Javascript passing arrays to functions by value, leaving original array unaltered - javascript

I've read many answers here relating to 'by value' and 'by reference' passing for sending arrays to javascript functions. I am however having a problem sending an array to a function and leaving the original array unaltered. This example llustrates the problem:
function myFunction(someArray)
{
// any function that makes an array based on a passed array;
// someArray has two dimensions;
// I've tried copying the passed array to a new array like this (I've also used 'someArray' directly in the code);
funcArray = new Array();
funcArray = someArray;
var i = 0;
for(i=0; i<funcArray.length; i++)
{
funcArray[i].reverse;
}
return funcArray;
}
I can't understand why anything in this function should alter the original array.
calling this function directly changes the original array if the function call is assigned to a new array:
myArray = [["A","B","C"],["D","E","F"],["G","H","I"]];
anotherArray = new Array();
anotherArray = myFunction(myArray);
// myArray gets modified!;
I tried using .valueOf() to send the primitive:
anotherArray = myFunction(myArray.valueOf());
// myArray gets modified!;
I have even tried breaking the array down element by element and sub-element by sub-element and assigning all to a new 2-d array and the original array still gets modified.
I have also joined the sub-elements to a string, processed them, split them back into arrays and the original array still gets modified.
Please, does any one know how I can pass the array values to a function and not have the passed array change?

Inside your function there's this:
funcArray = new Array();
funcArray = someArray;
This won't actually copy someArray but instead reference it, which is why the original array is modified.
You can use Array.slice() to create a so-called shallow copy of the array.
var funcArray = someArray.slice(0);
Modern versions of ES also support destructuring expressions, which make it look like this:
const funcArray = [...someArray];
The original array will be unaltered, but each of its elements would still reference their corresponding entries in the original array. For "deep cloning" you need to do this recursively; the most efficient way is discussed in the following question:
What is the most efficient way to deep clone an object in JavaScript?
Btw, I've added var before funcArray. Doing so makes it local to the function instead of being a global variable.

Make a copy of the array that you can use.
A simple way to do this is by using var clone = original.slice(0);

With ES6, you can use the rest element syntax (...) within a destructuring expression to perform a shallow copy directly in the parameter list, allowing you to keep the original array unaltered.
See example below:
const arr = [1, 2, 3, 4, 5];
function timesTen([...arr]) { // [...arr] shallow copies the array
for(let i = 0; i < arr.length; i++) {
arr[i] *= 10; // this would usually change the `arr` reference (but in our case doesn't)
}
return arr;
}
console.log(timesTen(arr));
console.log(arr); // unaltered
The above is useful if you have an array of primitives (strings, booleans, numbers, null, undefined, big ints, or symbols) because it does a shallow copy. If you have an array of objects, or an array of arrays (which are also technically just objects), you will want to perform a deep clone to avoid modifying the array and its references. The way to do that in modern-day JS is to use a structured clone, which helps deal with many of the shortcomings of deep cloning with the JSON.stringify() technique as previously used:
function myFunction(someArray) {
const funcArray = structuredClone(someArray);
...
}

What about destructuring assignment (ES6+, check compatibility)? Nice and clean solution.
function myFunction(someArray) {
for(let i = 0; i < someArray.length; i++)
{
someArray[i].reverse();
}
return someArray;
}
let myArray = [["A","B","C"],["D","E","F"],["G","H","I"]];
// Using destructuring assignment.
// NOTE: We can't just use `[...myArray]` because nested arrays will still be copied by reference.
let anotherArray = myFunction([...myArray.map(nested => [...nested])]);
console.log({original: myArray, copy: anotherArray});

A variable pointing to an array is a reference to it. When you pass an array, you're copying this reference.
You can make a shallow copy with slice(). If you want a full depth copy, then recurse in sub objects, keeping in mind the caveats when copying some objects.

