Note: this isn't one of the millions dups of the common array copy 'problem' where using arr.slice(0) fixes this 'problem'
That said, I want to understand why I am getting this unexpected result:
var oldArr = [[1,2],[3,4]];
var find = oldArr[1];
var newArr = oldArr.slice(0);
console.log(newArr.indexOf(find)); //1?
//proof that newArr is NOT referenced to oldArr
newArr[0] = "Hi";
newArr[1] = "How are you?";
console.log(oldArr+" "+newArr); //"1,2,3,4 Hi,How are you?"
jsFildde Demo
If you replace find with any of the following alternatives, it returns the expected -1:
Use [3,4] directly
Use a variable holding [3,4]
Use a variable with reference of another array holding [3,4]
I can't find any explanation on why there is any difference between these last three methods and the first example. As far as I know, there shouldn't be any.
Any ideas?
In:
[[1,2],[3,4]]
There are three array objects created.
Only the outer one is being slice'd. This results in a "shallow copy".
Consider this:
var a = [1,2]
var b = [3,4]
var c = [a,b]
var d = c.slice(0)
d[0] === a // true, which means it is the /same/ object
d[0][0] = "Hi!"
a // ["Hi!", 2]
Happy coding!
Everything is working as expected
indexOf searches by reference and not value, therefore if two objects do not === each other, indexOf will not find them
In your case:
var oldArr = [[1,2],[3,4]];
var find = oldArr[1];
var newArr = oldArr.slice(0);
find points to the array [3,4];
so newArray contains 2 objects (the two arrays).
If you want to see what is really going on try doing find.push(5)
Related
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));
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));
I am making an app in which an algorithm is run on an array. Because the contents of the array will change during the execution of the algorithm, I'm storing the array contents to another array beforehand - performing the 'if' statements on the source array but updating the temporary array, then equating them afterwards.
The problem is that after running the algorithm, the two arrays are still identical. It seems that updating the temporary array automatically updates the source array.
I've created this jsfiddle to demonstrate:
var a = new Array( 0 , 1 , 2 );
var b = a;
b[1]=3;
document.write( (a[1]==b[1]) );
//Should show 'false' as this will not be correct
The code above returns "True". Is this normal behaviour? How can I overcome this?
That is normal behaviour, when copying array you are actually making a reference, so when you do var b = a that means instead of copying value you are just copying the reference
if you want to shallow copy an array (array with only one depth level) then you can use simple method like this:
var b = a.slice(0);
But if you want to deep copy an array (array with 2 or more depth level) then you can use below method for that:
var objClone = function(srcObj) {
var destObj = (this instanceof Array) ? [] : {};
for (var i in srcObj)
{
if (srcObj[i] && typeof srcObj[i] == "object") {
destObj[i] = objClone(srcObj[i]);
} else {
destObj[i] = srcObj[i];
}
}
return destObj;
};
For use of both of these methods check this jsFiddle
b = a doesn't copy the array, but a reference. Use b = a.slice() to make a copy.
See also ...
As others have mentioned a = b does not copy the array. However, if you do end up using the .slice() method to copy arrays please note that objects within an array will be copied by reference.
Take the following example
var obj = { greeting: "hello world" };
var a = [ obj ];
var b = a;
a[0].greeting = "foo bar";
b[0].greeting // => "foo bar";
Other than that you are good to go :)
Can some one explain the conceptual difference between both of them. Read somewhere that the second one creates a new array by destroying all references to the existing array and the .length=0 just empties the array. But it didn't work in my case
//Declaration
var arr = new Array();
The below one is the looping code that executes again and again.
$("#dummy").load("something.php",function(){
arr.length =0;// expected to empty the array
$("div").each(function(){
arr = arr + $(this).html();
});
});
But if I replace the code with arr =[] in place of arr.length=0 it works fine. Can anyone explain what's happening here.
foo = [] creates a new array and assigns a reference to it to a variable. Any other references are unaffected and still point to the original array.
foo.length = 0 modifies the array itself. If you access it via a different variable, then you still get the modified array.
Read somewhere that the second one creates a new array by destroying all references to the existing array
That is backwards. It creates a new array and doesn't destroy other references.
var foo = [1,2,3];
var bar = [1,2,3];
var foo2 = foo;
var bar2 = bar;
foo = [];
bar.length = 0;
console.log(foo, bar, foo2, bar2);
gives:
[] [] [1, 2, 3] []
arr.length =0;// expected to empty the array
and it does empty the array, at least the first time. After the first time you do this:
arr = arr + $(this).html();
… which overwrites the array with a string.
The length property of a string is read-only, so assigning 0 to it has no effect.
The difference here is best demonstrated in the following example:
var arrayA = [1,2,3,4,5];
function clearUsingLength (ar) {
ar.length = 0;
}
function clearByOverwriting(ar) {
ar = [];
}
alert("Original Length: " + arrayA.length);
clearByOverwriting(arrayA);
alert("After Overwriting: " + arrayA.length);
clearUsingLength(arrayA);
alert("After Using Length: " + arrayA.length);
Of which a live demo can be seen here: http://www.jsfiddle.net/8Yn7e/
When you set a variable that points to an existing array to point to a new array, all you are doing is breaking the link the variable has to that original array.
When you use array.length = 0 (and other methods like array.splice(0, array.length) for instance), you are actually emptying the original array.
Are you sure it really works?
I did a little experiment here, and trying to "add" an Array with a String resulted in a string.
function xyz(){
var a = [];
alert(typeof(a+$("#first").html()));
// shows "string"
}
http://www.jsfiddle.net/4nKCF/
(tested in Opera 11)
I have some js arrays and I want to find the arrays by name. for example:
arr1 = [1,2,3];
arr2 = [3,4,5];
And access them like this:
var intNum = 2;
var arrName = 'arr' + intNum;
arrName[0]; // equals 3
Is this possible?
Thanks,
Kevin
It is possible but I would suggest you place these arrays as properties of an object, that makes it a lot easier
var arrays {
arr1 : [1,2,3],
arr2 : [4,5,6]
}
var arrNum = 2;
var arr = arrays["arr" + arrNum] // arrays.arr2
Properties of objects can be accessed both using the . operator and as named items using the ["propname"] notation.
Using eval or resorting to using the above trick on window is not advised.
Eval'ing is normally a sign of ill constructed code, and using window relies on window being the Variable Object of the global scope - this is not part of any spec and will not necessarily work across browsers.
window['arr'+intNum]
so
arr1 = [1,2,3];
arr2 = [3,4,5];
intNum=2;
alert(window['arr'+intNum][1]); //will alert 4