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
|
show 1 more 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
1
What did you assign the result of? Because what you wrote is valid you are taking thesomeObject
member of theindexOf
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
15 hours ago
9
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
15 hours ago
7
@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 anumber
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
14 hours ago
4
@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.
– Bergi
14 hours 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
13 hours ago
|
show 1 more 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
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 14 hours ago
Bergi
366k58546872
366k58546872
asked 15 hours ago
De Wet van AsDe Wet van As
1098
1098
1
What did you assign the result of? Because what you wrote is valid you are taking thesomeObject
member of theindexOf
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
15 hours ago
9
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
15 hours ago
7
@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 anumber
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
14 hours ago
4
@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.
– Bergi
14 hours 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
13 hours ago
|
show 1 more comment
1
What did you assign the result of? Because what you wrote is valid you are taking thesomeObject
member of theindexOf
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
15 hours ago
9
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
15 hours ago
7
@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 anumber
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
14 hours ago
4
@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.
– Bergi
14 hours 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
13 hours 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
15 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
15 hours ago
9
9
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
15 hours ago
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
15 hours ago
7
7
@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
14 hours 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
14 hours ago
4
4
@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.
– Bergi
14 hours 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
14 hours 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
13 hours 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
13 hours ago
|
show 1 more 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']
By using the 'static' way to access a property, of course, TypeScript will raise an error!
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, since the value between bracket will be evaluated at runtime, after TypeScript transpiling.
That's why it will be implicitly marked as any
.
As David Sherret mentioned in his answer, you can force TypeScript to raise an error by adding the flag --noImplicitAny
, please refer to his answer for more details about this!
Hoping this helps ;)
4
I think answers here are missing theTypeScript
tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you haveindex: number
andindex = arr.indexOf[obj]
then that should be a compilation error. Butindex: any
wouldn't throw a compilation error.
– vlaz
15 hours ago
@vlaz +1. sjahan gives OP a quick explanation of theundefined
result but the main question remains...
– Florian
15 hours ago
1
"Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.
– vlaz
12 hours ago
@vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.
– sjahan
12 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 an element access expression returns an object typed as any
when the type has no index signature defined (this is an implicit any
).
So again, since --noImplicitAny
is not enabled, it does not error. I highly recommend turning this compiler option on.
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])
// undefined
1
Array.indexOf
isundefined
,Array.prototype.indexOf
, on the other hand, is a function.
– Pavlo
14 hours ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
14 hours ago
add a comment |
The only real issue here is that you expected Typescript to throw an error that exposed the problem in your logic. The intended logic was to use curly braces and utilise the someArray.indexOf(someObject)
function.
What happened when you used square braces, someArray.indexOf[someObject]
, was that the JS runtime first converted your object someObject
into a string by calling the function someObject.toString
, which most likely returned "[object object]"
. Then the someArray.indexOf
object was queried for the key "[object object]"
which wasn't present, returning undefined
. As far as Typescript goes, this is completely fine.
David Sherret pointed out that --noImplicitAny
would have pointed out the error, but it would only have pointed out a different error, as he explained, which would not directly have helped you to find the flaw in your logic.
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']
By using the 'static' way to access a property, of course, TypeScript will raise an error!
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, since the value between bracket will be evaluated at runtime, after TypeScript transpiling.
That's why it will be implicitly marked as any
.
As David Sherret mentioned in his answer, you can force TypeScript to raise an error by adding the flag --noImplicitAny
, please refer to his answer for more details about this!
Hoping this helps ;)
4
I think answers here are missing theTypeScript
tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you haveindex: number
andindex = arr.indexOf[obj]
then that should be a compilation error. Butindex: any
wouldn't throw a compilation error.
– vlaz
15 hours ago
@vlaz +1. sjahan gives OP a quick explanation of theundefined
result but the main question remains...
– Florian
15 hours ago
1
"Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.
– vlaz
12 hours ago
@vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.
– sjahan
12 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']
By using the 'static' way to access a property, of course, TypeScript will raise an error!
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, since the value between bracket will be evaluated at runtime, after TypeScript transpiling.
That's why it will be implicitly marked as any
.
As David Sherret mentioned in his answer, you can force TypeScript to raise an error by adding the flag --noImplicitAny
, please refer to his answer for more details about this!
