Why does accessing a property of indexOf still compile?
I made a typo in TypeScript which was picked up during code review.
I used someArray.indexOf[someObject] instead of someArray.indexOf(someObject).
I would expect an error from the IDE/Compiler. Instead, no errors were raised and the result was simply undefined.
Can anyone explain this?
typescript methods properties
edited 1 hour ago
Bergi
366k58546872
asked 2 hours ago
De Wet van AsDe Wet van As
917
add a comment |
I made a typo in TypeScript which was picked up during code review.
I used someArray.indexOf[someObject] instead of someArray.indexOf(someObject).
I would expect an error from the IDE/Compiler. Instead, no errors were raised and the result was simply undefined.
Can anyone explain this?
typescript methods properties
edited 1 hour ago
Bergi
366k58546872
asked 2 hours ago
De Wet van AsDe Wet van As
917
1
What did you assign the result of? Because what you wrote is valid you are taking thesomeObjectmember of theindexOfmethod. Well, trying to. The only error would come from TypeScript compilation and only if you try to assign the result to something that doesn't match the expected type.
– vlaz
2 hours ago
1
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
2 hours ago
2
@DeWetvanAs I am actually curious about your problem - this seems like a genuine bug/problem with TypeScript see example here. It seems that if you are trying to assign to a variable of typenumber, the result of.indexOf[someObject]shouldn't be considered anumberand thus the compilation would fail. That's the whole idea of TypeScript is - to enforce the types. The answers here focus on JS but ignore this.
– vlaz
1 hour ago
3
@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.
– Bergi
1 hour ago
@Bergi As the question has underlined more and more the TS side, I've edited my answer! I think this should explain it :)
– sjahan
53 mins ago
add a comment |
I made a typo in TypeScript which was picked up during code review.
I used someArray.indexOf[someObject] instead of someArray.indexOf(someObject).
I would expect an error from the IDE/Compiler. Instead, no errors were raised and the result was simply undefined.
Can anyone explain this?
typescript methods properties
edited 1 hour ago
Bergi
366k58546872
asked 2 hours ago
De Wet van AsDe Wet van As
917
I made a typo in TypeScript which was picked up during code review.
I used someArray.indexOf[someObject] instead of someArray.indexOf(someObject).
I would expect an error from the IDE/Compiler. Instead, no errors were raised and the result was simply undefined.
Can anyone explain this?
typescript methods properties
typescript methods properties
edited 1 hour ago
Bergi
366k58546872
asked 2 hours ago
De Wet van AsDe Wet van As
917
edited 1 hour ago
Bergi
366k58546872
asked 2 hours ago
De Wet van AsDe Wet van As
917
edited 1 hour ago
Bergi
366k58546872
edited 1 hour ago
Bergi
366k58546872
edited 1 hour ago
Bergi
366k58546872
366k58546872
asked 2 hours ago
De Wet van AsDe Wet van As
917
asked 2 hours ago
De Wet van AsDe Wet van As
917
asked 2 hours ago
De Wet van AsDe Wet van As
917
917
1
What did you assign the result of? Because what you wrote is valid you are taking thesomeObjectmember of theindexOfmethod. Well, trying to. The only error would come from TypeScript compilation and only if you try to assign the result to something that doesn't match the expected type.
– vlaz
2 hours ago
1
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
2 hours ago
2
@DeWetvanAs I am actually curious about your problem - this seems like a genuine bug/problem with TypeScript see example here. It seems that if you are trying to assign to a variable of typenumber, the result of.indexOf[someObject]shouldn't be considered anumberand thus the compilation would fail. That's the whole idea of TypeScript is - to enforce the types. The answers here focus on JS but ignore this.
– vlaz
1 hour ago
3
@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.