If you need to do this with an object, try this fancy trick...
MY_NEW_OBJECT = JSON.parse(JSON.stringify(MY_OBJECT));

A generic solution would be...
// Use the JSON parse to clone the data.
function cloneData(data) {
// Convert the data into a string first
var jsonString = JSON.stringify(data);
// Parse the string to create a new instance of the data
return JSON.parse(jsonString);
}
// An array with data
var original = [1, 2, 3, 4];
function mutate(data) {
// This function changes a value in the array
data[2] = 4;
}
// Mutate clone
mutate(cloneData(original));
// Mutate original
mutate(original);
This works for objects as well as arrays.
Very effective when you need deep cloning or you don't know what the type is.
Deep cloning example...
var arrayWithObjects = [ { id: 1 }, { id: 2 }, { id: 3 } ];
function mutate(data) {
// In this case a property of an object is changed!
data[1].id = 4;
}
// Mutates a (DEEP) cloned version of the array
mutate(cloneData(arrayWithObjects));
console.log(arrayWithObjects[1].id) // ==> 2
Warnings
Using the JSON parser to clone is not the most performant option!
It doesn't clone functions only JSON supported data types
Cannot clone circular references

by default in javascript except objects and arrays, everything is copy-by-value
but if you want to use copy-by-value for arrays: use [yourArray].slice(0)
and for objects use Object.assign(target, ...sources)

var aArray = [0.0, 1.0, 2.0];
var aArrayCopy = aArray.concat();
aArrayCopy[0] = "A changed value.";
console.log("aArray: "+aArray[0]+", "+aArray[1]+", "+aArray[2]);
console.log("aArrayCopy: "+aArrayCopy[0]+", "+aArrayCopy[1]+", "+aArrayCopy[2]);
This answer has been edited. Initially I put forth that the new operator handled the solution, but soon afterward recognized that error. Instead, I opted to use the concat() method to create a copy. The original answer did not show the entire array, so the error was inadvertently concealed. The new output shown below will prove that this answer works as expected.
aArray: 0, 1, 2
aArrayCopy: A changed value., 1, 2

Related

I should be getting [[1,0],[0,0]] why am I getting [[1,0],[1,0]]? [duplicate]

