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Why is the 'in' operator throwing an error with a string literal instead of logging false?

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10

As per MDN the in operator returns true if the property exists and accordingly the first example logs true. But when using a string literal, why is it throwing an error instead of logging false?

let let1 = new String('test');

console.log(let1.length);

console.log('length' in let1)

var let1 = 'test';

console.log(let1.length);

console.log('length' in let1);

javascript string

share|improve this question

edited 5 hours ago

Boann

37.4k1290122

asked 8 hours ago

brkbrk

29.9k32245

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'd assume the temporary wrapper object created for the string is not enumerable ..?
  
  – Teemu
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Teemu No. There is no temporary wrapper object created at all
  
  – Bergi
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bergi Well, that explains a lot.
  
  – Teemu
  8 hours ago
  

  
  
  
  

add a comment | 

10

As per MDN the in operator returns true if the property exists and accordingly the first example logs true. But when using a string literal, why is it throwing an error instead of logging false?

let let1 = new String('test');

console.log(let1.length);

console.log('length' in let1)

var let1 = 'test';

console.log(let1.length);

console.log('length' in let1);

javascript string

share|improve this question

edited 5 hours ago

Boann

37.4k1290122

asked 8 hours ago

brkbrk

29.9k32245

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'd assume the temporary wrapper object created for the string is not enumerable ..?
  
  – Teemu
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Teemu No. There is no temporary wrapper object created at all
  
  – Bergi
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bergi Well, that explains a lot.
  
  – Teemu
  8 hours ago
  

  
  
  
  

add a comment | 

As per MDN the in operator returns true if the property exists and accordingly the first example logs true. But when using a string literal, why is it throwing an error instead of logging false?

let let1 = new String('test');

console.log(let1.length);

console.log('length' in let1)

var let1 = 'test';

console.log(let1.length);

console.log('length' in let1);

javascript string

share|improve this question

edited 5 hours ago

Boann

37.4k1290122

asked 8 hours ago

brkbrk

29.9k32245

let let1 = new String('test');

console.log(let1.length);

console.log('length' in let1)

var let1 = 'test';

console.log(let1.length);

console.log('length' in let1);

let let1 = new String('test');

console.log(let1.length);

console.log('length' in let1)

let let1 = new String('test');

console.log(let1.length);

console.log('length' in let1)

var let1 = 'test';

console.log(let1.length);

console.log('length' in let1);

var let1 = 'test';

console.log(let1.length);

console.log('length' in let1);

share|improve this question

edited 5 hours ago

Boann

37.4k1290122

asked 8 hours ago

brkbrk

29.9k32245

share|improve this question

edited 5 hours ago

Boann

37.4k1290122

asked 8 hours ago

brkbrk

29.9k32245

edited 5 hours ago

Boann

37.4k1290122

edited 5 hours ago

Boann

37.4k1290122

asked 8 hours ago

brkbrk

29.9k32245

asked 8 hours ago

brkbrk

29.9k32245

5 Answers
5

active

oldest

votes

13

In a sense it is a matter of timing. String literals do not have any properties. The reason that you can call methods and lookup properties on primitive strings is because JavaScript automatically wraps the string primitive in a String object when a method call or property lookup is attempted. JavaScript does not interpret the in operator as a method call or property lookup so it does not wrap the primitive in an object and you get an error (because a string primitive is not an object).

See Distinction between string primitives and String objects

Also, the same docs referenced in your question specifically note that using in on a string primitive will throw an error.

You must specify an object on the right side of the in operator. For
example, you can specify a string created with the String constructor,
but you cannot specify a string literal.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

benvcbenvc

6,6381828

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  This is the shortest and most concise correct answer shown.
  
  – Scott Marcus
  8 hours ago
  

  
  
  
  

add a comment | 

6

It throws an error because in is an operator for objects:

prop in object

but when you declare a string as `` (` string literals) or "" '' (",' string literals) you don't create an object.

Check

typeof new String("x") ("object")

and

typeof `x` ("string").

Those are two different things in JavaScript.

share|improve this answer

edited 5 hours ago

Boann

37.4k1290122

answered 8 hours ago

SkillGGSkillGG

1839

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  actually my expectation was it will log false instead of throwing error.Actually I was doing if(!(prop in someObj))
  
  – brk
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

3

Because new creates an Object and string literal ('') is not an object. and in operator applicable only to an object instance.

console.log(typeof (new String('ddd')))

console.log(typeof ('ddd'))

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Stranger in the QStranger in the Q

7161618

add a comment | 

2

typeof('test') == string (string literal)

typof(new String('test')) == object (string object)

you can't use in with a string literal.

