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Strange behavior of an array filled by Array.prototype.fill()
Create ArrayList from arrayHow do I check if an array includes an object in JavaScript?How to append something to an array?Deleting an element from an array in PHPHow to insert an item into an array at a specific index (JavaScript)?How do I empty an array in JavaScript?Loop through an array in JavaScriptHow to check if an object is an array?How do I remove a particular element from an array in JavaScript?For-each over an array in JavaScript?
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I face something I don't understand with an array. Indeed, I created an array I have filled with empty subArrays to obtain a 2D Matrix.
But when I manipulate the array it doesn't behave as I expected.
var arr = new Array(5);
arr.fill([]);
arr[2].push("third rank item");
console.log(arr);
//[ [ 'third rank item' ],
// [ 'third rank item' ],
// [ 'third rank item' ],
// [ 'third rank item' ],
// [ 'third rank item' ] ]
Every lights on this matter will be welcomed
javascript arrays
add a comment |
I face something I don't understand with an array. Indeed, I created an array I have filled with empty subArrays to obtain a 2D Matrix.
But when I manipulate the array it doesn't behave as I expected.
var arr = new Array(5);
arr.fill([]);
arr[2].push("third rank item");
console.log(arr);
//[ [ 'third rank item' ],
// [ 'third rank item' ],
// [ 'third rank item' ],
// [ 'third rank item' ],
// [ 'third rank item' ] ]
Every lights on this matter will be welcomed
javascript arrays
How do you expect it to behave?
– Feathercrown
Dec 13 '16 at 13:12
2
You're filling with the same empty array. You could instead tryArray.from(new Array(5), () => [])
.
– user663031
Dec 13 '16 at 13:12
@torazaburo Can you explain it briefly? OP targeted the 3rd item in the outer array and pushed a string that results in pushing the string in all the internal arrays. Bit confusing.
– Rajaprabhu Aravindasamy
Dec 13 '16 at 13:14
All the internal arrays are the same array.
– user663031
Dec 13 '16 at 13:16
All the internal arrays are the same array.
– user663031
Dec 13 '16 at 13:16
add a comment |
I face something I don't understand with an array. Indeed, I created an array I have filled with empty subArrays to obtain a 2D Matrix.
But when I manipulate the array it doesn't behave as I expected.
var arr = new Array(5);
arr.fill([]);
arr[2].push("third rank item");
console.log(arr);
//[ [ 'third rank item' ],
// [ 'third rank item' ],
// [ 'third rank item' ],
// [ 'third rank item' ],
// [ 'third rank item' ] ]
Every lights on this matter will be welcomed
javascript arrays
I face something I don't understand with an array. Indeed, I created an array I have filled with empty subArrays to obtain a 2D Matrix.
But when I manipulate the array it doesn't behave as I expected.
var arr = new Array(5);
arr.fill([]);
arr[2].push("third rank item");
console.log(arr);
//[ [ 'third rank item' ],
// [ 'third rank item' ],
// [ 'third rank item' ],
// [ 'third rank item' ],
// [ 'third rank item' ] ]
Every lights on this matter will be welcomed
javascript arrays
javascript arrays
asked Dec 13 '16 at 13:09
kevin ternetkevin ternet
2,634717
2,634717
How do you expect it to behave?
– Feathercrown
Dec 13 '16 at 13:12
2
You're filling with the same empty array. You could instead tryArray.from(new Array(5), () => [])
.
– user663031
Dec 13 '16 at 13:12
@torazaburo Can you explain it briefly? OP targeted the 3rd item in the outer array and pushed a string that results in pushing the string in all the internal arrays. Bit confusing.
– Rajaprabhu Aravindasamy
Dec 13 '16 at 13:14
All the internal arrays are the same array.
– user663031
Dec 13 '16 at 13:16
All the internal arrays are the same array.
– user663031
Dec 13 '16 at 13:16
add a comment |
How do you expect it to behave?
– Feathercrown
Dec 13 '16 at 13:12
2
You're filling with the same empty array. You could instead tryArray.from(new Array(5), () => [])
.
– user663031
Dec 13 '16 at 13:12
@torazaburo Can you explain it briefly? OP targeted the 3rd item in the outer array and pushed a string that results in pushing the string in all the internal arrays. Bit confusing.
– Rajaprabhu Aravindasamy
Dec 13 '16 at 13:14
All the internal arrays are the same array.