– Bergi
1 hour ago
@Bergi As the question has underlined more and more the TS side, I've edited my answer! I think this should explain it :)
– sjahan
53 mins ago
add a comment |
1
What did you assign the result of? Because what you wrote is valid you are taking thesomeObjectmember of theindexOfmethod. Well, trying to. The only error would come from TypeScript compilation and only if you try to assign the result to something that doesn't match the expected type.
– vlaz
2 hours ago
1
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
2 hours ago
2
@DeWetvanAs I am actually curious about your problem - this seems like a genuine bug/problem with TypeScript see example here. It seems that if you are trying to assign to a variable of typenumber, the result of.indexOf[someObject]shouldn't be considered anumberand thus the compilation would fail. That's the whole idea of TypeScript is - to enforce the types. The answers here focus on JS but ignore this.
– vlaz
1 hour ago
3
@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.
– Bergi
1 hour ago
@Bergi As the question has underlined more and more the TS side, I've edited my answer! I think this should explain it :)
– sjahan
53 mins ago
1
1
What did you assign the result of? Because what you wrote is valid you are taking the
someObject member of the indexOf method. Well, trying to. The only error would come from TypeScript compilation and only if you try to assign the result to something that doesn't match the expected type.– vlaz
2 hours ago
What did you assign the result of? Because what you wrote is valid you are taking the
someObject member of the indexOf method. Well, trying to. The only error would come from TypeScript compilation and only if you try to assign the result to something that doesn't match the expected type.– vlaz
2 hours ago
1
1
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
2 hours ago
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
2 hours ago
2
2
@DeWetvanAs I am actually curious about your problem - this seems like a genuine bug/problem with TypeScript see example here. It seems that if you are trying to assign to a variable of type
number, the result of .indexOf[someObject] shouldn't be considered a number and thus the compilation would fail. That's the whole idea of TypeScript is - to enforce the types. The answers here focus on JS but ignore this.– vlaz
1 hour ago
@DeWetvanAs I am actually curious about your problem - this seems like a genuine bug/problem with TypeScript see example here. It seems that if you are trying to assign to a variable of type
number, the result of .indexOf[someObject] shouldn't be considered a number and thus the compilation would fail. That's the whole idea of TypeScript is - to enforce the types. The answers here focus on JS but ignore this.– vlaz
1 hour ago
3
3
@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.
– Bergi
1 hour ago
@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.
– Bergi
1 hour ago
@Bergi As the question has underlined more and more the TS side, I've edited my answer! I think this should explain it :)
– sjahan
53 mins ago
@Bergi As the question has underlined more and more the TS side, I've edited my answer! I think this should explain it :)
– sjahan
53 mins ago
add a comment |
4 Answers
4
active
oldest
votes
Quite easy.
someArray.indexOf you know that this is a function, which is also an object and can have properties.
By doing someArray.indexOf[someObject], you are trying to reach the property with the key valued to the value of someObject.
Of course, it is not defined on the indexOf function, so it returns undefined.
Quick example that illustrates the syntax and the fact that a function can have properties ;) :
const array = ;
array.indexOf['anyValue'] = 'test';
console.log(array.indexOf.anyValue);EDIT
Here is an attempt of an answer for the TypeScript side of the question.
As you already know, TypeScript is designed to be compatible with JavaScript. Therefore, as in JS, you can access a property of an object by the following ways:
- 'Statically':
obj.property
- 'Dynamically':
obj['property']
Both ways will perform the same, but with the dynamic way of accessing the property, there is no way TypeScript compiler can determine the type of it or if it exists or not. That's why it doesn't raise an error.
(Of course, someone could say the compiler could understand that here, it could read the string 'property', but don't forget that you could pass anything between the brackets and its value could change at run time, far outside of the TypeScript scope.)
By using the 'static' way to access a property, of course, TypeScript will raise an error!
Hoping this helps ;)
answered 2 hours ago
sjahansjahan
3,2251827
3
I think answers here are missing theTypeScripttag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you haveindex: numberandindex = arr.indexOf[obj]then that should be a compilation error. Butindex: anywouldn't throw a compilation error.