The in operator returns true if the specified property is in the specified object or its prototype chain.

share|improve this answer

answered 8 hours ago

FedeScFedeSc

877923

add a comment | 

1

The in operator can only be used to check if a property is in an
object. You can't search in strings, or in numbers, or other primitive
types.

The first example works and prints 'true' because length is a property of a string object.

The second example doesn't work and gives you an error because you are trying to look for a property length in something (a string) that is not an object.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

VHSVHS

7,22431128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Notice, that let1.length works in the snippet.
  
  – Teemu
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Right. But let1 is a string, not an object in the second example.
  
  – VHS
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Umh ... the second example works as well.
  
  – Teemu
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The second example wouldn't work because let1 is not an object.
  
  – VHS
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Just run the second snippet, the first console.log shows 4.
  
  – Teemu
  8 hours ago
  
  
  

  
  
  
  

 | 
show 3 more comments

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13

In a sense it is a matter of timing. String literals do not have any properties. The reason that you can call methods and lookup properties on primitive strings is because JavaScript automatically wraps the string primitive in a String object when a method call or property lookup is attempted. JavaScript does not interpret the in operator as a method call or property lookup so it does not wrap the primitive in an object and you get an error (because a string primitive is not an object).

See Distinction between string primitives and String objects

Also, the same docs referenced in your question specifically note that using in on a string primitive will throw an error.

You must specify an object on the right side of the in operator. For
example, you can specify a string created with the String constructor,
but you cannot specify a string literal.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

benvcbenvc

6,6381828

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  This is the shortest and most concise correct answer shown.
  
  – Scott Marcus
  8 hours ago
  

  
  
  
  

add a comment | 

13

In a sense it is a matter of timing. String literals do not have any properties. The reason that you can call methods and lookup properties on primitive strings is because JavaScript automatically wraps the string primitive in a String object when a method call or property lookup is attempted. JavaScript does not interpret the in operator as a method call or property lookup so it does not wrap the primitive in an object and you get an error (because a string primitive is not an object).

See Distinction between string primitives and String objects

Also, the same docs referenced in your question specifically note that using in on a string primitive will throw an error.

You must specify an object on the right side of the in operator. For
example, you can specify a string created with the String constructor,
but you cannot specify a string literal.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

benvcbenvc

6,6381828

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  This is the shortest and most concise correct answer shown.
  
  – Scott Marcus
  8 hours ago
  

  
  
  
  

add a comment | 

In a sense it is a matter of timing. String literals do not have any properties. The reason that you can call methods and lookup properties on primitive strings is because JavaScript automatically wraps the string primitive in a String object when a method call or property lookup is attempted. JavaScript does not interpret the in operator as a method call or property lookup so it does not wrap the primitive in an object and you get an error (because a string primitive is not an object).

See Distinction between string primitives and String objects

Also, the same docs referenced in your question specifically note that using in on a string primitive will throw an error.

You must specify an object on the right side of the in operator. For
example, you can specify a string created with the String constructor,
but you cannot specify a string literal.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

benvcbenvc

6,6381828

share|improve this answer

edited 8 hours ago

answered 8 hours ago

benvcbenvc

6,6381828

answered 8 hours ago

benvcbenvc

6,6381828

answered 8 hours ago

benvcbenvc

6,6381828

6

It throws an error because in is an operator for objects:

prop in object

but when you declare a string as `` (` string literals) or "" '' (",' string literals) you don't create an object.

Check

typeof new String("x") ("object")

and

typeof `x` ("string").

Those are two different things in JavaScript.

share|improve this answer

edited 5 hours ago

Boann

37.4k1290122

answered 8 hours ago

SkillGGSkillGG

1839

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  actually my expectation was it will log false instead of throwing error.Actually I was doing if(!(prop in someObj))
  
  – brk
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

6

It throws an error because in is an operator for objects:

prop in object

but when you declare a string as `` (` string literals) or "" '' (",' string literals) you don't create an object.

Check

typeof new String("x") ("object")

and

typeof `x` ("string").

Those are two different things in JavaScript.

share|improve this answer

edited 5 hours ago

Boann

37.4k1290122

answered 8 hours ago

SkillGGSkillGG

1839

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  actually my expectation was it will log false instead of throwing error.Actually I was doing if(!(prop in someObj))
  
  – brk
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

It throws an error because in is an operator for objects:

prop in object

but when you declare a string as `` (` string literals) or "" '' (",' string literals) you don't create an object.

Check

typeof new String("x") ("object")

and

typeof `x` ("string").