– user663031
Dec 13 '16 at 13:16
All the internal arrays are the same array.
– user663031
Dec 13 '16 at 13:16
How do you expect it to behave?
– Feathercrown
Dec 13 '16 at 13:12
How do you expect it to behave?
– Feathercrown
Dec 13 '16 at 13:12
2
2
You're filling with the same empty array. You could instead try
Array.from(new Array(5), () => [])
.– user663031
Dec 13 '16 at 13:12
You're filling with the same empty array. You could instead try
Array.from(new Array(5), () => [])
.– user663031
Dec 13 '16 at 13:12
@torazaburo Can you explain it briefly? OP targeted the 3rd item in the outer array and pushed a string that results in pushing the string in all the internal arrays. Bit confusing.
– Rajaprabhu Aravindasamy
Dec 13 '16 at 13:14
@torazaburo Can you explain it briefly? OP targeted the 3rd item in the outer array and pushed a string that results in pushing the string in all the internal arrays. Bit confusing.
– Rajaprabhu Aravindasamy
Dec 13 '16 at 13:14
All the internal arrays are the same array.
– user663031
Dec 13 '16 at 13:16
All the internal arrays are the same array.
– user663031
Dec 13 '16 at 13:16
All the internal arrays are the same array.
– user663031
Dec 13 '16 at 13:16
All the internal arrays are the same array.
– user663031
Dec 13 '16 at 13:16
add a comment |
7 Answers
7
active
oldest
votes
This is the same old problem with arrays (and objects in general) being references rather than values.
Specifically, when you do arr.fill([])
, you are taking that one single empty array and using that to fill the parent one.
It's like saying:
var arr = new Array(5);
arr[0] = arr[1] = arr[2] = arr[3] = arr[4] = [];
They all refer to the same array! So when you then go on to modify one of them, it looks like they're all modified (but really it's still the same one)
Unfortunately there's no simple way to assign an empty array to each one. You could do something like:
Array.apply(null, Array(5)).map(function() return [];);
Essentially, create an (initialised) empty array of length 5 and map each (empty) value to a new []
.
EDIT: Seems like I'm stuck in old times. As per @torazaburo's comment, you can use Array.from
instead of Array.apply(null, Array(5)).map
, like so:
Array.from( new Array(5), function() return []; );
Thank's for the clear explanation and smart solution
– kevin ternet
Dec 13 '16 at 13:56
add a comment |
As you can notice using array.fill you're filling the array with a reference to the same array,
if you want to instantiate each array index to an empty array a normal while loop will do:
var arr = [];
var n = 5
while(n--)
arr[n] = []
arr[2].push("third rank item");
console.log(arr);
Option 2:
if you have lodash package available, you can also use _.map as this is specificaly designed to loop through a sparse array (native map will skip non init values)
var arr =_.map(new Array(5), (x => []))
arr[2].push("third rank item");
console.log(arr)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.2/lodash.min.js"></script>
Thank's for giving an extra solution with lodash
– kevin ternet
Dec 13 '16 at 13:54
add a comment |
The eleventh line of the ECMA doc of Array.prototype.fill is clearly giving the reason for the mystery.
Repeat, while k < final
Let Pk be ToString(k).
Let setStatus be Set(O, Pk, value, true).
ReturnIfAbrupt(setStatus).
Increase k by 1.
Here "value" is just a reference received. And they are setting it as a property to array directly. That means all the filled arrays are just reference to a single array object.
1
Thank's. You make me very well aware of the importance of reading documentation
– kevin ternet
Dec 13 '16 at 13:52
add a comment |
It's happens cause of reference. Array is a type of object and object works on their references when you fill your array with [] or new Array()
fill run only ones and put the same array in all indexes that's why when you update an sub-array all are updated.
Solution:
let arr = new Array(5).fill(0).map(ele => ele = []);
arr[2].push("something");
OR
let arr = Array.of([], [], [], []);
arr[2].push("something");
Result: as expected only 2 index of arr
is updated.
Quick to write and easy to understand. Thanks
– kevin ternet
Dec 13 '16 at 13:51
you are welcome.