Hoping this helps ;)
4
I think answers here are missing theTypeScript
tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you haveindex: number
andindex = arr.indexOf[obj]
then that should be a compilation error. Butindex: any
wouldn't throw a compilation error.
– vlaz
15 hours ago
@vlaz +1. sjahan gives OP a quick explanation of theundefined
result but the main question remains...
– Florian
15 hours ago
1
"Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.
– vlaz
12 hours ago
@vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.
– sjahan
12 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']
By using the 'static' way to access a property, of course, TypeScript will raise an error!
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, since the value between bracket will be evaluated at runtime, after TypeScript transpiling.
That's why it will be implicitly marked as any
.
As David Sherret mentioned in his answer, you can force TypeScript to raise an error by adding the flag --noImplicitAny
, please refer to his answer for more details about this!
Hoping this helps ;)
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']
By using the 'static' way to access a property, of course, TypeScript will raise an error!
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, since the value between bracket will be evaluated at runtime, after TypeScript transpiling.
That's why it will be implicitly marked as any
.
As David Sherret mentioned in his answer, you can force TypeScript to raise an error by adding the flag --noImplicitAny
, please refer to his answer for more details about this!
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);
edited 12 hours ago
answered 15 hours ago
sjahansjahan
3,2501828
3,2501828
4
I think answers here are missing theTypeScript
tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you haveindex: number
andindex = arr.indexOf[obj]
then that should be a compilation error. Butindex: any
wouldn't throw a compilation error.
– vlaz
15 hours ago
@vlaz +1. sjahan gives OP a quick explanation of theundefined
result but the main question remains...
– Florian
15 hours ago
1
"Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.
– vlaz
12 hours ago
@vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.
– sjahan
12 hours ago
add a comment |
4
I think answers here are missing theTypeScript
tag. It's entirely reasonable to expect a compilation error in TS. Then again, it depends if you haveindex: number
andindex = arr.indexOf[obj]
then that should be a compilation error. Butindex: any
wouldn't throw a compilation error.
– vlaz
15 hours ago
@vlaz +1. sjahan gives OP a quick explanation of theundefined
result but the main question remains...
– Florian
15 hours ago
1
"Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.
– vlaz
12 hours ago
@vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.
– sjahan
12 hours ago
4
4
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
15 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
15 hours ago
@vlaz +1. sjahan gives OP a quick explanation of the
undefined
result but the main question remains...– Florian
15 hours ago
@vlaz +1. sjahan gives OP a quick explanation of the
undefined
result but the main question remains...– Florian
15 hours ago
1
1
"Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.
– vlaz
12 hours ago
"Of course, someone could say the compiler could understand" Yes, the compiler would understand. That is the entire reason for the compiler to be there and if TypeScript can't figure out that type is undetermined, then what good is TS at all? That's just a logical examination of the statement - David Sherret's answer points out that TS does indeed understand undetermined types and can raise an error for it. So your edit dispenses incorrect assumptions and information.
– vlaz
12 hours ago
@vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.
– sjahan
12 hours ago
@vlaz I think I didn't express myself correctly and my point could get misinterpreted. I'm going to edit the last part, thank you.
– sjahan
12 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 an element access expression returns an object typed as any
when the type has no index signature defined (this is an implicit any
).
So again, since --noImplicitAny
is not enabled, it does not error. I highly recommend turning this compiler option on.
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 an element access expression returns an object typed as any
when the type has no index signature defined (this is an implicit any
).
So again, since --noImplicitAny
is not enabled, it does not error. I highly recommend turning this compiler option on.
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 an element access expression returns an object typed as any
when the type has no index signature defined (this is an implicit any
).
So again, since --noImplicitAny
is not enabled, it does not error. I highly recommend turning this compiler option on.
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 an element access expression returns an object typed as any
when the type has no index signature defined (this is an implicit any
).
So again, since --noImplicitAny
is not enabled, it does not error. I highly recommend turning this compiler option on.
edited 9 hours ago
answered 13 hours ago
David SherretDavid Sherret
51k16120123
51k16120123
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])
// undefined
1
Array.indexOf
isundefined
,Array.prototype.indexOf
, on the other hand, is a function.
– Pavlo
14 hours ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
14 hours 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])
// undefined
1
Array.indexOf
isundefined
,Array.prototype.indexOf
, on the other hand, is a function.
