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
Apr 4 at 18:49
@Teemu No. There is no temporary wrapper object created at all
– Bergi
Apr 4 at 18:51
@Bergi Well, that explains a lot.
– Teemu
Apr 4 at 18:52
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 Apr 4 at 21:55
Boann
37.5k1290122
37.5k1290122
asked Apr 4 at 18:44
brkbrk
30.2k32246
30.2k32246
I'd assume the temporary wrapper object created for the string is not enumerable ..?
– Teemu
Apr 4 at 18:49
@Teemu No. There is no temporary wrapper object created at all
– Bergi
Apr 4 at 18:51
@Bergi Well, that explains a lot.
– Teemu
Apr 4 at 18:52
add a comment |
I'd assume the temporary wrapper object created for the string is not enumerable ..?
– Teemu
Apr 4 at 18:49
@Teemu No. There is no temporary wrapper object created at all
– Bergi
Apr 4 at 18:51
@Bergi Well, that explains a lot.
– Teemu
Apr 4 at 18:52
I'd assume the temporary wrapper object created for the string is not enumerable ..?
– Teemu
Apr 4 at 18:49
I'd assume the temporary wrapper object created for the string is not enumerable ..?
– Teemu
Apr 4 at 18:49
@Teemu No. There is no temporary wrapper object created at all
– Bergi
Apr 4 at 18:51
@Teemu No. There is no temporary wrapper object created at all
– Bergi
Apr 4 at 18:51
@Bergi Well, that explains a lot.
– Teemu
Apr 4 at 18:52
@Bergi Well, that explains a lot.
– Teemu
Apr 4 at 18:52
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.
3
This is the shortest and most concise correct answer shown.
– Scott Marcus
Apr 4 at 19:05
Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.
– ChrisM
Apr 5 at 16:12
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(template) 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
Apr 4 at 18:51
add a comment |
JavaScript operator in
only applicable to an Object
s instances.
When you using constructor new String('abc')
this will causing creating of a String
object instance.
In other side, when you using only string literals or call function String('abc')
without new
it creates an string primitive. (like Number
and Boolen
)
Some behaviour of primitives and objects is differrent, look at this simple example's output:
console.log(typeof (new String('ddd'))) // "object"
console.log(typeof ('ddd')) // "string"
console.log(eval('1 + 2')) // 3
console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}
In code where you use methods on string primitives javascript engine automatically wraps primitives with corresponding objects to perform methods call.
But in
it is not an method call, its language operator an in this case wrapping is not applied.
PS: Sorry for my english.
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.
1
Notice, thatlet1.length
works in the snippet.
– Teemu
Apr 4 at 18:53
Right. Butlet1
is a string, not an object in the second example.
– VHS
Apr 4 at 18:54
1
Umh ... the second example works as well.
– Teemu
Apr 4 at 19:01
1
Just run the second snippet, the first console.log shows4
.
– Teemu
Apr 4 at 19:04
1
Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs4
, the length oflet1
. Take a look at benvc's answer on this post.
– Teemu
Apr 4 at 19:06
|
show 4 more comments
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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.
3
This is the shortest and most concise correct answer shown.
– Scott Marcus
Apr 4 at 19:05
Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.
– ChrisM
Apr 5 at 16:12
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.
3
This is the shortest and most concise correct answer shown.
– Scott Marcus
Apr 4 at 19:05
Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.
– ChrisM
Apr 5 at 16:12
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 Apr 4 at 19:16
answered Apr 4 at 18:56
benvcbenvc
6,9031928
6,9031928
3
This is the shortest and most concise correct answer shown.
– Scott Marcus
Apr 4 at 19:05
Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.
– ChrisM
Apr 5 at 16:12
add a comment |
3
This is the shortest and most concise correct answer shown.
– Scott Marcus
Apr 4 at 19:05
Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.
– ChrisM
Apr 5 at 16:12
3
3
This is the shortest and most concise correct answer shown.
– Scott Marcus
Apr 4 at 19:05
This is the shortest and most concise correct answer shown.