– Umair Ahmed
Dec 13 '16 at 14:18
add a comment |
Try this ,this is quick solution for you in one line.
var arr = new Array(5);
arr = Array.from(arr, x => []);
arr[2].push("third rank item");
console.log(arr);
add a comment |
You can try this,
var arr = new Array(5);
var i = 0;
while (i < arr.length)
arr.fill([], i++);
arr[2].push("third rank item");
console.log(arr);
add a comment |
With ES6 I recommend this method to create 2 or multidimensional arrays:
// create an M x N dimension grid and fill it with 0's
const myArray = [...Array(M)].map(r => [...Array(N)].map(r => 0));
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is the same old problem with arrays (and objects in general) being references rather than values.
Specifically, when you do arr.fill([])
, you are taking that one single empty array and using that to fill the parent one.
It's like saying:
var arr = new Array(5);
arr[0] = arr[1] = arr[2] = arr[3] = arr[4] = [];
They all refer to the same array! So when you then go on to modify one of them, it looks like they're all modified (but really it's still the same one)
Unfortunately there's no simple way to assign an empty array to each one. You could do something like:
Array.apply(null, Array(5)).map(function() return [];);
Essentially, create an (initialised) empty array of length 5 and map each (empty) value to a new []
.
EDIT: Seems like I'm stuck in old times. As per @torazaburo's comment, you can use Array.from
instead of Array.apply(null, Array(5)).map
, like so:
Array.from( new Array(5), function() return []; );
Thank's for the clear explanation and smart solution
– kevin ternet
Dec 13 '16 at 13:56
add a comment |
This is the same old problem with arrays (and objects in general) being references rather than values.
Specifically, when you do arr.fill([])
, you are taking that one single empty array and using that to fill the parent one.
It's like saying:
var arr = new Array(5);
arr[0] = arr[1] = arr[2] = arr[3] = arr[4] = [];
They all refer to the same array! So when you then go on to modify one of them, it looks like they're all modified (but really it's still the same one)
Unfortunately there's no simple way to assign an empty array to each one. You could do something like:
Array.apply(null, Array(5)).map(function() return [];);
Essentially, create an (initialised) empty array of length 5 and map each (empty) value to a new []
.
EDIT: Seems like I'm stuck in old times. As per @torazaburo's comment, you can use Array.from
instead of Array.apply(null, Array(5)).map
, like so:
Array.from( new Array(5), function() return []; );
Thank's for the clear explanation and smart solution
– kevin ternet
Dec 13 '16 at 13:56
add a comment |
This is the same old problem with arrays (and objects in general) being references rather than values.
Specifically, when you do arr.fill([])
, you are taking that one single empty array and using that to fill the parent one.
It's like saying:
var arr = new Array(5);
arr[0] = arr[1] = arr[2] = arr[3] = arr[4] = [];
They all refer to the same array! So when you then go on to modify one of them, it looks like they're all modified (but really it's still the same one)
Unfortunately there's no simple way to assign an empty array to each one. You could do something like:
Array.apply(null, Array(5)).map(function() return [];);
Essentially, create an (initialised) empty array of length 5 and map each (empty) value to a new []
.
EDIT: Seems like I'm stuck in old times. As per @torazaburo's comment, you can use Array.from
instead of Array.apply(null, Array(5)).map
, like so:
Array.from( new Array(5), function() return []; );
This is the same old problem with arrays (and objects in general) being references rather than values.
Specifically, when you do arr.fill([])
, you are taking that one single empty array and using that to fill the parent one.
It's like saying:
var arr = new Array(5);
arr[0] = arr[1] = arr[2] = arr[3] = arr[4] = [];
They all refer to the same array! So when you then go on to modify one of them, it looks like they're all modified (but really it's still the same one)
Unfortunately there's no simple way to assign an empty array to each one. You could do something like:
Array.apply(null, Array(5)).map(function() return [];);
Essentially, create an (initialised) empty array of length 5 and map each (empty) value to a new []
.