I've written the following JavaScript:
var myArray = ['a', 'b', 'c'];
var copyOfMyArray = myArray;
copyOfMyArray.splice(0, 1);
alert(myArray); // alerts ['b','c']
alert(copyOfMyArray); // alerts ['b','c']
var myNumber = 5;
var copyOfMyNumber = myNumber;
copyOfMyNumber = copyOfMyNumber - 1;
alert(myNumber); // alerts 5
alert(copyOfMyNumber); // alerts 4
This code declares a variable myArray and sets it to an array value. It then declares a second variable copyOfMyArray and sets it to myArray.
It performs an operation on copyOfMyArray and then alerts both myArray and copyOfMyArray. Somehow, when I perform an operation on copyOfMyArray, it appears that the same operation is performed on myArray.
The code then does the same thing with a number value: It declares a variable myNumber and sets it to a number value. It then declares a second variable copyOfMyNumber and sets it to myNumber. It performs an operation on copyOfMyNumber and then alerts both myNumber and copyOfMyNumber. Here, I get the expected behavior: different values for myNumber and copyOfMyNumber.
What is the difference between an array and a number in JavaScript that it seems changing an array changes the value of a copy of the array, where as changing a number does not change the value of a copy of the number?
I'm guessing that for some reason, the array is referred to by reference and the number by value, but why? How can I know what behavior to expect with other objects?
An array in JavaScript is also an object and variables only hold a reference to an object, not the object itself. Thus both variables have a reference to the same object.
Your comparison with the number example is not correct btw. You assign a new value to copyOfMyNumber. If you assign a new value to copyOfMyArray it will not change myArray either.
You can create a copy of an array using slice [docs]:
var copyOfMyArray = myArray.slice(0);
But note that this only returns a shallow copy, i.e. objects inside the array will not be cloned.
Well, the only possible answer — and the correct one — is that you're not actually copying the array. When you write
var copyOfArray = array;
you're assigning a reference to the same array into another variable. They're both pointing at the same object, in other words.
So everyone here has done a great job of explaining why this is happening - I just wanted to drop a line and let you know how I was able to fix this - pretty easily:
thingArray = ['first_thing', 'second_thing', 'third_thing']
function removeFirstThingAndPreserveArray(){
var copyOfThingArray = [...thingArray]
copyOfThingArray.shift();
return copyOfThingArray;
}
This is using the ... spread syntax.
Spread Syntax Source
EDIT: As to the why of this, and to answer your question:
What is the difference between an array and a number in JavaScript that it seems changing an array changes the value of a copy of the array, where as changing a number does not change the value of a copy of the number?
The answer is that in JavaScript, arrays and objects are mutable, while strings and numbers and other primitives are immutable. When we do an assignment like:
var myArray = ['a', 'b', 'c'];
var copyOfMyArray = myArray;
copyOfMyArray is really just a reference to myArray, not an actual copy.
I would recommend this article, What are immutable and mutable data structures?, to dig deeper into the subject.
MDN Glossary: Mutable
Cloning objects -
A loop / array.push produces a similar result to array.slice(0) or array.clone(). Values are all passed by reference, but since most primitive data types are immutable, subsequent operations produce the desired result - a 'clone'. This is not true of objects and arrays, of course, which allow for modification of the original reference (they are mutable types).
Take the following example:
const originalArray = [1, 'a', false, {foor: 'bar'}]
const newArray = [];
originalArray.forEach((v, i) => {
newArray.push(originalArray[i]);
});
newArray[0] = newArray[0] + 1;
newArray[1] = 'b';
newArray[2] = true;
newArray[3] = Object.assign(newArray[3], {bar: 'foo'});
The operations run on the newArray indices all produce the desired result, except the final (object), which, because it is copied by reference, will mutate the originalArray[3] as well.
https://jsfiddle.net/7ajz2m6w/
Note that array.slice(0) and array.clone() suffers from this same limitation.
One way to solve this is by effectively cloning the object during the push sequence:
originalArray.forEach((v, i) => {
const val = (typeof v === 'object') ? Object.assign({}, v) : v;
newArray.push(val);
});
https://jsfiddle.net/e5hmnjp0/
cheers
The issue with shallow copy is that all the objects aren't cloned, instead it get reference.So array.slice(0) will work fine only with literal array, but it will not do shallow copy with object array. In that case one way is..
var firstArray = [{name: 'foo', id: 121}, {name: 'zoo', id: 321}];
var clonedArray = firstArray.map((_arrayElement) => Object.assign({}, _arrayElement));
console.log(clonedArray);
// [{name: 'foo', id: 121}, {name: 'zoo', id: 321}] // shallow copy
In JS, operator "=" copy the pointer to the memory area of the array.
If you want to copy an array into another you have to use the Clone function.
For integers is different because they are a primitive type.
S.
Create a filter of the original array in the arrayCopy. So that changes to the new array won't affect original array.
var myArray = ['a', 'b', 'c'];
var arrayCopy = myArray.filter(function(f){return f;})
arrayCopy.splice(0, 1);
alert(myArray); // alerts ['a','b','c']
alert(arrayCopy); // alerts ['b','c']
Hope it helps.
Everything is copied by reference except primitive data types (strings and numbers IIRC).
You don't have any copies.
You have multiple variables holding the same array.
Similarly, you have multiple variables holding the same number.
When you write copyOfMyNumber = ..., you're putting a new number into the variable.
That's like writing copyOfMyArray = ....
When you write copyOfMyArray.splice, you're modifying the original array.
That isn't possible with numbers because numbers are immutable and cannot be modified,
You can add some error handling depending on your cases and use something similar to the following function to solve the issue. Please comment for any bugs / issues / efficiency ideas.
function CopyAnArray (ari1) {
var mxx4 = [];
for (var i=0;i<ari1.length;i++) {
var nads2 = [];
for (var j=0;j<ari1[0].length;j++) {
nads2.push(ari1[i][j]);
}
mxx4.push(nads2);
}
return mxx4;
}
An array, or an object in javascript always holds the same reference unless you clone or copy. Here is an exmaple:
http://plnkr.co/edit/Bqvsiddke27w9nLwYhcl?p=preview
// for showing that objects in javascript shares the same reference
var obj = {
"name": "a"
}
var arr = [];
//we push the same object
arr.push(obj);
arr.push(obj);
//if we change the value for one object
arr[0].name = "b";
//the other object also changes
alert(arr[1].name);
For object clone, we can use .clone() in jquery and angular.copy(), these functions will create new object with other reference. If you know more functions to do that, please tell me, thanks!
For Arrays with objects you can change the use JSON.parse and JSON.strinigfy to change the type of the array to an object refrence to a string and then back to a array without having to worry about the original array
var array = [{name:'John', age:34, logs:'[]'}, {name:'David', age:43, logs:'[3]'}];
var array2 = JSON.parse(JSON.stringify(array)); // turn object to function output
array2[0].age++;
alert(JSON.stringify(array));
alert(JSON.stringify(array2));