Those are two different things in JavaScript.

share|improve this answer

edited 5 hours ago

Boann

37.4k1290122

answered 8 hours ago

SkillGGSkillGG

1839

share|improve this answer

edited 5 hours ago

Boann

37.4k1290122

answered 8 hours ago

SkillGGSkillGG

1839

edited 5 hours ago

Boann

37.4k1290122

edited 5 hours ago

Boann

37.4k1290122

answered 8 hours ago

SkillGGSkillGG

1839

answered 8 hours ago

SkillGGSkillGG

1839

3

Because new creates an Object and string literal ('') is not an object. and in operator applicable only to an object instance.

console.log(typeof (new String('ddd')))

console.log(typeof ('ddd'))

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Stranger in the QStranger in the Q

7161618

add a comment | 

3

Because new creates an Object and string literal ('') is not an object. and in operator applicable only to an object instance.

console.log(typeof (new String('ddd')))

console.log(typeof ('ddd'))

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Stranger in the QStranger in the Q

7161618

add a comment | 

Because new creates an Object and string literal ('') is not an object. and in operator applicable only to an object instance.

console.log(typeof (new String('ddd')))

console.log(typeof ('ddd'))

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Stranger in the QStranger in the Q

7161618

console.log(typeof (new String('ddd')))

console.log(typeof ('ddd'))

console.log(typeof (new String('ddd')))

console.log(typeof ('ddd'))

console.log(typeof (new String('ddd')))

console.log(typeof ('ddd'))

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Stranger in the QStranger in the Q

7161618

answered 8 hours ago

Stranger in the QStranger in the Q

7161618

answered 8 hours ago

Stranger in the QStranger in the Q

7161618

2

typeof('test') == string (string literal)

typof(new String('test')) == object (string object)

you can't use in with a string literal.

The in operator returns true if the specified property is in the specified object or its prototype chain.

share|improve this answer

answered 8 hours ago

FedeScFedeSc

877923

add a comment | 

2

typeof('test') == string (string literal)

typof(new String('test')) == object (string object)

you can't use in with a string literal.

The in operator returns true if the specified property is in the specified object or its prototype chain.

share|improve this answer

answered 8 hours ago

FedeScFedeSc

877923

add a comment | 

typeof('test') == string (string literal)

typof(new String('test')) == object (string object)

you can't use in with a string literal.

The in operator returns true if the specified property is in the specified object or its prototype chain.

share|improve this answer

answered 8 hours ago

FedeScFedeSc

877923

share|improve this answer

answered 8 hours ago

FedeScFedeSc

877923

answered 8 hours ago

FedeScFedeSc

877923

answered 8 hours ago

FedeScFedeSc

877923

1

The in operator can only be used to check if a property is in an
object. You can't search in strings, or in numbers, or other primitive
types.

The first example works and prints 'true' because length is a property of a string object.

The second example doesn't work and gives you an error because you are trying to look for a property length in something (a string) that is not an object.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

VHSVHS

7,22431128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Notice, that let1.length works in the snippet.
  
  – Teemu
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Right. But let1 is a string, not an object in the second example.
  
  – VHS
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Umh ... the second example works as well.
  
  – Teemu
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The second example wouldn't work because let1 is not an object.
  
  – VHS
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Just run the second snippet, the first console.log shows 4.
  
  – Teemu
  8 hours ago
  
  
  

  
  
  
  

 | 
show 3 more comments

1

The in operator can only be used to check if a property is in an
object. You can't search in strings, or in numbers, or other primitive
types.

The first example works and prints 'true' because length is a property of a string object.

The second example doesn't work and gives you an error because you are trying to look for a property length in something (a string) that is not an object.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

VHSVHS

7,22431128

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Notice, that let1.length works in the snippet.
  
  – Teemu
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Right. But let1 is a string, not an object in the second example.
  
  – VHS
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Umh ... the second example works as well.
  
  – Teemu
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The second example wouldn't work because let1 is not an object.
  
  – VHS
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Just run the second snippet, the first console.log shows 4.
  
  – Teemu
  8 hours ago
  
  
  

  
  
  
  

 | 
show 3 more comments

The in operator can only be used to check if a property is in an
object. You can't search in strings, or in numbers, or other primitive
types.

The first example works and prints 'true' because length is a property of a string object.

The second example doesn't work and gives you an error because you are trying to look for a property length in something (a string) that is not an object.

share|improve this answer

edited 8 hours ago

answered 8 hours ago

VHSVHS

7,22431128

share|improve this answer

edited 8 hours ago

answered 8 hours ago

VHSVHS

7,22431128

answered 8 hours ago

VHSVHS

7,22431128

answered 8 hours ago

VHSVHS

7,22431128