– vlaz
2 hours ago
@vlaz +1. sjahan gives OP a quick explanation of theundefinedresult but the main question remains...
– Florian
2 hours ago
add a comment |
It does not error because the --noImplicitAny compiler option is not enabled. With that compiler option enabled you will get an error as expected:
The reason is that using an element access expression creates an implicit any type when the type has no index signature defined.
So again, since --noImplicitAny is not enabled, it does not error. I highly recommend turning this compiler option on.
answered 35 mins ago
David SherretDavid Sherret
50.9k16120122
add a comment |
array.indexOf is a function.
Functions are objects.
You were accessing the someObject property of the array.indexOf function.
You would have got undefined.
const array = [1, 2, 3]
const someObject = 'asdasd'
console.log(array.indexOf[someObject])
// undefinedanswered 2 hours ago
0xc14m1z0xc14m1z
1,419512
1
Array.indexOfisundefined,Array.prototype.indexOf, on the other hand, is a function.
– Pavlo
1 hour ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
1 hour ago
add a comment |
Functions in JavaScript are first class objects.
When you access function Array.indexOf() via bracket notation Array.indexOf['prop'] you actually trying to access a property which does not exist on indexOf so you get undefined.
answered 2 hours ago
GibboKGibboK
34.3k107317542
1
What about TypeScript?
– vlaz
2 hours ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Quite easy.
someArray.indexOf you know that this is a function, which is also an object and can have properties.
By doing someArray.indexOf[someObject], you are trying to reach the property with the key valued to the value of someObject.
Of course, it is not defined on the indexOf function, so it returns undefined.
Quick example that illustrates the syntax and the fact that a function can have properties ;) :
const array = ;
array.indexOf['anyValue'] = 'test';
console.log(array.indexOf.anyValue);EDIT
Here is an attempt of an answer for the TypeScript side of the question.
As you already know, TypeScript is designed to be compatible with JavaScript. Therefore, as in JS, you can access a property of an object by the following ways:
- 'Statically':
obj.property
- 'Dynamically':
obj['property']
Both ways will perform the same, but with the dynamic way of accessing the property, there is no way TypeScript compiler can determine the type of it or if it exists or not. That's why it doesn't raise an error.
(Of course, someone could say the compiler could understand that here, it could read the string 'property', but don't forget that you could pass anything between the brackets and its value could change at run time, far outside of the TypeScript scope.)
By using the 'static' way to access a property, of course, TypeScript will raise an error!
Hoping this helps ;)
answered 2 hours ago
sjahansjahan
3,2251827
3
I think answers here are missing theTypeScripttag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you haveindex: numberandindex = arr.indexOf[obj]then that should be a compilation error. Butindex: anywouldn't throw a compilation error.
– vlaz
2 hours ago
@vlaz +1. sjahan gives OP a quick explanation of theundefinedresult but the main question remains...
– Florian
2 hours ago
add a comment |
Quite easy.
someArray.indexOf you know that this is a function, which is also an object and can have properties.
By doing someArray.indexOf[someObject], you are trying to reach the property with the key valued to the value of someObject.
Of course, it is not defined on the indexOf function, so it returns undefined.
Quick example that illustrates the syntax and the fact that a function can have properties ;) :
const array = ;
array.indexOf['anyValue'] = 'test';
console.log(array.indexOf.anyValue);EDIT
Here is an attempt of an answer for the TypeScript side of the question.
As you already know, TypeScript is designed to be compatible with JavaScript. Therefore, as in JS, you can access a property of an object by the following ways:
- 'Statically':
obj.property
- 'Dynamically':
obj['property']
Both ways will perform the same, but with the dynamic way of accessing the property, there is no way TypeScript compiler can determine the type of it or if it exists or not. That's why it doesn't raise an error.
(Of course, someone could say the compiler could understand that here, it could read the string 'property', but don't forget that you could pass anything between the brackets and its value could change at run time, far outside of the TypeScript scope.)