How to get only value from global variable? [duplicate]

I've written the following JavaScript:
var myArray = ['a', 'b', 'c'];
var copyOfMyArray = myArray;
copyOfMyArray.splice(0, 1);
alert(myArray); // alerts ['b','c']
alert(copyOfMyArray); // alerts ['b','c']
var myNumber = 5;
var copyOfMyNumber = myNumber;
copyOfMyNumber = copyOfMyNumber - 1;
alert(myNumber); // alerts 5
alert(copyOfMyNumber); // alerts 4
This code declares a variable myArray and sets it to an array value. It then declares a second variable copyOfMyArray and sets it to myArray.
It performs an operation on copyOfMyArray and then alerts both myArray and copyOfMyArray. Somehow, when I perform an operation on copyOfMyArray, it appears that the same operation is performed on myArray.
The code then does the same thing with a number value: It declares a variable myNumber and sets it to a number value. It then declares a second variable copyOfMyNumber and sets it to myNumber. It performs an operation on copyOfMyNumber and then alerts both myNumber and copyOfMyNumber. Here, I get the expected behavior: different values for myNumber and copyOfMyNumber.
What is the difference between an array and a number in JavaScript that it seems changing an array changes the value of a copy of the array, where as changing a number does not change the value of a copy of the number?
I'm guessing that for some reason, the array is referred to by reference and the number by value, but why? How can I know what behavior to expect with other objects?
An array in JavaScript is also an object and variables only hold a reference to an object, not the object itself. Thus both variables have a reference to the same object.
Your comparison with the number example is not correct btw. You assign a new value to copyOfMyNumber. If you assign a new value to copyOfMyArray it will not change myArray either.
You can create a copy of an array using slice [docs]:
var copyOfMyArray = myArray.slice(0);
But note that this only returns a shallow copy, i.e. objects inside the array will not be cloned.
Well, the only possible answer — and the correct one — is that you're not actually copying the array. When you write
var copyOfArray = array;
you're assigning a reference to the same array into another variable. They're both pointing at the same object, in other words.
So everyone here has done a great job of explaining why this is happening - I just wanted to drop a line and let you know how I was able to fix this - pretty easily:
thingArray = ['first_thing', 'second_thing', 'third_thing']
function removeFirstThingAndPreserveArray(){
var copyOfThingArray = [...thingArray]
copyOfThingArray.shift();
return copyOfThingArray;
}
This is using the ... spread syntax.
Spread Syntax Source
EDIT: As to the why of this, and to answer your question:
What is the difference between an array and a number in JavaScript that it seems changing an array changes the value of a copy of the array, where as changing a number does not change the value of a copy of the number?
The answer is that in JavaScript, arrays and objects are mutable, while strings and numbers and other primitives are immutable. When we do an assignment like:
var myArray = ['a', 'b', 'c'];
var copyOfMyArray = myArray;
copyOfMyArray is really just a reference to myArray, not an actual copy.
I would recommend this article, What are immutable and mutable data structures?, to dig deeper into the subject.
MDN Glossary: Mutable
Cloning objects -
A loop / array.push produces a similar result to array.slice(0) or array.clone(). Values are all passed by reference, but since most primitive data types are immutable, subsequent operations produce the desired result - a 'clone'. This is not true of objects and arrays, of course, which allow for modification of the original reference (they are mutable types).