EDIT: Seems like I'm stuck in old times. As per @torazaburo's comment, you can use Array.from
instead of Array.apply(null, Array(5)).map
, like so:
Array.from( new Array(5), function() return []; );
answered Dec 13 '16 at 13:16
Niet the Dark AbsolNiet the Dark Absol
262k57368480
262k57368480
Thank's for the clear explanation and smart solution
– kevin ternet
Dec 13 '16 at 13:56
add a comment |
Thank's for the clear explanation and smart solution
– kevin ternet
Dec 13 '16 at 13:56
Thank's for the clear explanation and smart solution
– kevin ternet
Dec 13 '16 at 13:56
Thank's for the clear explanation and smart solution
– kevin ternet
Dec 13 '16 at 13:56
add a comment |
As you can notice using array.fill you're filling the array with a reference to the same array,
if you want to instantiate each array index to an empty array a normal while loop will do:
var arr = [];
var n = 5
while(n--)
arr[n] = []
arr[2].push("third rank item");
console.log(arr);
Option 2:
if you have lodash package available, you can also use _.map as this is specificaly designed to loop through a sparse array (native map will skip non init values)
var arr =_.map(new Array(5), (x => []))
arr[2].push("third rank item");
console.log(arr)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.2/lodash.min.js"></script>
Thank's for giving an extra solution with lodash
– kevin ternet
Dec 13 '16 at 13:54
add a comment |
As you can notice using array.fill you're filling the array with a reference to the same array,
if you want to instantiate each array index to an empty array a normal while loop will do:
var arr = [];
var n = 5
while(n--)
arr[n] = []
arr[2].push("third rank item");
console.log(arr);
Option 2:
if you have lodash package available, you can also use _.map as this is specificaly designed to loop through a sparse array (native map will skip non init values)
var arr =_.map(new Array(5), (x => []))
arr[2].push("third rank item");
console.log(arr)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.2/lodash.min.js"></script>
Thank's for giving an extra solution with lodash
– kevin ternet
Dec 13 '16 at 13:54
add a comment |
As you can notice using array.fill you're filling the array with a reference to the same array,
if you want to instantiate each array index to an empty array a normal while loop will do:
var arr = [];
var n = 5
while(n--)
arr[n] = []
arr[2].push("third rank item");
console.log(arr);
Option 2:
if you have lodash package available, you can also use _.map as this is specificaly designed to loop through a sparse array (native map will skip non init values)
var arr =_.map(new Array(5), (x => []))
arr[2].push("third rank item");
console.log(arr)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.2/lodash.min.js"></script>
As you can notice using array.fill you're filling the array with a reference to the same array,
if you want to instantiate each array index to an empty array a normal while loop will do:
var arr = [];
var n = 5
while(n--)
arr[n] = []
arr[2].push("third rank item");
console.log(arr);
Option 2:
if you have lodash package available, you can also use _.map as this is specificaly designed to loop through a sparse array (native map will skip non init values)
var arr =_.map(new Array(5), (x => []))
arr[2].push("third rank item");
console.log(arr)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.2/lodash.min.js"></script>
var arr = [];
var n = 5
while(n--)
arr[n] = []
arr[2].push("third rank item");
console.log(arr);
var arr = [];
var n = 5
while(n--)
arr[n] = []
arr[2].push("third rank item");
console.log(arr);
var arr =_.map(new Array(5), (x => []))
arr[2].push("third rank item");
console.log(arr)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.2/lodash.min.js"></script>
var arr =_.map(new Array(5), (x => []))
arr[2].push("third rank item");
console.log(arr)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.2/lodash.min.js"></script>
edited Dec 13 '16 at 13:24
answered Dec 13 '16 at 13:17
maiomanmaioman
11.7k32034
11.7k32034
Thank's for giving an extra solution with lodash
– kevin ternet
Dec 13 '16 at 13:54
add a comment |
Thank's for giving an extra solution with lodash
– kevin ternet
Dec 13 '16 at 13:54
Thank's for giving an extra solution with lodash
– kevin ternet
Dec 13 '16 at 13:54
Thank's for giving an extra solution with lodash
– kevin ternet
Dec 13 '16 at 13:54
add a comment |
The eleventh line of the ECMA doc of Array.prototype.fill is clearly giving the reason for the mystery.
Repeat, while k < final
Let Pk be ToString(k).
Let setStatus be Set(O, Pk, value, true).
ReturnIfAbrupt(setStatus).
Increase k by 1.
Here "value" is just a reference received. And they are setting it as a property to array directly. That means all the filled arrays are just reference to a single array object.
1
Thank's. You make me very well aware of the importance of reading documentation
– kevin ternet
Dec 13 '16 at 13:52
add a comment |
The eleventh line of the ECMA doc of Array.prototype.fill is clearly giving the reason for the mystery.
Repeat, while k < final
Let Pk be ToString(k).
