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Typescript conditional return type based on string argument
The 2019 Stack Overflow Developer Survey Results Are InTypescript: declare return type to be dependent on argument value?How do I make the method return type generic?Passing arguments to C# generic new() of templated typeType definition in object literal in TypeScriptTypeScript Objects as Dictionary types as in C#Are strongly-typed functions as parameters possible in TypeScript?TypeScript Converting a String to a numberHow to convert string to enum in TypeScript?Call new Date with union type “number | string”Typescript: Interfaces vs TypesUsing string literals types in Typescript
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When using union of string literals as input argument, what would it take to remove the casts and put the type into the function header:
const get = <T extends "barcode" | "mqtt">(s: T) =>
s === "barcode" ?
<T extends "barcode" ? scan: () => string : pan: () => string>scan: () => "we are scanning" :
<T extends "barcode" ? scan: () => string : pan: () => string>pan: () => "we are panning"
get("barcode").scan() // OK
get("mqtt").pan() // OK
get("barcode").pan() // Error
I ran into this trying to answer someone else's question: https://stackoverflow.com/a/55059318/2684980.
typescript generics
asked Mar 8 at 8:36
Jørgen TvedtJørgen Tvedt
4501416
add a comment |
When using union of string literals as input argument, what would it take to remove the casts and put the type into the function header:
const get = <T extends "barcode" | "mqtt">(s: T) =>
s === "barcode" ?
<T extends "barcode" ? scan: () => string : pan: () => string>scan: () => "we are scanning" :
<T extends "barcode" ? scan: () => string : pan: () => string>pan: () => "we are panning"
get("barcode").scan() // OK
get("mqtt").pan() // OK
get("barcode").pan() // Error
I ran into this trying to answer someone else's question: https://stackoverflow.com/a/55059318/2684980.
typescript generics
asked Mar 8 at 8:36
Jørgen TvedtJørgen Tvedt
4501416
add a comment |
When using union of string literals as input argument, what would it take to remove the casts and put the type into the function header:
const get = <T extends "barcode" | "mqtt">(s: T) =>
s === "barcode" ?
<T extends "barcode" ? scan: () => string : pan: () => string>scan: () => "we are scanning" :
<T extends "barcode" ? scan: () => string : pan: () => string>pan: () => "we are panning"
get("barcode").scan() // OK
get("mqtt").pan() // OK
get("barcode").pan() // Error
I ran into this trying to answer someone else's question: https://stackoverflow.com/a/55059318/2684980.
typescript generics
asked Mar 8 at 8:36
Jørgen TvedtJørgen Tvedt
4501416
When using union of string literals as input argument, what would it take to remove the casts and put the type into the function header:
const get = <T extends "barcode" | "mqtt">(s: T) =>
s === "barcode" ?
<T extends "barcode" ? scan: () => string : pan: () => string>scan: () => "we are scanning" :
<T extends "barcode" ? scan: () => string : pan: () => string>pan: () => "we are panning"
get("barcode").scan() // OK
get("mqtt").pan() // OK
get("barcode").pan() // Error
I ran into this trying to answer someone else's question: https://stackoverflow.com/a/55059318/2684980.
typescript generics
typescript generics
asked Mar 8 at 8:36
Jørgen TvedtJørgen Tvedt
4501416
asked Mar 8 at 8:36
Jørgen TvedtJørgen Tvedt
4501416
asked Mar 8 at 8:36
Jørgen TvedtJørgen Tvedt
4501416
asked Mar 8 at 8:36
Jørgen TvedtJørgen Tvedt
4501416
asked Mar 8 at 8:36
Jørgen TvedtJørgen Tvedt
4501416
4501416
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The cleanest solution is such cases (although not any more type safe than the type assertions) is to use overloads instead. You can use conditional types in the public signature, and a simple union in the implementation signature. You will need to switch to a function declaration as function expressions (arrow or regular) don't easily suport overloads:
function get<T extends "barcode" | "mqtt">(s: T): T extends "barcode" ? scan: () => string : pan: () => string
function get(s: "barcode" | "mqtt"): scan: () => string | pan: () => string
return s === "barcode" ?
scan: () => "we are scanning" :
pan: () => "we are panning"
get("barcode").scan() // OK
get("mqtt").pan() // OK
get("barcode").pan() // Error
answered Mar 8 at 9:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
73.3k35370
This would be better, although not exactly what I was hoping for. Do you have to specify the return type in the implementation?
– Jørgen Tvedt
Mar 8 at 13:33
@JørgenTvedt I think you can get away with removing it from the implementation. You will get an error id the conditional branches are not in the returned union
– Titian Cernicova-Dragomir
Mar 8 at 14:38
Ok, but I still don't understand why my code does not work without the casts, that's why I hesitate to mark this as the answer.