Take the following example:
const originalArray = [1, 'a', false, {foor: 'bar'}]
const newArray = [];
originalArray.forEach((v, i) => {
newArray.push(originalArray[i]);
});
newArray[0] = newArray[0] + 1;
newArray[1] = 'b';
newArray[2] = true;
newArray[3] = Object.assign(newArray[3], {bar: 'foo'});
The operations run on the newArray indices all produce the desired result, except the final (object), which, because it is copied by reference, will mutate the originalArray[3] as well.
https://jsfiddle.net/7ajz2m6w/
Note that array.slice(0) and array.clone() suffers from this same limitation.
One way to solve this is by effectively cloning the object during the push sequence:
originalArray.forEach((v, i) => {
const val = (typeof v === 'object') ? Object.assign({}, v) : v;
newArray.push(val);
});
https://jsfiddle.net/e5hmnjp0/
cheers
The issue with shallow copy is that all the objects aren't cloned, instead it get reference.So array.slice(0) will work fine only with literal array, but it will not do shallow copy with object array. In that case one way is..
var firstArray = [{name: 'foo', id: 121}, {name: 'zoo', id: 321}];
var clonedArray = firstArray.map((_arrayElement) => Object.assign({}, _arrayElement));
console.log(clonedArray);
// [{name: 'foo', id: 121}, {name: 'zoo', id: 321}] // shallow copy
In JS, operator "=" copy the pointer to the memory area of the array.
If you want to copy an array into another you have to use the Clone function.
For integers is different because they are a primitive type.
S.
Create a filter of the original array in the arrayCopy. So that changes to the new array won't affect original array.
var myArray = ['a', 'b', 'c'];
var arrayCopy = myArray.filter(function(f){return f;})
arrayCopy.splice(0, 1);
alert(myArray); // alerts ['a','b','c']
alert(arrayCopy); // alerts ['b','c']
Hope it helps.
Everything is copied by reference except primitive data types (strings and numbers IIRC).
You don't have any copies.
You have multiple variables holding the same array.
Similarly, you have multiple variables holding the same number.
When you write copyOfMyNumber = ..., you're putting a new number into the variable.
That's like writing copyOfMyArray = ....
When you write copyOfMyArray.splice, you're modifying the original array.
That isn't possible with numbers because numbers are immutable and cannot be modified,
You can add some error handling depending on your cases and use something similar to the following function to solve the issue. Please comment for any bugs / issues / efficiency ideas.
function CopyAnArray (ari1) {
var mxx4 = [];
for (var i=0;i<ari1.length;i++) {
var nads2 = [];
for (var j=0;j<ari1[0].length;j++) {
nads2.push(ari1[i][j]);
}
mxx4.push(nads2);
}
return mxx4;
}
An array, or an object in javascript always holds the same reference unless you clone or copy. Here is an exmaple:
http://plnkr.co/edit/Bqvsiddke27w9nLwYhcl?p=preview
// for showing that objects in javascript shares the same reference
var obj = {
"name": "a"
}
var arr = [];
//we push the same object
arr.push(obj);
arr.push(obj);
//if we change the value for one object
arr[0].name = "b";
//the other object also changes
alert(arr[1].name);
For object clone, we can use .clone() in jquery and angular.copy(), these functions will create new object with other reference. If you know more functions to do that, please tell me, thanks!
For Arrays with objects you can change the use JSON.parse and JSON.strinigfy to change the type of the array to an object refrence to a string and then back to a array without having to worry about the original array
var array = [{name:'John', age:34, logs:'[]'}, {name:'David', age:43, logs:'[3]'}];
var array2 = JSON.parse(JSON.stringify(array)); // turn object to function output
array2[0].age++;
alert(JSON.stringify(array));
alert(JSON.stringify(array2));

