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JavaScript Hoisting: Function Can Refer to Another Function Declared Below It?

How can I merge properties of two JavaScript objects dynamically?How can I convert a string to boolean in JavaScript?What's the difference between using “let” and “var”?Set a default parameter value for a JavaScript functionHow can I get query string values in JavaScript?How do I include a JavaScript file in another JavaScript file?How can I pretty-print JSON using JavaScript?Is there a standard function to check for null, undefined, or blank variables in JavaScript?Javascript function scoping and hoistingJavaScript 'hoisting'

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1

I have:

// Shouldn't we have the following hoisted?
// var multiply; (undefined)
// var add; (undefined)

var multiply = function(num) 
 return add(num) * 2;
;

var add = function(num) 
 return num + 1;
;

console.log(multiply(1)); // No error, somehow "multiply" calls "add"!

I thought that JavaScript variables are hoisted to the top, but not the values that they are assigned to. Somehow multiply calling add doesn't return an error, even though add is declared below multiply.

javascript function scope hoisting lexical

share|improve this question

edited Mar 9 at 1:40

Hadoren

asked Mar 9 at 1:01

HadorenHadoren

82210

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think so, based on my admittedly flawed memory of reading this blog.bitsrc.io/…, which might teach you more than you wanted to know. The basics I remember include that each function gets its own execution context which includes a couple of internal lexical environments (one of which is specific to those identifiers declared with var) and these lexical environments can access the outer environment. Also, I think var and function declarations get hoisted but let and const declarations don't...
  
  – Cat
  Mar 9 at 1:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The explanation is pretty simple. JS lets your functions reference variables that don't exist. So the potential for error only comes after the function is invoked. At the time of invocation, if the referenced variable doesn't exist, you get an error. As a test, remove the add() function, and update multiply() so that it uses a try/catch, where the try tries to call add(), and the catch creates it at window.add = function() .... You'll see that the ReferenceError is caught on the first invocation, but subsequent invocations succeed.
  
  – ziggy wiggy
  Mar 9 at 1:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  ...like this: jsfiddle.net/cpzj309w
  
  – ziggy wiggy
  Mar 9 at 1:53
  

  
  
  
  

add a comment | 

1

I have:

// Shouldn't we have the following hoisted?
// var multiply; (undefined)
// var add; (undefined)

var multiply = function(num) 
 return add(num) * 2;
;

var add = function(num) 
 return num + 1;
;

console.log(multiply(1)); // No error, somehow "multiply" calls "add"!

I thought that JavaScript variables are hoisted to the top, but not the values that they are assigned to. Somehow multiply calling add doesn't return an error, even though add is declared below multiply.

javascript function scope hoisting lexical

share|improve this question

edited Mar 9 at 1:40

Hadoren

asked Mar 9 at 1:01

HadorenHadoren

82210

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think so, based on my admittedly flawed memory of reading this blog.bitsrc.io/…, which might teach you more than you wanted to know. The basics I remember include that each function gets its own execution context which includes a couple of internal lexical environments (one of which is specific to those identifiers declared with var) and these lexical environments can access the outer environment. Also, I think var and function declarations get hoisted but let and const declarations don't...
  
  – Cat
  Mar 9 at 1:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The explanation is pretty simple. JS lets your functions reference variables that don't exist. So the potential for error only comes after the function is invoked. At the time of invocation, if the referenced variable doesn't exist, you get an error. As a test, remove the add() function, and update multiply() so that it uses a try/catch, where the try tries to call add(), and the catch creates it at window.add = function() .... You'll see that the ReferenceError is caught on the first invocation, but subsequent invocations succeed.
  
  – ziggy wiggy
  Mar 9 at 1:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  ...like this: jsfiddle.net/cpzj309w
  
  – ziggy wiggy
  Mar 9 at 1:53
  

  
  
  
  

add a comment | 

I have:

// Shouldn't we have the following hoisted?
// var multiply; (undefined)
// var add; (undefined)

var multiply = function(num) 
 return add(num) * 2;
;

var add = function(num) 
 return num + 1;
;

console.log(multiply(1)); // No error, somehow "multiply" calls "add"!

I thought that JavaScript variables are hoisted to the top, but not the values that they are assigned to. Somehow multiply calling add doesn't return an error, even though add is declared below multiply.

javascript function scope hoisting lexical

share|improve this question

edited Mar 9 at 1:40

Hadoren

asked Mar 9 at 1:01

HadorenHadoren

82210

// Shouldn't we have the following hoisted?
// var multiply; (undefined)
// var add; (undefined)

var multiply = function(num) 
 return add(num) * 2;
;

var add = function(num) 
 return num + 1;
;

console.log(multiply(1)); // No error, somehow "multiply" calls "add"!

share|improve this question

edited Mar 9 at 1:40

Hadoren

asked Mar 9 at 1:01

HadorenHadoren

82210

share|improve this question

edited Mar 9 at 1:40

Hadoren

asked Mar 9 at 1:01

HadorenHadoren

82210

asked Mar 9 at 1:01

HadorenHadoren

82210

asked Mar 9 at 1:01

HadorenHadoren

82210

1 Answer
1

active

oldest

votes

1

You are correct in that the declarations are hoisted to the top, while the assignments are not.