– Scott Marcus
Apr 4 at 19:05
Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.
– ChrisM
Apr 5 at 16:12
Nice. I think I know a lot about JS after working with it for a few years now but I'm still finding out interesting quirks like this.
– ChrisM
Apr 5 at 16:12
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(template) 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
Apr 4 at 18:51
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(template) 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
Apr 4 at 18:51
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(template) 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(template) 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 Apr 5 at 18:06
answered Apr 4 at 18:49
SkillGGSkillGG
334112
334112
actually my expectation was it will logfalse
instead of throwing error.Actually I was doingif(!(prop in someObj))
– brk
Apr 4 at 18:51
add a comment |
actually my expectation was it will logfalse
instead of throwing error.Actually I was doingif(!(prop in someObj))
– brk
Apr 4 at 18:51
actually my expectation was it will log
false
instead of throwing error.Actually I was doing if(!(prop in someObj))
– brk
Apr 4 at 18:51
actually my expectation was it will log
false
instead of throwing error.Actually I was doing if(!(prop in someObj))
– brk
Apr 4 at 18:51
add a comment |
JavaScript operator in
only applicable to an Object
s instances.
When you using constructor new String('abc')
this will causing creating of a String
object instance.
In other side, when you using only string literals or call function String('abc')
without new
it creates an string primitive. (like Number
and Boolen
)
Some behaviour of primitives and objects is differrent, look at this simple example's output:
console.log(typeof (new String('ddd'))) // "object"
console.log(typeof ('ddd')) // "string"
console.log(eval('1 + 2')) // 3
console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}
In code where you use methods on string primitives javascript engine automatically wraps primitives with corresponding objects to perform methods call.
But in
it is not an method call, its language operator an in this case wrapping is not applied.
PS: Sorry for my english.
add a comment |
JavaScript operator in
only applicable to an Object
s instances.
When you using constructor new String('abc')
this will causing creating of a String
object instance.
In other side, when you using only string literals or call function String('abc')
without new
it creates an string primitive. (like Number
and Boolen
)
Some behaviour of primitives and objects is differrent, look at this simple example's output:
console.log(typeof (new String('ddd'))) // "object"
console.log(typeof ('ddd')) // "string"
console.log(eval('1 + 2')) // 3
console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}
In code where you use methods on string primitives javascript engine automatically wraps primitives with corresponding objects to perform methods call.
But in
it is not an method call, its language operator an in this case wrapping is not applied.
PS: Sorry for my english.
add a comment |
JavaScript operator in
only applicable to an Object
s instances.
When you using constructor new String('abc')
this will causing creating of a String
object instance.
In other side, when you using only string literals or call function String('abc')
without new
it creates an string primitive. (like Number
and Boolen
)
Some behaviour of primitives and objects is differrent, look at this simple example's output:
console.log(typeof (new String('ddd'))) // "object"
console.log(typeof ('ddd')) // "string"
console.log(eval('1 + 2')) // 3
console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}
In code where you use methods on string primitives javascript engine automatically wraps primitives with corresponding objects to perform methods call.
But in
it is not an method call, its language operator an in this case wrapping is not applied.
PS: Sorry for my english.
JavaScript operator in
only applicable to an Object
s instances.
When you using constructor new String('abc')
this will causing creating of a String
object instance.
In other side, when you using only string literals or call function String('abc')
without new
it creates an string primitive. (like Number
and Boolen
)
Some behaviour of primitives and objects is differrent, look at this simple example's output:
console.log(typeof (new String('ddd'))) // "object"
console.log(typeof ('ddd')) // "string"
console.log(eval('1 + 2')) // 3
console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}
In code where you use methods on string primitives javascript engine automatically wraps primitives with corresponding objects to perform methods call.
But in
it is not an method call, its language operator an in this case wrapping is not applied.