Array Contains Multiple Instances of Same Array When Pushing Array Javascript [duplicate]

I've written the following JavaScript:
var myArray = ['a', 'b', 'c'];
var copyOfMyArray = myArray;
copyOfMyArray.splice(0, 1);
alert(myArray); // alerts ['b','c']
alert(copyOfMyArray); // alerts ['b','c']
var myNumber = 5;
var copyOfMyNumber = myNumber;
copyOfMyNumber = copyOfMyNumber - 1;
alert(myNumber); // alerts 5
alert(copyOfMyNumber); // alerts 4
This code declares a variable myArray and sets it to an array value. It then declares a second variable copyOfMyArray and sets it to myArray.
It performs an operation on copyOfMyArray and then alerts both myArray and copyOfMyArray. Somehow, when I perform an operation on copyOfMyArray, it appears that the same operation is performed on myArray.
The code then does the same thing with a number value: It declares a variable myNumber and sets it to a number value. It then declares a second variable copyOfMyNumber and sets it to myNumber. It performs an operation on copyOfMyNumber and then alerts both myNumber and copyOfMyNumber. Here, I get the expected behavior: different values for myNumber and copyOfMyNumber.
What is the difference between an array and a number in JavaScript that it seems changing an array changes the value of a copy of the array, where as changing a number does not change the value of a copy of the number?
I'm guessing that for some reason, the array is referred to by reference and the number by value, but why? How can I know what behavior to expect with other objects?
An array in JavaScript is also an object and variables only hold a reference to an object, not the object itself. Thus both variables have a reference to the same object.
Your comparison with the number example is not correct btw. You assign a new value to copyOfMyNumber. If you assign a new value to copyOfMyArray it will not change myArray either.
You can create a copy of an array using slice [docs]:
var copyOfMyArray = myArray.slice(0);
But note that this only returns a shallow copy, i.e. objects inside the array will not be cloned.
Well, the only possible answer — and the correct one — is that you're not actually copying the array. When you write
var copyOfArray = array;
you're assigning a reference to the same array into another variable. They're both pointing at the same object, in other words.
So everyone here has done a great job of explaining why this is happening - I just wanted to drop a line and let you know how I was able to fix this - pretty easily:
thingArray = ['first_thing', 'second_thing', 'third_thing']
function removeFirstThingAndPreserveArray(){
var copyOfThingArray = [...thingArray]
copyOfThingArray.shift();
return copyOfThingArray;
}
This is using the ... spread syntax.
Spread Syntax Source
EDIT: As to the why of this, and to answer your question:
What is the difference between an array and a number in JavaScript that it seems changing an array changes the value of a copy of the array, where as changing a number does not change the value of a copy of the number?
The answer is that in JavaScript, arrays and objects are mutable, while strings and numbers and other primitives are immutable. When we do an assignment like:
var myArray = ['a', 'b', 'c'];
var copyOfMyArray = myArray;
copyOfMyArray is really just a reference to myArray, not an actual copy.
I would recommend this article, What are immutable and mutable data structures?, to dig deeper into the subject.
MDN Glossary: Mutable
Cloning objects -
A loop / array.push produces a similar result to array.slice(0) or array.clone(). Values are all passed by reference, but since most primitive data types are immutable, subsequent operations produce the desired result - a 'clone'. This is not true of objects and arrays, of course, which allow for modification of the original reference (they are mutable types).