By using the 'static' way to access a property, of course, TypeScript will raise an error!
Hoping this helps ;)
answered 2 hours ago
sjahansjahan
3,2251827
3
I think answers here are missing theTypeScripttag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you haveindex: numberandindex = arr.indexOf[obj]then that should be a compilation error. Butindex: anywouldn't throw a compilation error.
– vlaz
2 hours ago
@vlaz +1. sjahan gives OP a quick explanation of theundefinedresult but the main question remains...
– Florian
2 hours ago
add a comment |
Quite easy.
someArray.indexOf you know that this is a function, which is also an object and can have properties.
By doing someArray.indexOf[someObject], you are trying to reach the property with the key valued to the value of someObject.
Of course, it is not defined on the indexOf function, so it returns undefined.
Quick example that illustrates the syntax and the fact that a function can have properties ;) :
const array = ;
array.indexOf['anyValue'] = 'test';
console.log(array.indexOf.anyValue);EDIT
Here is an attempt of an answer for the TypeScript side of the question.
As you already know, TypeScript is designed to be compatible with JavaScript. Therefore, as in JS, you can access a property of an object by the following ways:
- 'Statically':
obj.property
- 'Dynamically':
obj['property']
Both ways will perform the same, but with the dynamic way of accessing the property, there is no way TypeScript compiler can determine the type of it or if it exists or not. That's why it doesn't raise an error.
(Of course, someone could say the compiler could understand that here, it could read the string 'property', but don't forget that you could pass anything between the brackets and its value could change at run time, far outside of the TypeScript scope.)
By using the 'static' way to access a property, of course, TypeScript will raise an error!
Hoping this helps ;)
answered 2 hours ago
sjahansjahan
3,2251827
Quite easy.
someArray.indexOf you know that this is a function, which is also an object and can have properties.
By doing someArray.indexOf[someObject], you are trying to reach the property with the key valued to the value of someObject.
Of course, it is not defined on the indexOf function, so it returns undefined.
Quick example that illustrates the syntax and the fact that a function can have properties ;) :
const array = ;
array.indexOf['anyValue'] = 'test';
console.log(array.indexOf.anyValue);EDIT
Here is an attempt of an answer for the TypeScript side of the question.
As you already know, TypeScript is designed to be compatible with JavaScript. Therefore, as in JS, you can access a property of an object by the following ways:
- 'Statically':
obj.property
- 'Dynamically':
obj['property']
Both ways will perform the same, but with the dynamic way of accessing the property, there is no way TypeScript compiler can determine the type of it or if it exists or not. That's why it doesn't raise an error.
(Of course, someone could say the compiler could understand that here, it could read the string 'property', but don't forget that you could pass anything between the brackets and its value could change at run time, far outside of the TypeScript scope.)
By using the 'static' way to access a property, of course, TypeScript will raise an error!
Hoping this helps ;)
const array = ;
array.indexOf['anyValue'] = 'test';
console.log(array.indexOf.anyValue);const array = ;
array.indexOf['anyValue'] = 'test';
console.log(array.indexOf.anyValue);answered 2 hours ago
sjahansjahan
3,2251827
edited 54 mins ago
answered 2 hours ago
sjahansjahan
3,2251827
answered 2 hours ago
sjahansjahan
3,2251827
answered 2 hours ago
sjahansjahan
3,2251827
3,2251827
3
I think answers here are missing theTypeScripttag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you haveindex: numberandindex = arr.indexOf[obj]then that should be a compilation error. Butindex: anywouldn't throw a compilation error.
– vlaz
2 hours ago
@vlaz +1. sjahan gives OP a quick explanation of theundefinedresult but the main question remains...
– Florian
2 hours ago
add a comment |
3
I think answers here are missing theTypeScripttag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you haveindex: numberandindex = arr.indexOf[obj]then that should be a compilation error. Butindex: anywouldn't throw a compilation error.
