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11

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 8 hours ago

Boann

37.4k1290122

asked 11 hours ago

brkbrk

29.9k32245

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

11

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 8 hours ago

Boann

37.4k1290122

asked 11 hours ago

brkbrk

29.9k32245

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bergi Well, that explains a lot.
  
  – Teemu
  11 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 8 hours ago

Boann

37.4k1290122

asked 11 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 8 hours ago

Boann

37.4k1290122

asked 11 hours ago

brkbrk

29.9k32245

share|improve this question

edited 8 hours ago

Boann

37.4k1290122

asked 11 hours ago

brkbrk

29.9k32245

edited 8 hours ago

Boann

37.4k1290122

edited 8 hours ago

Boann

37.4k1290122

asked 11 hours ago

brkbrk

29.9k32245

asked 11 hours ago

brkbrk

29.9k32245

5 Answers
5

active

oldest

votes

15

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 11 hours ago

answered 11 hours ago

benvcbenvc

6,6731828

• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

8

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 8 hours ago

Boann

37.4k1290122

answered 11 hours ago

SkillGGSkillGG

30319

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

4

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 11 hours ago

answered 11 hours ago

Stranger in the QStranger in the Q

7261618

add a comment | 

3

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 11 hours ago

FedeScFedeSc

887923

add a comment | 

2

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 11 hours ago

answered 11 hours ago

VHSVHS

7,23431128

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 3 more comments

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15

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 11 hours ago

answered 11 hours ago

benvcbenvc

6,6731828

• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

15

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 11 hours ago

answered 11 hours ago

benvcbenvc

6,6731828

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  This is the shortest and most concise correct answer shown.
  
  – Scott Marcus
  11 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 11 hours ago

answered 11 hours ago

benvcbenvc

6,6731828

share|improve this answer

edited 11 hours ago

answered 11 hours ago

benvcbenvc

6,6731828

answered 11 hours ago

benvcbenvc

6,6731828

answered 11 hours ago

benvcbenvc

6,6731828

8

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 8 hours ago

Boann

37.4k1290122

answered 11 hours ago

SkillGGSkillGG

30319

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

8

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 8 hours ago

Boann

37.4k1290122

answered 11 hours ago

SkillGGSkillGG

30319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  actually my expectation was it will log false instead of throwing error.Actually I was doing if(!(prop in someObj))
  
  – brk
  11 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 8 hours ago

Boann

37.4k1290122

answered 11 hours ago

SkillGGSkillGG

30319

share|improve this answer

edited 8 hours ago

Boann

37.4k1290122

answered 11 hours ago

SkillGGSkillGG

30319

edited 8 hours ago

Boann

37.4k1290122

edited 8 hours ago

Boann

37.4k1290122

answered 11 hours ago

SkillGGSkillGG

30319

answered 11 hours ago

SkillGGSkillGG

30319

4

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 11 hours ago

answered 11 hours ago

Stranger in the QStranger in the Q

7261618

add a comment | 

4

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 11 hours ago

answered 11 hours ago

Stranger in the QStranger in the Q

7261618

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 11 hours ago

answered 11 hours ago

Stranger in the QStranger in the Q

7261618

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 11 hours ago

answered 11 hours ago

Stranger in the QStranger in the Q

7261618

answered 11 hours ago

Stranger in the QStranger in the Q

7261618

answered 11 hours ago

Stranger in the QStranger in the Q

7261618

3

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 11 hours ago

FedeScFedeSc

887923

add a comment | 

3

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 11 hours ago

FedeScFedeSc

887923

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 11 hours ago

FedeScFedeSc

887923

share|improve this answer

answered 11 hours ago

FedeScFedeSc

887923

answered 11 hours ago

FedeScFedeSc

887923

answered 11 hours ago

FedeScFedeSc

887923

2

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 11 hours ago

answered 11 hours ago

VHSVHS

7,23431128

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 3 more comments

2

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 11 hours ago

answered 11 hours ago

VHSVHS

7,23431128

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Just run the second snippet, the first console.log shows 4.
  
  – Teemu
  11 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 11 hours ago

answered 11 hours ago

VHSVHS

7,23431128

share|improve this answer

edited 11 hours ago

answered 11 hours ago

VHSVHS

7,23431128

answered 11 hours ago

VHSVHS

7,23431128

answered 11 hours ago

VHSVHS

7,23431128