– Jørgen Tvedt
Mar 9 at 16:37
1
@JørgenTvedt because Typescript does not reason about conditional types that contain unresolved type parameters. Except for very limited circumstances, it will just check for an exact type match.
– Titian Cernicova-Dragomir
Mar 9 at 17:25
1
@JørgenTvedt Typescript does not take flow analysis to this level. Flow analysis will impact just the types of the variables inside the guarded block. It will not help the compiler infer any conditional types.
– Titian Cernicova-Dragomir
Mar 10 at 11:04
|
show 1 more comment
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
The cleanest solution is such cases (although not any more type safe than the type assertions) is to use overloads instead. You can use conditional types in the public signature, and a simple union in the implementation signature. You will need to switch to a function declaration as function expressions (arrow or regular) don't easily suport overloads:
function get<T extends "barcode" | "mqtt">(s: T): T extends "barcode" ? scan: () => string : pan: () => string
function get(s: "barcode" | "mqtt"): scan: () => string | pan: () => string
return s === "barcode" ?
scan: () => "we are scanning" :
pan: () => "we are panning"
get("barcode").scan() // OK
get("mqtt").pan() // OK
get("barcode").pan() // Error
answered Mar 8 at 9:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
73.3k35370
This would be better, although not exactly what I was hoping for. Do you have to specify the return type in the implementation?
– Jørgen Tvedt
Mar 8 at 13:33
@JørgenTvedt I think you can get away with removing it from the implementation. You will get an error id the conditional branches are not in the returned union
– Titian Cernicova-Dragomir
Mar 8 at 14:38
Ok, but I still don't understand why my code does not work without the casts, that's why I hesitate to mark this as the answer.
– Jørgen Tvedt
Mar 9 at 16:37
1
@JørgenTvedt because Typescript does not reason about conditional types that contain unresolved type parameters. Except for very limited circumstances, it will just check for an exact type match.
– Titian Cernicova-Dragomir
Mar 9 at 17:25
1
@JørgenTvedt Typescript does not take flow analysis to this level. Flow analysis will impact just the types of the variables inside the guarded block. It will not help the compiler infer any conditional types.
– Titian Cernicova-Dragomir
Mar 10 at 11:04
|
show 1 more comment
The cleanest solution is such cases (although not any more type safe than the type assertions) is to use overloads instead. You can use conditional types in the public signature, and a simple union in the implementation signature. You will need to switch to a function declaration as function expressions (arrow or regular) don't easily suport overloads:
function get<T extends "barcode" | "mqtt">(s: T): T extends "barcode" ? scan: () => string : pan: () => string
function get(s: "barcode" | "mqtt"): scan: () => string | pan: () => string
return s === "barcode" ?
scan: () => "we are scanning" :
pan: () => "we are panning"
get("barcode").scan() // OK
get("mqtt").pan() // OK
get("barcode").pan() // Error
answered Mar 8 at 9:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
73.3k35370
This would be better, although not exactly what I was hoping for. Do you have to specify the return type in the implementation?
– Jørgen Tvedt
Mar 8 at 13:33
@JørgenTvedt I think you can get away with removing it from the implementation. You will get an error id the conditional branches are not in the returned union
– Titian Cernicova-Dragomir
Mar 8 at 14:38
Ok, but I still don't understand why my code does not work without the casts, that's why I hesitate to mark this as the answer.
– Jørgen Tvedt
Mar 9 at 16:37
1
@JørgenTvedt because Typescript does not reason about conditional types that contain unresolved type parameters. Except for very limited circumstances, it will just check for an exact type match.
– Titian Cernicova-Dragomir
Mar 9 at 17:25
1
@JørgenTvedt Typescript does not take flow analysis to this level. Flow analysis will impact just the types of the variables inside the guarded block. It will not help the compiler infer any conditional types.
– Titian Cernicova-Dragomir
Mar 10 at 11:04
|
show 1 more comment
The cleanest solution is such cases (although not any more type safe than the type assertions) is to use overloads instead. You can use conditional types in the public signature, and a simple union in the implementation signature. You will need to switch to a function declaration as function expressions (arrow or regular) don't easily suport overloads:
function get<T extends "barcode" | "mqtt">(s: T): T extends "barcode" ? scan: () => string : pan: () => string
function get(s: "barcode" | "mqtt"): scan: () => string | pan: () => string
return s === "barcode" ?
