"<br>");
arr = fiddle(arr);
$('body').append("After function call arr ="+arr+"<br>"); // Why this hasn't changed?
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Related

Why object method is rewrites the original objects? [duplicate]

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);

How to use apply with currying?

I have code that is using currying to get the average on an array that results from concatenating two arrays: an n size array and an m size array.
var avg = function(...n){
let tot=0;
for(let i=0; i<n.length; i++){
tot += n[i];
}
return tot/n.length;
};
var spiceUp = function(fn, ...n){
return function(...m){
return fn.apply(this, n.concat(m));
}
};
var doAvg = spiceUp(avg, 1,2,3);
console.log(doAvg(4,5,6));
In this line return fn.apply(this, n.concat(m));, I don't understand why do we need to use apply. What is the object we are binding with the average function and why does just normal calling (return fn(n.concat(m));) not work?
In that example, this is not that important. It would also work if instead of this you would pass an empty object instead. It's just an example on how to use apply.
What you need to focus is on the second parameter n.concat(m). They key concept here is that passing an array as a second argument you are calling that function (fn) passing each value in the array as an argument.
About your second question: no, it won't work because fn expects several arguments (one per value to calculate the average) while by doing return fn(n.concat(m)); you are just passing one argument, an array containing all values
Maybe you would understand it better with a simpler example:
function sum3params(a,b,c){
return a+b+c;
}
console.log(sum3params([3,4,2])) // won't work
console.log(sum3params.apply(this, [3,4,2])) // will work ('this' is not important here)
For this use case, it does not. But consider the following:
var foo = {
bar: 3
};
var addBar = function(a, b) { return a + b + this.bar };
foo.add3AndBar = spiceUp(addBar, 3);
foo.add3AndBar(1); // 7
Using apply means that your spiceUp function can be applied to methods as well as normal functions. For more likely example, consider partially applying when defining a method on a prototype:
const ENV = "linux";
DoesSomePlatformSpecificStuff.prototype.getPath = spiceUp(ENV);
apply also will spread the gathered array of arguments back out into positional arguments which can also be done like so:
return fn(...n.concat(m));
Which can be simplified as
return fn(...n, ...m);
Which is equivalent to
return fn.apply(undefined, n.concat(m));

Is there a form of C++ references in Javascript? [duplicate]

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);

How to pass parameter to the function with apply in javascript?

I have functions, to pass the parameters from a to b.
but it seems the apply just access 2 parameters
function a(){
var p1=1;
var p2=2;
b.apply(this,arguments);
}
function b(){
//so how to get p1,p2 in this function?
console.log(arguments);
}
a(3,4);
so how to pass the p1 p2 in function a and get 1,2,3,4 in functionb?
You create an array with all the parameters you need to pass:
var combinedArray = [p1, p2].concat(Array.prototype.slice.call(arguments));
this function:
Creates an array with first 2 elements as p1 and p2
And appends the arguments array to it.
Then you can call it as b.apply(this, combinedArray)
PS:
The Array.prototype.slice.call(arguments) is used to convert arguments object to a real array.
In order to pass your variables to another function you have to add them to the array you provide to apply. But you can not add them directly to the arguments variable since it's not the real array. So first of all you have to build a new array which you would pass to apply. You can transform arguments to array by var newArray = Array.prototype.slice.call(arguments); - this creates new array from arguments. Now you can act with it as with regular array (newArray.push(p1); or newArray = newArray.concat([p1, p2]); and so on). And then just pass this array to apply instead of arguments.
So your code would change this way:
function a(){
var p1=1;
var p2=2;
var argumentsArray = Array.prototype.slice.call(arguments);
b.apply(this, argumentsArray.concat([p1, p2]));
}
function b(){
console.log(arguments);
}
a(3,4); // output: [3, 4, 1, 2];
If you need to prepend p1 and p2 you can use unshift
Try this code.
void temp() {
var animals = ["cat", "dog", "horse", "cow"];
animals.forEach(function (eachName, index) {
console.log(index + 1 + ". " + eachName);
});
}

What is this javascript code doing?

this.String = {
Get : function (val) {
return function() {
return val;
}
}
};
What is the ':' doing?
this.String = {} specifies an object. Get is a property of that object. In javascript, object properties and their values are separated by a colon ':'.
So, per the example, you would call the function like this
this.String.Get('some string');
More examples:
var foo = {
bar : 'foobar',
other : {
a : 'wowza'
}
}
alert(foo.bar); //alerts 'foobar'
alert(foo.other.a) //alerts 'wowza'
Others have already explained what this code does. It creates an object (called this.String) that contains a single function (called Get). I'd like to explain when you could use this function.
This function can be useful in cases where you need a higher order function (that is a function that expects another function as its argument).
Say you have a function that does something to each element of an Array, lets call it map. You could use this function like so:
function inc (x)
{
return x + 1;
}
var arr = [1, 2, 3];
var newArr = arr.map(inc);
What the map function will do, is create a new array containing the values [2, 3, 4]. It will do this by calling the function inc with each element of the array.
Now, if you use this method a lot, you might continuously be calling map with all sorts of arguments:
arr.map(inc); // to increase each element
arr.map(even); // to create a list of booleans (even or odd)
arr.map(toString); // to create a list of strings
If for some reason you'd want to replace the entire array with the same string (but keeping the array of the same size), you could call it like so:
arr.map(this.String.Get("my String"));
This will create a new array of the same size as arr, but just containing the string "my String" over and over again.
Note that in some languages, this function is predefined and called const or constant (since it will always return the same value, each time you call it, no matter what its arguments are).
Now, if you think that this example isn't very useful, I would agree with you. But there are cases, when programming with higher order functions, when this technique is used.
For example, it can be useful if you have a tree you want to 'clear' of its values but keep the structure of the tree. You could do tree.map(this.String.Get("default value")) and get a whole new tree is created that has the exact same shape as the original, but none of its values.
It assigns an object that has a property "Get" to this.String. "Get" is assigned an anonymous function, which will return a function that just returns the argument that was given to the first returning function. Sounds strange, but here is how it can be used:
var ten = this.String["Get"](10)();
ten will then contain a 10. Instead, you could have written the equivalent
var ten = this.String.Get(10)();
// saving the returned function can have more use:
var generatingFunction = this.String.Get("something");
alert(generatingFunction()); // displays "something"
That is, : just assigns some value to a property.
This answer may be a bit superflous since Tom's is a good answer but just to boil it down and be complete:-
this.String = {};
Adds an object to the current object with the property name of String.
var fn = function(val) {
return function() { return(val); }
}
Returns a function from a closure which in turn returns the parameter used in creating the closure. Hence:-
var fnInner = fn("Hello World!");
alert(fnInner()); // Displays Hello World!
In combination then:-
this.String = { Get: function(val) {
return function() { return(val); }
}
Adds an object to the current object with the property name of String that has a method called Get that returns a function from a closure which in turn returns the parameter used in creating the closure.
var fnInner = this.String.Get("Yasso!");
alert(fnInner()); //displays Yasso!

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