Why is the 'in' operator throwing an error with a string literal instead of logging false?Why can't I use...
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Why is the 'in' operator throwing an error with a string literal instead of logging false?
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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
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
I'd assume the temporary wrapper object created for the string is not enumerable ..?
– Teemu
6 hours ago
@Teemu No. There is no temporary wrapper object created at all
– Bergi
6 hours ago
@Bergi Well, that explains a lot.
– Teemu
6 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
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);
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);
javascript string
javascript string
edited 3 hours ago
Boann
37.4k1290122
37.4k1290122
asked 6 hours ago
brkbrk
29.8k32244
29.8k32244
I'd assume the temporary wrapper object created for the string is not enumerable ..?
– Teemu
6 hours ago
@Teemu No. There is no temporary wrapper object created at all
– Bergi
6 hours ago
@Bergi Well, that explains a lot.
– Teemu
6 hours ago
add a comment |
I'd assume the temporary wrapper object created for the string is not enumerable ..?
– Teemu
6 hours ago
@Teemu No. There is no temporary wrapper object created at all
– Bergi
6 hours ago
@Bergi Well, that explains a lot.
– Teemu
6 hours ago
I'd assume the temporary wrapper object created for the string is not enumerable ..?
– Teemu
6 hours ago
I'd assume the temporary wrapper object created for the string is not enumerable ..?
– Teemu
6 hours ago
@Teemu No. There is no temporary wrapper object created at all
– Bergi
6 hours ago
@Teemu No. There is no temporary wrapper object created at all
– Bergi
6 hours ago
@Bergi Well, that explains a lot.
– Teemu
6 hours ago
@Bergi Well, that explains a lot.
– Teemu
6 hours ago
add a comment |
5 Answers
5
active
oldest
votes
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 theString
constructor,
but you cannot specify a string literal.
2
This is the shortest and most concise correct answer shown.
– Scott Marcus
6 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.
actually my expectation was it will logfalse
instead of throwing error.Actually I was doingif(!(prop in someObj))
– brk
6 hours ago
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'))
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.
add a comment |
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.
Notice, thatlet1.length
works in the snippet.
– Teemu
6 hours ago
Right. Butlet1
is a string, not an object in the second example.
– VHS
6 hours ago
Umh ... the second example works as well.
– Teemu
6 hours ago
The second example wouldn't work becauselet1
is not an object.
– VHS
6 hours ago
Just run the second snippet, the first console.log shows4
.
– Teemu
6 hours ago
|
show 3 more comments
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 theString
constructor,
but you cannot specify a string literal.
2
This is the shortest and most concise correct answer shown.
– Scott Marcus
6 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 theString
constructor,
but you cannot specify a string literal.
2
This is the shortest and most concise correct answer shown.
– Scott Marcus
6 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 theString
constructor,
but you cannot specify a string literal.
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 theString
constructor,
but you cannot specify a string literal.
edited 6 hours ago
answered 6 hours ago
benvcbenvc
6,5981827
6,5981827
2
This is the shortest and most concise correct answer shown.
– Scott Marcus
6 hours ago
add a comment |
2
This is the shortest and most concise correct answer shown.
– Scott Marcus
6 hours ago
2
2
This is the shortest and most concise correct answer shown.
– Scott Marcus
6 hours ago
This is the shortest and most concise correct answer shown.
– Scott Marcus
6 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.
actually my expectation was it will logfalse
instead of throwing error.Actually I was doingif(!(prop in someObj))
– brk
6 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.
actually my expectation was it will logfalse
instead of throwing error.Actually I was doingif(!(prop in someObj))
– brk
6 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.
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.
edited 3 hours ago
Boann
37.4k1290122
37.4k1290122
answered 6 hours ago
SkillGGSkillGG
1839
1839
actually my expectation was it will logfalse
instead of throwing error.Actually I was doingif(!(prop in someObj))
– brk
6 hours ago
add a comment |
actually my expectation was it will logfalse
instead of throwing error.Actually I was doingif(!(prop in someObj))
– brk
6 hours ago
actually my expectation was it will log
false
instead of throwing error.Actually I was doing if(!(prop in someObj))
– brk
6 hours ago
actually my expectation was it will log
false
instead of throwing error.Actually I was doing if(!(prop in someObj))
– brk
6 hours ago
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'))
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'))
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'))
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'))
console.log(typeof (new String('ddd')))
console.log(typeof ('ddd'))
console.log(typeof (new String('ddd')))
console.log(typeof ('ddd'))
edited 6 hours ago
answered 6 hours ago
Stranger in the QStranger in the Q
7161618
7161618
add a comment |
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.
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.
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.
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.
answered 6 hours ago
FedeScFedeSc
877923
877923
add a comment |
add a comment |
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.
Notice, thatlet1.length
works in the snippet.
– Teemu
6 hours ago
Right. Butlet1
is a string, not an object in the second example.
– VHS
6 hours ago
Umh ... the second example works as well.
– Teemu
6 hours ago
The second example wouldn't work becauselet1
is not an object.
– VHS
6 hours ago
Just run the second snippet, the first console.log shows4
.
– Teemu
6 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.
Notice, thatlet1.length
works in the snippet.
– Teemu
6 hours ago
Right. Butlet1
is a string, not an object in the second example.
– VHS
6 hours ago
Umh ... the second example works as well.
– Teemu
6 hours ago
The second example wouldn't work becauselet1
is not an object.
– VHS
6 hours ago
Just run the second snippet, the first console.log shows4
.
– Teemu
6 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.
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.
edited 6 hours ago
answered 6 hours ago
VHSVHS
7,22431128
7,22431128
Notice, thatlet1.length
works in the snippet.
– Teemu
6 hours ago
Right. Butlet1
is a string, not an object in the second example.
– VHS
6 hours ago
Umh ... the second example works as well.
– Teemu
6 hours ago
The second example wouldn't work becauselet1
is not an object.
– VHS
6 hours ago
Just run the second snippet, the first console.log shows4
.
– Teemu
6 hours ago
|
show 3 more comments
Notice, thatlet1.length
works in the snippet.
– Teemu
6 hours ago
Right. Butlet1
is a string, not an object in the second example.
– VHS
6 hours ago
Umh ... the second example works as well.
– Teemu
6 hours ago
The second example wouldn't work becauselet1
is not an object.
– VHS
6 hours ago
Just run the second snippet, the first console.log shows4
.
– Teemu
6 hours ago
Notice, that
let1.length
works in the snippet.– Teemu
6 hours ago
Notice, that
let1.length
works in the snippet.– Teemu
6 hours ago
Right. But
let1
is a string, not an object in the second example.– VHS
6 hours ago
Right. But
let1
is a string, not an object in the second example.– VHS
6 hours ago
Umh ... the second example works as well.
– Teemu
6 hours ago
Umh ... the second example works as well.
– Teemu
6 hours ago
The second example wouldn't work because
let1
is not an object.– VHS
6 hours ago
The second example wouldn't work because
let1
is not an object.– VHS
6 hours ago
Just run the second snippet, the first console.log shows
4
.– Teemu
6 hours ago
Just run the second snippet, the first console.log shows
4
.– Teemu
6 hours ago
|
show 3 more comments
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I'd assume the temporary wrapper object created for the string is not enumerable ..?
– Teemu
6 hours ago
@Teemu No. There is no temporary wrapper object created at all
– Bergi
6 hours ago
@Bergi Well, that explains a lot.
– Teemu
6 hours ago