– vlaz
2 hours ago
@vlaz +1. sjahan gives OP a quick explanation of theundefinedresult but the main question remains...
– Florian
2 hours ago
3
3
I think answers here are missing the
TypeScript tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you have index: number and index = arr.indexOf[obj] then that should be a compilation error. But index: any wouldn't throw a compilation error.– vlaz
2 hours ago
I think answers here are missing the
TypeScript tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you have index: number and index = arr.indexOf[obj] then that should be a compilation error. But index: any wouldn't throw a compilation error.– vlaz
2 hours ago
@vlaz +1. sjahan gives OP a quick explanation of the
undefined result but the main question remains...– Florian
2 hours ago
@vlaz +1. sjahan gives OP a quick explanation of the
undefined result but the main question remains...– Florian
2 hours ago
add a comment |
It does not error because the --noImplicitAny compiler option is not enabled. With that compiler option enabled you will get an error as expected:
The reason is that using an element access expression creates an implicit any type when the type has no index signature defined.
So again, since --noImplicitAny is not enabled, it does not error. I highly recommend turning this compiler option on.
answered 35 mins ago
David SherretDavid Sherret
50.9k16120122
add a comment |
It does not error because the --noImplicitAny compiler option is not enabled. With that compiler option enabled you will get an error as expected:
The reason is that using an element access expression creates an implicit any type when the type has no index signature defined.
So again, since --noImplicitAny is not enabled, it does not error. I highly recommend turning this compiler option on.
answered 35 mins ago
David SherretDavid Sherret
50.9k16120122
add a comment |
It does not error because the --noImplicitAny compiler option is not enabled. With that compiler option enabled you will get an error as expected:
The reason is that using an element access expression creates an implicit any type when the type has no index signature defined.
So again, since --noImplicitAny is not enabled, it does not error. I highly recommend turning this compiler option on.
answered 35 mins ago
David SherretDavid Sherret
50.9k16120122
It does not error because the --noImplicitAny compiler option is not enabled. With that compiler option enabled you will get an error as expected:
The reason is that using an element access expression creates an implicit any type when the type has no index signature defined.
So again, since --noImplicitAny is not enabled, it does not error. I highly recommend turning this compiler option on.
answered 35 mins ago
David SherretDavid Sherret
50.9k16120122
edited 6 mins ago
answered 35 mins ago
David SherretDavid Sherret
50.9k16120122
answered 35 mins ago
David SherretDavid Sherret
50.9k16120122
answered 35 mins ago
David SherretDavid Sherret
50.9k16120122
50.9k16120122
add a comment |
add a comment |
array.indexOf is a function.
Functions are objects.
You were accessing the someObject property of the array.indexOf function.
You would have got undefined.
const array = [1, 2, 3]
const someObject = 'asdasd'
console.log(array.indexOf[someObject])
// undefinedanswered 2 hours ago
0xc14m1z0xc14m1z
1,419512
1
Array.indexOfisundefined,Array.prototype.indexOf, on the other hand, is a function.
– Pavlo
1 hour ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
1 hour ago
add a comment |
array.indexOf is a function.
Functions are objects.
You were accessing the someObject property of the array.indexOf function.
You would have got undefined.
const array = [1, 2, 3]
const someObject = 'asdasd'
console.log(array.indexOf[someObject])
// undefinedanswered 2 hours ago
0xc14m1z0xc14m1z
1,419512
1
Array.indexOfisundefined,Array.prototype.indexOf, on the other hand, is a function.
– Pavlo
1 hour ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
1 hour ago
add a comment |
array.indexOf is a function.
Functions are objects.
You were accessing the someObject property of the array.indexOf function.
You would have got undefined.
const array = [1, 2, 3]
const someObject = 'asdasd'
console.log(array.indexOf[someObject])
// undefinedanswered 2 hours ago
0xc14m1z0xc14m1z
1,419512
array.indexOf is a function.