However, functions don't keep the values of any variables outside of them from when they are created. Instead, they use whatever is in them when they are called. In this case, add is undefined when the multiply function is created, but is assigned a function before multiply is called, so multiply uses the new assigned function.

To see this more clearly, consider this code:

var multiply = function(num) 
 return add(num) * 2;
;

// Would be an error
// console.log(multiply(1));

var add = function(num) 
 return num + 1;
;

console.log(multiply(1)); // Prints 4

add = function(num) 
 return num + 2;
;

console.log(multiply(1)); // Prints 6

The last console.log prints 6 because multiply used the new function in add instead of keeping the one it had before.

share|improve this answer

edited Mar 9 at 1:54

VLAZ

5,10742235

answered Mar 9 at 1:08

NicholasNicholas

30619

add a comment | 

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1

You are correct in that the declarations are hoisted to the top, while the assignments are not.

However, functions don't keep the values of any variables outside of them from when they are created. Instead, they use whatever is in them when they are called. In this case, add is undefined when the multiply function is created, but is assigned a function before multiply is called, so multiply uses the new assigned function.

To see this more clearly, consider this code:

var multiply = function(num) 
 return add(num) * 2;
;

// Would be an error
// console.log(multiply(1));

var add = function(num) 
 return num + 1;
;

console.log(multiply(1)); // Prints 4

add = function(num) 
 return num + 2;
;

console.log(multiply(1)); // Prints 6

The last console.log prints 6 because multiply used the new function in add instead of keeping the one it had before.

share|improve this answer

edited Mar 9 at 1:54

VLAZ

5,10742235

answered Mar 9 at 1:08

NicholasNicholas

30619

add a comment | 

1

You are correct in that the declarations are hoisted to the top, while the assignments are not.

However, functions don't keep the values of any variables outside of them from when they are created. Instead, they use whatever is in them when they are called. In this case, add is undefined when the multiply function is created, but is assigned a function before multiply is called, so multiply uses the new assigned function.

To see this more clearly, consider this code:

var multiply = function(num) 
 return add(num) * 2;
;

// Would be an error
// console.log(multiply(1));

var add = function(num) 
 return num + 1;
;

console.log(multiply(1)); // Prints 4

add = function(num) 
 return num + 2;
;

console.log(multiply(1)); // Prints 6

The last console.log prints 6 because multiply used the new function in add instead of keeping the one it had before.

share|improve this answer

edited Mar 9 at 1:54

VLAZ

5,10742235

answered Mar 9 at 1:08

NicholasNicholas

30619

add a comment | 

You are correct in that the declarations are hoisted to the top, while the assignments are not.

However, functions don't keep the values of any variables outside of them from when they are created. Instead, they use whatever is in them when they are called. In this case, add is undefined when the multiply function is created, but is assigned a function before multiply is called, so multiply uses the new assigned function.

To see this more clearly, consider this code:

var multiply = function(num) 
 return add(num) * 2;
;

// Would be an error
// console.log(multiply(1));

var add = function(num) 
 return num + 1;
;

console.log(multiply(1)); // Prints 4

add = function(num) 
 return num + 2;
;

console.log(multiply(1)); // Prints 6

The last console.log prints 6 because multiply used the new function in add instead of keeping the one it had before.

share|improve this answer

edited Mar 9 at 1:54

VLAZ

5,10742235

answered Mar 9 at 1:08

NicholasNicholas

30619

var multiply = function(num) 
 return add(num) * 2;
;

// Would be an error
// console.log(multiply(1));

var add = function(num) 
 return num + 1;
;

console.log(multiply(1)); // Prints 4

add = function(num) 
 return num + 2;
;

console.log(multiply(1)); // Prints 6

var multiply = function(num) 
 return add(num) * 2;
;

// Would be an error
// console.log(multiply(1));

var add = function(num) 
 return num + 1;
;

console.log(multiply(1)); // Prints 4

add = function(num) 
 return num + 2;
;

console.log(multiply(1)); // Prints 6

var multiply = function(num) 
 return add(num) * 2;
;

// Would be an error
// console.log(multiply(1));

var add = function(num) 
 return num + 1;
;

console.log(multiply(1)); // Prints 4

add = function(num) 
 return num + 2;
;

console.log(multiply(1)); // Prints 6

share|improve this answer

edited Mar 9 at 1:54

VLAZ

5,10742235

answered Mar 9 at 1:08

NicholasNicholas

30619

edited Mar 9 at 1:54

VLAZ

5,10742235

edited Mar 9 at 1:54

VLAZ

5,10742235

answered Mar 9 at 1:08

NicholasNicholas

30619

answered Mar 9 at 1:08

NicholasNicholas

30619