– Pavlo
14 hours ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
14 hours 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])
// undefined
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])
// undefined
const array = [1, 2, 3]
const someObject = 'asdasd'
console.log(array.indexOf[someObject])
// undefined
const array = [1, 2, 3]
const someObject = 'asdasd'
console.log(array.indexOf[someObject])
// undefined
edited 14 hours ago
answered 15 hours ago
0xc14m1z0xc14m1z
1,469513
1,469513
1
Array.indexOf
isundefined
,Array.prototype.indexOf
, on the other hand, is a function.
– Pavlo
14 hours ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
14 hours ago
add a comment |
1
Array.indexOf
isundefined
,Array.prototype.indexOf
, on the other hand, is a function.
– Pavlo
14 hours ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
14 hours ago
1
1
Array.indexOf
is undefined
, Array.prototype.indexOf
, on the other hand, is a function.– Pavlo
14 hours ago
Array.indexOf
is undefined
, Array.prototype.indexOf
, on the other hand, is a function.– Pavlo
14 hours ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
14 hours ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
14 hours ago
add a comment |
The only real issue here is that you expected Typescript to throw an error that exposed the problem in your logic. The intended logic was to use curly braces and utilise the someArray.indexOf(someObject)
function.
What happened when you used square braces, someArray.indexOf[someObject]
, was that the JS runtime first converted your object someObject
into a string by calling the function someObject.toString
, which most likely returned "[object object]"
. Then the someArray.indexOf
object was queried for the key "[object object]"
which wasn't present, returning undefined
. As far as Typescript goes, this is completely fine.
David Sherret pointed out that --noImplicitAny
would have pointed out the error, but it would only have pointed out a different error, as he explained, which would not directly have helped you to find the flaw in your logic.
add a comment |
The only real issue here is that you expected Typescript to throw an error that exposed the problem in your logic. The intended logic was to use curly braces and utilise the someArray.indexOf(someObject)
function.
What happened when you used square braces, someArray.indexOf[someObject]
, was that the JS runtime first converted your object someObject
into a string by calling the function someObject.toString
, which most likely returned "[object object]"
. Then the someArray.indexOf
object was queried for the key "[object object]"
which wasn't present, returning undefined
. As far as Typescript goes, this is completely fine.
David Sherret pointed out that --noImplicitAny
would have pointed out the error, but it would only have pointed out a different error, as he explained, which would not directly have helped you to find the flaw in your logic.
add a comment |
The only real issue here is that you expected Typescript to throw an error that exposed the problem in your logic. The intended logic was to use curly braces and utilise the someArray.indexOf(someObject)
function.
What happened when you used square braces, someArray.indexOf[someObject]
, was that the JS runtime first converted your object someObject
into a string by calling the function someObject.toString
, which most likely returned "[object object]"
. Then the someArray.indexOf
object was queried for the key "[object object]"
which wasn't present, returning undefined
. As far as Typescript goes, this is completely fine.
David Sherret pointed out that --noImplicitAny
would have pointed out the error, but it would only have pointed out a different error, as he explained, which would not directly have helped you to find the flaw in your logic.
The only real issue here is that you expected Typescript to throw an error that exposed the problem in your logic. The intended logic was to use curly braces and utilise the someArray.indexOf(someObject)
function.
What happened when you used square braces, someArray.indexOf[someObject]
, was that the JS runtime first converted your object someObject
into a string by calling the function someObject.toString
, which most likely returned "[object object]"
. Then the someArray.indexOf
object was queried for the key "[object object]"
which wasn't present, returning undefined
. As far as Typescript goes, this is completely fine.
David Sherret pointed out that --noImplicitAny
would have pointed out the error, but it would only have pointed out a different error, as he explained, which would not directly have helped you to find the flaw in your logic.
answered 5 hours ago
Alex SteinbergAlex Steinberg
419411
419411
add a comment |
add a comment |
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1
What did you assign the result of? Because what you wrote is valid you are taking the
someObject
member of theindexOf
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
15 hours ago
9
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
15 hours ago
7
@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 anumber
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
14 hours ago
4
@DeWetvanAs You might want to not accept an answer yet if none of them explains why you are not getting a TypeScript error.
– Bergi
14 hours 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
13 hours ago