How come if a = b, then b = a? [duplicate]

-4

This question already has an answer here:

• 
  Copy array by value
  
  33 answers
  
  

Let me explain. I have the following piece of code :

let a = [1, 2, 3];

let b = a;

b.splice(2, 1);

console.log("a: " + a);

console.log("b: " + b);

I would have expected to get something like :

a = [1, 2, 3];

b = [1, 2];

However, after running the code, it turns out both "a" and "b" equal [1, 2]. I am really confused, since "b" should only have been assigned a copy of "a", instead of acting as some sort of a pointer to "a". I wonder if it is because JS handles arrays (objects) differently of if it is something specific to the splice function. I would also like to know how you would bypass that odd behavior.

Thanks for your time.

javascript arrays

share|improve this question

asked Jan 2 at 15:12

John KrakovJohn Krakov

705

marked as duplicate by Denys Séguret javascript
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Jan 2 at 15:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  b references a, it is not a copy. Someone will find the dupe.
  
  – epascarello
  Jan 2 at 15:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  "since "b" should only have been assigned a copy of "a" Why ? Where are you copying ?
  
  – Denys Séguret
  Jan 2 at 15:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  BTW copying would have been let b = a.slice();
  
  – Denys Séguret
  Jan 2 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  duplicate stackoverflow.com/a/6605700/4267015
  
  – Naor Tedgi
  Jan 2 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DenysSéguret what do you mean?
  
  – John Krakov
  Jan 2 at 15:32
  

  
  
  
  

 | 
show 4 more comments

-4

This question already has an answer here:

• 
  Copy array by value
  
  33 answers
  
  

Let me explain. I have the following piece of code :

let a = [1, 2, 3];

let b = a;

b.splice(2, 1);

console.log("a: " + a);

console.log("b: " + b);

I would have expected to get something like :

a = [1, 2, 3];

b = [1, 2];

However, after running the code, it turns out both "a" and "b" equal [1, 2]. I am really confused, since "b" should only have been assigned a copy of "a", instead of acting as some sort of a pointer to "a". I wonder if it is because JS handles arrays (objects) differently of if it is something specific to the splice function. I would also like to know how you would bypass that odd behavior.

Thanks for your time.

javascript arrays

share|improve this question

asked Jan 2 at 15:12

John KrakovJohn Krakov

705

marked as duplicate by Denys Séguret javascript
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Jan 2 at 15:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  b references a, it is not a copy. Someone will find the dupe.
  
  – epascarello
  Jan 2 at 15:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  "since "b" should only have been assigned a copy of "a" Why ? Where are you copying ?
  
  – Denys Séguret
  Jan 2 at 15:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  BTW copying would have been let b = a.slice();
  
  – Denys Séguret
  Jan 2 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  duplicate stackoverflow.com/a/6605700/4267015
  
  – Naor Tedgi
  Jan 2 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DenysSéguret what do you mean?
  
  – John Krakov
  Jan 2 at 15:32
  

  
  
  
  

 | 
show 4 more comments

-4

-4

-4

0

This question already has an answer here:

• 
  Copy array by value
  
  33 answers
  
  

Let me explain. I have the following piece of code :

let a = [1, 2, 3];

let b = a;

b.splice(2, 1);

console.log("a: " + a);

console.log("b: " + b);

I would have expected to get something like :

a = [1, 2, 3];

b = [1, 2];

However, after running the code, it turns out both "a" and "b" equal [1, 2]. I am really confused, since "b" should only have been assigned a copy of "a", instead of acting as some sort of a pointer to "a". I wonder if it is because JS handles arrays (objects) differently of if it is something specific to the splice function. I would also like to know how you would bypass that odd behavior.

Thanks for your time.

javascript arrays

share|improve this question

asked Jan 2 at 15:12

John KrakovJohn Krakov

705

This question already has an answer here:

• 
  Copy array by value
  
  33 answers
  
  

Let me explain. I have the following piece of code :

let a = [1, 2, 3];

let b = a;

b.splice(2, 1);

console.log("a: " + a);

console.log("b: " + b);

I would have expected to get something like :

a = [1, 2, 3];

b = [1, 2];

However, after running the code, it turns out both "a" and "b" equal [1, 2]. I am really confused, since "b" should only have been assigned a copy of "a", instead of acting as some sort of a pointer to "a". I wonder if it is because JS handles arrays (objects) differently of if it is something specific to the splice function. I would also like to know how you would bypass that odd behavior.

Thanks for your time.