Functions are objects.
You were accessing the someObject property of the array.indexOf function.
You would have got undefined.
const array = [1, 2, 3]
const someObject = 'asdasd'
console.log(array.indexOf[someObject])
// undefinedconst array = [1, 2, 3]
const someObject = 'asdasd'
console.log(array.indexOf[someObject])
// undefinedconst array = [1, 2, 3]
const someObject = 'asdasd'
console.log(array.indexOf[someObject])
// undefinedanswered 2 hours ago
0xc14m1z0xc14m1z
1,419512
edited 1 hour ago
answered 2 hours ago
0xc14m1z0xc14m1z
1,419512
answered 2 hours ago
0xc14m1z0xc14m1z
1,419512
answered 2 hours ago
0xc14m1z0xc14m1z
1,419512
1,419512
1
Array.indexOfisundefined,Array.prototype.indexOf, on the other hand, is a function.
– Pavlo
1 hour ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
1 hour ago
add a comment |
1
Array.indexOfisundefined,Array.prototype.indexOf, on the other hand, is a function.
– Pavlo
1 hour ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
1 hour ago
1
1
Array.indexOf is undefined, Array.prototype.indexOf, on the other hand, is a function.– Pavlo
1 hour ago
Array.indexOf is undefined, Array.prototype.indexOf, on the other hand, is a function.– Pavlo
1 hour ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
1 hour ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
1 hour ago
add a comment |
Functions in JavaScript are first class objects.
When you access function Array.indexOf() via bracket notation Array.indexOf['prop'] you actually trying to access a property which does not exist on indexOf so you get undefined.
answered 2 hours ago
GibboKGibboK
34.3k107317542
1
What about TypeScript?
– vlaz
2 hours ago
add a comment |
Functions in JavaScript are first class objects.
When you access function Array.indexOf() via bracket notation Array.indexOf['prop'] you actually trying to access a property which does not exist on indexOf so you get undefined.
answered 2 hours ago
GibboKGibboK
34.3k107317542
1
What about TypeScript?
– vlaz
2 hours ago
add a comment |
Functions in JavaScript are first class objects.
When you access function Array.indexOf() via bracket notation Array.indexOf['prop'] you actually trying to access a property which does not exist on indexOf so you get undefined.
answered 2 hours ago
GibboKGibboK
34.3k107317542
Functions in JavaScript are first class objects.
When you access function Array.indexOf() via bracket notation Array.indexOf['prop'] you actually trying to access a property which does not exist on indexOf so you get undefined.
answered 2 hours ago
GibboKGibboK
34.3k107317542
answered 2 hours ago
GibboKGibboK
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answered 2 hours ago
GibboKGibboK
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answered 2 hours ago
GibboKGibboK
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34.3k107317542
1
What about TypeScript?
– vlaz
2 hours ago
add a comment |
1
What about TypeScript?
– vlaz
2 hours ago
1
1
What about TypeScript?
– vlaz
2 hours ago
What about TypeScript?
– vlaz
2 hours ago
add a comment |
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What did you assign the result of? Because what you wrote is valid you are taking the
someObjectmember of theindexOfmethod. Well, trying to. The only error would come from TypeScript compilation and only if you try to assign the result to something that doesn't match the expected type.– vlaz
2 hours ago
1
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
2 hours ago
2
@DeWetvanAs I am actually curious about your problem - this seems like a genuine bug/problem with TypeScript see example here. It seems that if you are trying to assign to a variable of type
number, the result of.indexOf[someObject]shouldn't be considered anumberand thus the compilation would fail. That's the whole idea of TypeScript is - to enforce the types. The answers here focus on JS but ignore this.– vlaz
1 hour ago
3
@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.
– Bergi
1 hour ago
@Bergi As the question has underlined more and more the TS side, I've edited my answer! I think this should explain it :)
– sjahan
53 mins ago