PS: Sorry for my english.
console.log(typeof (new String('ddd'))) // "object"
console.log(typeof ('ddd')) // "string"
console.log(eval('1 + 2')) // 3
console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}
console.log(typeof (new String('ddd'))) // "object"
console.log(typeof ('ddd')) // "string"
console.log(eval('1 + 2')) // 3
console.log(eval(new String('1 + 2'))) // {"0":"1","1":" ","2":"+","3":" ","4":"2"}
edited Apr 5 at 12:06
answered Apr 4 at 18:49
Stranger in the QStranger in the Q
1,1061820
1,1061820
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 Apr 4 at 18:48
FedeScFedeSc
917923
917923
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.
1
Notice, thatlet1.length
works in the snippet.
– Teemu
Apr 4 at 18:53
Right. Butlet1
is a string, not an object in the second example.
– VHS
Apr 4 at 18:54
1
Umh ... the second example works as well.
– Teemu
Apr 4 at 19:01
1
Just run the second snippet, the first console.log shows4
.
– Teemu
Apr 4 at 19:04
1
Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs4
, the length oflet1
. Take a look at benvc's answer on this post.
– Teemu
Apr 4 at 19:06
|
show 4 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.
1
Notice, thatlet1.length
works in the snippet.
– Teemu
Apr 4 at 18:53
Right. Butlet1
is a string, not an object in the second example.
– VHS
Apr 4 at 18:54
1
Umh ... the second example works as well.
– Teemu
Apr 4 at 19:01
1
Just run the second snippet, the first console.log shows4
.
– Teemu
Apr 4 at 19:04
1
Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs4
, the length oflet1
. Take a look at benvc's answer on this post.
– Teemu
Apr 4 at 19:06
|
show 4 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 Apr 4 at 18:55
answered Apr 4 at 18:53
VHSVHS
7,37631128
7,37631128
1
Notice, thatlet1.length
works in the snippet.
– Teemu
Apr 4 at 18:53
Right. Butlet1
is a string, not an object in the second example.
– VHS
Apr 4 at 18:54
1
Umh ... the second example works as well.
– Teemu
Apr 4 at 19:01
1
Just run the second snippet, the first console.log shows4
.
– Teemu
Apr 4 at 19:04
1
Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs4
, the length oflet1
. Take a look at benvc's answer on this post.
– Teemu
Apr 4 at 19:06
|
show 4 more comments
1
Notice, thatlet1.length
works in the snippet.
– Teemu
Apr 4 at 18:53
Right. Butlet1
is a string, not an object in the second example.
– VHS
Apr 4 at 18:54
1
Umh ... the second example works as well.
– Teemu
Apr 4 at 19:01
1
Just run the second snippet, the first console.log shows4
.
– Teemu
Apr 4 at 19:04
1
Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs4
, the length oflet1
. Take a look at benvc's answer on this post.
– Teemu
Apr 4 at 19:06
1
1
Notice, that
let1.length
works in the snippet.– Teemu
Apr 4 at 18:53
Notice, that
let1.length
works in the snippet.– Teemu
Apr 4 at 18:53
Right. But
let1
is a string, not an object in the second example.– VHS
Apr 4 at 18:54
Right. But
let1
is a string, not an object in the second example.– VHS
Apr 4 at 18:54
1
1
Umh ... the second example works as well.
– Teemu
Apr 4 at 19:01
Umh ... the second example works as well.
– Teemu
Apr 4 at 19:01
1
1
Just run the second snippet, the first console.log shows
4
.– Teemu
Apr 4 at 19:04
Just run the second snippet, the first console.log shows
4
.– Teemu
Apr 4 at 19:04
1
1
Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs
4
, the length of let1
. Take a look at benvc's answer on this post.– Teemu
Apr 4 at 19:06
Yes, but you're saying retrieving the length woud throw the error. On the very first line it logs
4
, the length of let1
. Take a look at benvc's answer on this post.– Teemu
Apr 4 at 19:06
|
show 4 more comments
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I'd assume the temporary wrapper object created for the string is not enumerable ..?
– Teemu
Apr 4 at 18:49
@Teemu No. There is no temporary wrapper object created at all
– Bergi
Apr 4 at 18:51
@Bergi Well, that explains a lot.
– Teemu
Apr 4 at 18:52