Let setStatus be Set(O, Pk, value, true).
ReturnIfAbrupt(setStatus).
Increase k by 1.
Here "value" is just a reference received. And they are setting it as a property to array directly. That means all the filled arrays are just reference to a single array object.
1
Thank's. You make me very well aware of the importance of reading documentation
– kevin ternet
Dec 13 '16 at 13:52
add a comment |
The eleventh line of the ECMA doc of Array.prototype.fill is clearly giving the reason for the mystery.
Repeat, while k < final
Let Pk be ToString(k).
Let setStatus be Set(O, Pk, value, true).
ReturnIfAbrupt(setStatus).
Increase k by 1.
Here "value" is just a reference received. And they are setting it as a property to array directly. That means all the filled arrays are just reference to a single array object.
The eleventh line of the ECMA doc of Array.prototype.fill is clearly giving the reason for the mystery.
Repeat, while k < final
Let Pk be ToString(k).
Let setStatus be Set(O, Pk, value, true).
ReturnIfAbrupt(setStatus).
Increase k by 1.
Here "value" is just a reference received. And they are setting it as a property to array directly. That means all the filled arrays are just reference to a single array object.
answered Dec 13 '16 at 13:26
Rajaprabhu AravindasamyRajaprabhu Aravindasamy
56.3k1269104
56.3k1269104
1
Thank's. You make me very well aware of the importance of reading documentation
– kevin ternet
Dec 13 '16 at 13:52
add a comment |
1
Thank's. You make me very well aware of the importance of reading documentation
– kevin ternet
Dec 13 '16 at 13:52
1
1
Thank's. You make me very well aware of the importance of reading documentation
– kevin ternet
Dec 13 '16 at 13:52
Thank's. You make me very well aware of the importance of reading documentation
– kevin ternet
Dec 13 '16 at 13:52
add a comment |
It's happens cause of reference. Array is a type of object and object works on their references when you fill your array with [] or new Array()
fill run only ones and put the same array in all indexes that's why when you update an sub-array all are updated.
Solution:
let arr = new Array(5).fill(0).map(ele => ele = []);
arr[2].push("something");
OR
let arr = Array.of([], [], [], []);
arr[2].push("something");
Result: as expected only 2 index of arr
is updated.
Quick to write and easy to understand. Thanks
– kevin ternet
Dec 13 '16 at 13:51
you are welcome.
– Umair Ahmed
Dec 13 '16 at 14:18
add a comment |
It's happens cause of reference. Array is a type of object and object works on their references when you fill your array with [] or new Array()
fill run only ones and put the same array in all indexes that's why when you update an sub-array all are updated.
Solution:
let arr = new Array(5).fill(0).map(ele => ele = []);
arr[2].push("something");
OR
let arr = Array.of([], [], [], []);
arr[2].push("something");
Result: as expected only 2 index of arr
is updated.
Quick to write and easy to understand. Thanks
– kevin ternet
Dec 13 '16 at 13:51
you are welcome.
– Umair Ahmed
Dec 13 '16 at 14:18
add a comment |
It's happens cause of reference. Array is a type of object and object works on their references when you fill your array with [] or new Array()
fill run only ones and put the same array in all indexes that's why when you update an sub-array all are updated.
Solution:
let arr = new Array(5).fill(0).map(ele => ele = []);
arr[2].push("something");
OR
let arr = Array.of([], [], [], []);
arr[2].push("something");
Result: as expected only 2 index of arr
is updated.
It's happens cause of reference. Array is a type of object and object works on their references when you fill your array with [] or new Array()
fill run only ones and put the same array in all indexes that's why when you update an sub-array all are updated.
Solution:
let arr = new Array(5).fill(0).map(ele => ele = []);
arr[2].push("something");
OR
let arr = Array.of([], [], [], []);
arr[2].push("something");
Result: as expected only 2 index of arr
is updated.
edited Jan 20 '17 at 12:40
answered Dec 13 '16 at 13:45
Umair AhmedUmair Ahmed
2,64021727
2,64021727
Quick to write and easy to understand. Thanks
– kevin ternet
Dec 13 '16 at 13:51
you are welcome.
– Umair Ahmed
Dec 13 '16 at 14:18
add a comment |
Quick to write and easy to understand. Thanks
– kevin ternet
Dec 13 '16 at 13:51
you are welcome.