Take the following example:
const originalArray = [1, 'a', false, {foor: 'bar'}]
const newArray = [];
originalArray.forEach((v, i) => {
newArray.push(originalArray[i]);
});
newArray[0] = newArray[0] + 1;
newArray[1] = 'b';
newArray[2] = true;
newArray[3] = Object.assign(newArray[3], {bar: 'foo'});
The operations run on the newArray indices all produce the desired result, except the final (object), which, because it is copied by reference, will mutate the originalArray[3] as well.
https://jsfiddle.net/7ajz2m6w/
Note that array.slice(0) and array.clone() suffers from this same limitation.
One way to solve this is by effectively cloning the object during the push sequence:
originalArray.forEach((v, i) => {
const val = (typeof v === 'object') ? Object.assign({}, v) : v;
newArray.push(val);
});
https://jsfiddle.net/e5hmnjp0/
cheers
The issue with shallow copy is that all the objects aren't cloned, instead it get reference.So array.slice(0) will work fine only with literal array, but it will not do shallow copy with object array. In that case one way is..
var firstArray = [{name: 'foo', id: 121}, {name: 'zoo', id: 321}];
var clonedArray = firstArray.map((_arrayElement) => Object.assign({}, _arrayElement));
console.log(clonedArray);
// [{name: 'foo', id: 121}, {name: 'zoo', id: 321}] // shallow copy
In JS, operator "=" copy the pointer to the memory area of the array.
If you want to copy an array into another you have to use the Clone function.
For integers is different because they are a primitive type.
S.
Create a filter of the original array in the arrayCopy. So that changes to the new array won't affect original array.
var myArray = ['a', 'b', 'c'];
var arrayCopy = myArray.filter(function(f){return f;})
arrayCopy.splice(0, 1);
alert(myArray); // alerts ['a','b','c']
alert(arrayCopy); // alerts ['b','c']
Hope it helps.
Everything is copied by reference except primitive data types (strings and numbers IIRC).
You don't have any copies.
You have multiple variables holding the same array.
Similarly, you have multiple variables holding the same number.
When you write copyOfMyNumber = ..., you're putting a new number into the variable.
That's like writing copyOfMyArray = ....
When you write copyOfMyArray.splice, you're modifying the original array.
That isn't possible with numbers because numbers are immutable and cannot be modified,
You can add some error handling depending on your cases and use something similar to the following function to solve the issue. Please comment for any bugs / issues / efficiency ideas.
function CopyAnArray (ari1) {
var mxx4 = [];
for (var i=0;i<ari1.length;i++) {
var nads2 = [];
for (var j=0;j<ari1[0].length;j++) {
nads2.push(ari1[i][j]);
}
mxx4.push(nads2);
}
return mxx4;
}
An array, or an object in javascript always holds the same reference unless you clone or copy. Here is an exmaple:
http://plnkr.co/edit/Bqvsiddke27w9nLwYhcl?p=preview
// for showing that objects in javascript shares the same reference
var obj = {
"name": "a"
}
var arr = [];
//we push the same object
arr.push(obj);
arr.push(obj);
//if we change the value for one object
arr[0].name = "b";
//the other object also changes
alert(arr[1].name);
For object clone, we can use .clone() in jquery and angular.copy(), these functions will create new object with other reference. If you know more functions to do that, please tell me, thanks!
For Arrays with objects you can change the use JSON.parse and JSON.strinigfy to change the type of the array to an object refrence to a string and then back to a array without having to worry about the original array
var array = [{name:'John', age:34, logs:'[]'}, {name:'David', age:43, logs:'[3]'}];
var array2 = JSON.parse(JSON.stringify(array)); // turn object to function output
array2[0].age++;
alert(JSON.stringify(array));
alert(JSON.stringify(array2));