This question already has an answer here:

• 
  Copy array by value
  
  33 answers
  
  

let a = [1, 2, 3];

let b = a;

b.splice(2, 1);

console.log("a: " + a);

console.log("b: " + b);

let a = [1, 2, 3];

let b = a;

b.splice(2, 1);

console.log("a: " + a);

console.log("b: " + b);

javascript arrays

javascript arrays

share|improve this question

asked Jan 2 at 15:12

John KrakovJohn Krakov

705

share|improve this question

asked Jan 2 at 15:12

John KrakovJohn Krakov

705

share|improve this question

share|improve this question

asked Jan 2 at 15:12

John KrakovJohn Krakov

705

asked Jan 2 at 15:12

John KrakovJohn Krakov

705

asked Jan 2 at 15:12

John KrakovJohn Krakov

705

705

marked as duplicate by Denys Séguret javascript
Users with the  javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.

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Jan 2 at 15:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Denys Séguret javascript
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Jan 2 at 15:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  b references a, it is not a copy. Someone will find the dupe.
  
  – epascarello
  Jan 2 at 15:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  "since "b" should only have been assigned a copy of "a" Why ? Where are you copying ?
  
  – Denys Séguret
  Jan 2 at 15:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  BTW copying would have been let b = a.slice();
  
  – Denys Séguret
  Jan 2 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  duplicate stackoverflow.com/a/6605700/4267015
  
  – Naor Tedgi
  Jan 2 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DenysSéguret what do you mean?
  
  – John Krakov
  Jan 2 at 15:32
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  b references a, it is not a copy. Someone will find the dupe.
  
  – epascarello
  Jan 2 at 15:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  "since "b" should only have been assigned a copy of "a" Why ? Where are you copying ?
  
  – Denys Séguret
  Jan 2 at 15:14
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  BTW copying would have been let b = a.slice();
  
  – Denys Séguret
  Jan 2 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  duplicate stackoverflow.com/a/6605700/4267015
  
  – Naor Tedgi
  Jan 2 at 15:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DenysSéguret what do you mean?
  
  – John Krakov
  Jan 2 at 15:32
  

  
  
  
  

6

6

b references a, it is not a copy. Someone will find the dupe.

– epascarello
Jan 2 at 15:14

b references a, it is not a copy. Someone will find the dupe.

– epascarello
Jan 2 at 15:14

"since "b" should only have been assigned a copy of "a" Why ? Where are you copying ?

– Denys Séguret
Jan 2 at 15:14

"since "b" should only have been assigned a copy of "a" Why ? Where are you copying ?

– Denys Séguret
Jan 2 at 15:14

2

2

BTW copying would have been let b = a.slice();

– Denys Séguret
Jan 2 at 15:15

BTW copying would have been let b = a.slice();

– Denys Séguret
Jan 2 at 15:15

duplicate stackoverflow.com/a/6605700/4267015

– Naor Tedgi
Jan 2 at 15:16

duplicate stackoverflow.com/a/6605700/4267015

– Naor Tedgi
Jan 2 at 15:16

@DenysSéguret what do you mean?

– John Krakov
Jan 2 at 15:32

@DenysSéguret what do you mean?

– John Krakov
Jan 2 at 15:32

 | 
show 4 more comments

1 Answer
1

active

oldest

votes

2

In the first line you are creating the array and pointing a variable to that array.
In the second line let b = a;, here you are pointing b variable to the a array again.

So both a and b variables will be pointing to the same array. When you do a change to the array, both a and b values will be changed.

share|improve this answer

answered Jan 2 at 15:16

SandSand

1,7213822

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

In the first line you are creating the array and pointing a variable to that array.
In the second line let b = a;, here you are pointing b variable to the a array again.

So both a and b variables will be pointing to the same array. When you do a change to the array, both a and b values will be changed.

share|improve this answer

answered Jan 2 at 15:16

SandSand

1,7213822

add a comment | 

2

In the first line you are creating the array and pointing a variable to that array.
In the second line let b = a;, here you are pointing b variable to the a array again.

So both a and b variables will be pointing to the same array. When you do a change to the array, both a and b values will be changed.

share|improve this answer

answered Jan 2 at 15:16

SandSand

1,7213822

add a comment | 

2

2

2

In the first line you are creating the array and pointing a variable to that array.
In the second line let b = a;, here you are pointing b variable to the a array again.

So both a and b variables will be pointing to the same array. When you do a change to the array, both a and b values will be changed.

share|improve this answer

answered Jan 2 at 15:16

SandSand

1,7213822

In the first line you are creating the array and pointing a variable to that array.
In the second line let b = a;, here you are pointing b variable to the a array again.

So both a and b variables will be pointing to the same array. When you do a change to the array, both a and b values will be changed.

share|improve this answer

answered Jan 2 at 15:16

SandSand

1,7213822

share|improve this answer

share|improve this answer

answered Jan 2 at 15:16

SandSand

1,7213822

answered Jan 2 at 15:16

SandSand

1,7213822

answered Jan 2 at 15:16

SandSand

1,7213822

1,7213822

add a comment | 

add a comment | 

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