– Umair Ahmed
Dec 13 '16 at 14:18
Quick to write and easy to understand. Thanks
– kevin ternet
Dec 13 '16 at 13:51
Quick to write and easy to understand. Thanks
– kevin ternet
Dec 13 '16 at 13:51
you are welcome.
– Umair Ahmed
Dec 13 '16 at 14:18
you are welcome.
– Umair Ahmed
Dec 13 '16 at 14:18
add a comment |
Try this ,this is quick solution for you in one line.
var arr = new Array(5);
arr = Array.from(arr, x => []);
arr[2].push("third rank item");
console.log(arr);
add a comment |
Try this ,this is quick solution for you in one line.
var arr = new Array(5);
arr = Array.from(arr, x => []);
arr[2].push("third rank item");
console.log(arr);
add a comment |
Try this ,this is quick solution for you in one line.
var arr = new Array(5);
arr = Array.from(arr, x => []);
arr[2].push("third rank item");
console.log(arr);
Try this ,this is quick solution for you in one line.
var arr = new Array(5);
arr = Array.from(arr, x => []);
arr[2].push("third rank item");
console.log(arr);
var arr = new Array(5);
arr = Array.from(arr, x => []);
arr[2].push("third rank item");
console.log(arr);
var arr = new Array(5);
arr = Array.from(arr, x => []);
arr[2].push("third rank item");
console.log(arr);
answered Dec 31 '16 at 6:34
vijayvijay
881515
881515
add a comment |
add a comment |
You can try this,
var arr = new Array(5);
var i = 0;
while (i < arr.length)
arr.fill([], i++);
arr[2].push("third rank item");
console.log(arr);
add a comment |
You can try this,
var arr = new Array(5);
var i = 0;
while (i < arr.length)
arr.fill([], i++);
arr[2].push("third rank item");
console.log(arr);
add a comment |
You can try this,
var arr = new Array(5);
var i = 0;
while (i < arr.length)
arr.fill([], i++);
arr[2].push("third rank item");
console.log(arr);
You can try this,
var arr = new Array(5);
var i = 0;
while (i < arr.length)
arr.fill([], i++);
arr[2].push("third rank item");
console.log(arr);
var arr = new Array(5);
var i = 0;
while (i < arr.length)
arr.fill([], i++);
arr[2].push("third rank item");
console.log(arr);
var arr = new Array(5);
var i = 0;
while (i < arr.length)
arr.fill([], i++);
arr[2].push("third rank item");
console.log(arr);
edited Dec 31 '16 at 6:35
answered Dec 31 '16 at 6:08
vijayvijay
881515
881515
add a comment |
add a comment |
With ES6 I recommend this method to create 2 or multidimensional arrays:
// create an M x N dimension grid and fill it with 0's
const myArray = [...Array(M)].map(r => [...Array(N)].map(r => 0));
add a comment |
With ES6 I recommend this method to create 2 or multidimensional arrays:
// create an M x N dimension grid and fill it with 0's
const myArray = [...Array(M)].map(r => [...Array(N)].map(r => 0));
add a comment |
With ES6 I recommend this method to create 2 or multidimensional arrays:
// create an M x N dimension grid and fill it with 0's
const myArray = [...Array(M)].map(r => [...Array(N)].map(r => 0));
With ES6 I recommend this method to create 2 or multidimensional arrays:
// create an M x N dimension grid and fill it with 0's
const myArray = [...Array(M)].map(r => [...Array(N)].map(r => 0));
edited Mar 23 at 1:28
Nino Filiu
3,35341631
3,35341631
answered Mar 23 at 1:06
Joseph WonJoseph Won
1
1
add a comment |
add a comment |
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How do you expect it to behave?
– Feathercrown
Dec 13 '16 at 13:12
2
You're filling with the same empty array. You could instead try
Array.from(new Array(5), () => [])
.– user663031
Dec 13 '16 at 13:12
@torazaburo Can you explain it briefly? OP targeted the 3rd item in the outer array and pushed a string that results in pushing the string in all the internal arrays. Bit confusing.
– Rajaprabhu Aravindasamy
Dec 13 '16 at 13:14
All the internal arrays are the same array.
– user663031
Dec 13 '16 at 13:16
All the internal arrays are the same array.
– user663031
Dec 13 '16 at 13:16