How to use JavaScript spread syntax on array and object together?

I'm solving a problem where the task is to merge multiple objects from the input array and return a single object with all the keys and values merged. In case a key exists in one or more objects in the input array, the most recent value for that key should be stored in the final returned object.
An example -
var a={1:'1',2:'2',3:'3'},
b={3:'4',5:'6',6:'7',7:'8'},
c={5:'9',8:'9',6:'12',23:'35'}
o=[a,b,c];
The returned object should be -
{ '1': '1','2': '2','3': '4','5': '9','6': '12','7': '8','8': '9','23':'35' }
As an example of the duplicate key case, key 3 exists in both objects a and b, so in the final result the value is {3:'4'} because that is the most recent.
I want to use the spread syntax for Objects in this problem, but that solves only a part of it. What I mean is, if I use spread for each object individually, that works, like -
function foo() {
var a={'1':'1','2':'2','3':'3'},
b={'3':'4','5':'6','6':'7','7':'8'},
c={'5':'9','8':'9','6':'12','23':'35'},
arr = [a,b,c];
return {...arr[0], ...arr[1], ...arr[2]}
}
console.log(foo());
For this to work, I need to write out each array element with spread, as in the above snippet - ...arr[0], ...arr[1], ...arr[2]. However, the input array can contain any number of objects, so writing out each element is not feasible.
Normally, using spread on an iterable like an array, allows you to expand the array elements, like so-
var parts = ['shoulders', 'knees'];
var lyrics = ['head', ...parts, 'and', 'toes'];
console.log(lyrics)
Is it possible to use spread on the input array to collect all the individual objects, on which spread can be applied again, to get the final object?
You could just use Object.assign():
return Object.assign(...arr);
Here's a complete snippet:
function foo() {
var a={'1':'1','2':'2','3':'3'},
b={'3':'4','5':'6','6':'7','7':'8'},
c={'5':'9','8':'9','6':'12','23':'35'},
arr = [a,b,c];
return Object.assign(...arr);
}
console.log(foo());
Note that this implicitly modifies the first object in your array. If you don't want that, pass a new empty object as the first argument to Object.assign():
return Object.assign({}, ...arr);
You can directly use spread syntax on object to merge them.
function foo() {
var a={'1':'1','2':'2','3':'3'},
b={'3':'4','5':'6','6':'7','7':'8'},
c={'5':'9','8':'9','6':'12','23':'35'};
return {...a, ...b, ...c};
}
console.log(foo());
If you have an array, then you can use array#reduce and Object#assign.
var a={'1':'1','2':'2','3':'3'},
b={'3':'4','5':'6','6':'7','7':'8'},
c={'5':'9','8':'9','6':'12','23':'35'},
arr = [a,b,c];
var result = arr.reduce(function(r,o){
return Object.assign(r,o);
},Object.create(null));
console.log(result);
NOTE : The Rest/Spread Properties for ECMAScript proposal (stage 3) adds spread properties to object literals. It copies own enumerable properties from a provided object onto a new object.

What's is the difference between Array.of(n) , Array(n) and array = [n]?

As the title, i'm wondering what's the difference between this 3 methods of initialization an array.
I'm actually more interested in the new Array.of() method provided from ES6, why they feel the needs of implements that?
The Array constructor can be called in two ways: a list of values to be used as values for array elements, or with a single numeric value giving the initial length:
var myArray = new Array("hello", "world"); // 2 elements
var otherArray = new Array(100); // 100 elements, all empty
Because there's an ambiguity when just one number is passed, that old API is considered badly designed. Thus, there's Array.of(), which is the same as the first option for the Array constructor:
var otherArray = Array.of(100); // 1 element
The third way to make an array is with an array initialization expression:
var otherArray = [100]; // 1 element
The array instances that are created by each of the above are functionally equivalent and completely interchangeable.
One more thing: why does Array.of() have to exist, given that we can use the array initialization expression? Well, Array.of() is a function, so it can be used as a value applied in functional-style programming. You can (as a slightly dumb example) copy an array with:
var copy = Array.of.apply(Array, original);
One reason that's dumb is that there's also (in ES2015) Array.from() to do the same thing:
var copy = Array.from(original);
That works on any sort of iterable original, so it's a good way to turn arguments or a NodeList into an array.
The MDN site has documentation on Array.of(). The constructor and the array initializer form have been around since forever, so any JavaScript reference will cover those (though possibly without reference to Array.of()).
Array.of(2) will create an array with the element 2.
var temp = Array.of(2); // temp = [2]
Array(2) will create an array of 2 elements.
var temp = new Array(2); // temp = [undefined, undefined]
temp = [2] will create an array with the element 2.
var temp = [2]; // temp = [2]

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