scan: () => "we are scanning" :
pan: () => "we are panning"
get("barcode").scan() // OK
get("mqtt").pan() // OK
get("barcode").pan() // Error
answered Mar 8 at 9:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
73.3k35370
The cleanest solution is such cases (although not any more type safe than the type assertions) is to use overloads instead. You can use conditional types in the public signature, and a simple union in the implementation signature. You will need to switch to a function declaration as function expressions (arrow or regular) don't easily suport overloads:
function get<T extends "barcode" | "mqtt">(s: T): T extends "barcode" ? scan: () => string : pan: () => string
function get(s: "barcode" | "mqtt"): scan: () => string | pan: () => string
return s === "barcode" ?
scan: () => "we are scanning" :
pan: () => "we are panning"
get("barcode").scan() // OK
get("mqtt").pan() // OK
get("barcode").pan() // Error
answered Mar 8 at 9:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
73.3k35370
answered Mar 8 at 9:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
73.3k35370
answered Mar 8 at 9:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
73.3k35370
answered Mar 8 at 9:09
Titian Cernicova-DragomirTitian Cernicova-Dragomir
73.3k35370
73.3k35370
This would be better, although not exactly what I was hoping for. Do you have to specify the return type in the implementation?
– Jørgen Tvedt
Mar 8 at 13:33
@JørgenTvedt I think you can get away with removing it from the implementation. You will get an error id the conditional branches are not in the returned union
– Titian Cernicova-Dragomir
Mar 8 at 14:38
Ok, but I still don't understand why my code does not work without the casts, that's why I hesitate to mark this as the answer.
– Jørgen Tvedt
Mar 9 at 16:37
1
@JørgenTvedt because Typescript does not reason about conditional types that contain unresolved type parameters. Except for very limited circumstances, it will just check for an exact type match.
– Titian Cernicova-Dragomir
Mar 9 at 17:25
1
@JørgenTvedt Typescript does not take flow analysis to this level. Flow analysis will impact just the types of the variables inside the guarded block. It will not help the compiler infer any conditional types.
– Titian Cernicova-Dragomir
Mar 10 at 11:04
|
show 1 more comment
This would be better, although not exactly what I was hoping for. Do you have to specify the return type in the implementation?
– Jørgen Tvedt
Mar 8 at 13:33
@JørgenTvedt I think you can get away with removing it from the implementation. You will get an error id the conditional branches are not in the returned union
– Titian Cernicova-Dragomir
Mar 8 at 14:38
Ok, but I still don't understand why my code does not work without the casts, that's why I hesitate to mark this as the answer.
– Jørgen Tvedt
Mar 9 at 16:37
1
@JørgenTvedt because Typescript does not reason about conditional types that contain unresolved type parameters. Except for very limited circumstances, it will just check for an exact type match.
– Titian Cernicova-Dragomir
Mar 9 at 17:25
1
@JørgenTvedt Typescript does not take flow analysis to this level. Flow analysis will impact just the types of the variables inside the guarded block. It will not help the compiler infer any conditional types.
– Titian Cernicova-Dragomir
Mar 10 at 11:04
This would be better, although not exactly what I was hoping for. Do you have to specify the return type in the implementation?
– Jørgen Tvedt
Mar 8 at 13:33
This would be better, although not exactly what I was hoping for. Do you have to specify the return type in the implementation?
– Jørgen Tvedt
Mar 8 at 13:33
@JørgenTvedt I think you can get away with removing it from the implementation. You will get an error id the conditional branches are not in the returned union
– Titian Cernicova-Dragomir
Mar 8 at 14:38
@JørgenTvedt I think you can get away with removing it from the implementation. You will get an error id the conditional branches are not in the returned union
– Titian Cernicova-Dragomir
Mar 8 at 14:38
Ok, but I still don't understand why my code does not work without the casts, that's why I hesitate to mark this as the answer.
– Jørgen Tvedt
Mar 9 at 16:37
Ok, but I still don't understand why my code does not work without the casts, that's why I hesitate to mark this as the answer.
– Jørgen Tvedt
Mar 9 at 16:37
1
1
@JørgenTvedt because Typescript does not reason about conditional types that contain unresolved type parameters. Except for very limited circumstances, it will just check for an exact type match.
– Titian Cernicova-Dragomir
Mar 9 at 17:25
@JørgenTvedt because Typescript does not reason about conditional types that contain unresolved type parameters. Except for very limited circumstances, it will just check for an exact type match.
– Titian Cernicova-Dragomir
Mar 9 at 17:25
1
1
@JørgenTvedt Typescript does not take flow analysis to this level. Flow analysis will impact just the types of the variables inside the guarded block. It will not help the compiler infer any conditional types.
– Titian Cernicova-Dragomir
Mar 10 at 11:04
@JørgenTvedt Typescript does not take flow analysis to this level. Flow analysis will impact just the types of the variables inside the guarded block. It will not help the compiler infer any conditional types.
– Titian Cernicova-Dragomir
Mar 10 at 11:04
|
show 1 more comment
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