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
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
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
2 days ago
11
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
2 days 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
2 days ago
@vlaz it was in fact a complex object with multiple properties.
– De Wet van As
2 days 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
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 2 days ago
Bergi
366k58546872
366k58546872
asked 2 days ago
De Wet van AsDe Wet van As
1318
1318
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
2 days ago
11
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
2 days 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
2 days ago
@vlaz it was in fact a complex object with multiple properties.
– De Wet van As
2 days ago
add a 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
2 days ago
11
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
2 days 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
2 days ago
@vlaz it was in fact a complex object with multiple properties.
– De Wet van As
2 days 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 days 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 days ago
11
11
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
2 days ago
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
2 days 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
2 days 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
2 days ago
@vlaz it was in fact a complex object with multiple properties.
– De Wet van As
2 days ago
@vlaz it was in fact a complex object with multiple properties.
– De Wet van As
2 days 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']
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
2 days ago
@vlaz +1. sjahan gives OP a quick explanation of theundefined
result but the main question remains...
– Florian
2 days 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
2 days 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
2 days 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
2 days ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
2 days 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
2 days ago
@vlaz +1. sjahan gives OP a quick explanation of theundefined
result but the main question remains...
– Florian
2 days 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
2 days 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
2 days 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
2 days ago
@vlaz +1. sjahan gives OP a quick explanation of theundefined
result but the main question remains...
– Florian
2 days 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
2 days 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
2 days 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 2 days ago
answered 2 days ago
sjahansjahan
3,2802928
3,2802928
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
2 days ago
@vlaz +1. sjahan gives OP a quick explanation of theundefined
result but the main question remains...
– Florian
2 days 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
2 days 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
2 days 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
2 days ago
@vlaz +1. sjahan gives OP a quick explanation of theundefined
result but the main question remains...
– Florian
2 days 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
2 days 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
2 days 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
2 days 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 days ago
@vlaz +1. sjahan gives OP a quick explanation of the
undefined
result but the main question remains...– Florian
2 days ago
@vlaz +1. sjahan gives OP a quick explanation of the
undefined
result but the main question remains...– Florian
2 days 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
2 days 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
2 days 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
2 days 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
2 days 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 2 days ago
answered 2 days ago
David SherretDavid Sherret
51.1k16120123
51.1k16120123
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
2 days ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
2 days 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
2 days ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
2 days 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 2 days ago
answered 2 days ago
0xc14m1z0xc14m1z
1,534613
1,534613
1
Array.indexOf
isundefined
,Array.prototype.indexOf
, on the other hand, is a function.
– Pavlo
2 days ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
2 days ago
add a comment |
1
Array.indexOf
isundefined
,Array.prototype.indexOf
, on the other hand, is a function.
– Pavlo
2 days ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
2 days ago
1
1
Array.indexOf
is undefined
, Array.prototype.indexOf
, on the other hand, is a function.– Pavlo
2 days ago
Array.indexOf
is undefined
, Array.prototype.indexOf
, on the other hand, is a function.– Pavlo
2 days ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
2 days ago
You’re right! I fixed the typo in my answer. Thanks
– 0xc14m1z
2 days 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 2 days ago
Alex SteinbergAlex Steinberg
429411
429411
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
2 days ago
11
Welcome to javascript, where everything is an object!
– Jean-Baptiste Yunès
2 days 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
2 days ago
@vlaz it was in fact a complex object with multiple properties.
– De Wet van As
2 days ago