Enforce key and value types of non-indexed members
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1
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I have a function which can take any object literal so long as all of the values in the object are strings:
function getFirstLetters(obj: { [key: string]: string }): string {
return Object.keys(obj).map(key => obj[key][0]);
}
This works well for any indexed type, the problem arises when I try to use non-indexed objects:
interface SomeData {
user: string;
loc: string;
}
const someData: SomeData = {
user: "coolGuy42",
loc: "New York",
};
function getFirstLetters(obj: { [key: string]: string }): string {
return Object.keys(obj).map(key => obj[key][0]);
}
// Argument of type 'SomeData' is not
// assignable to parameter of type
// '{ [key: string]: string; }'.
// Index signature is missing in type 'SomeData'.
getFirstLetters(someData);
The error is straightforward - I have specifically requested that the function validate obj
based on it having an index signature, NOT on the type of its values alone.
Is there any way to make my function work with all objects with a uniform value type without asking anyone who uses it to include an index signature in their interface?
typescript
add a comment |
up vote
1
down vote
favorite
I have a function which can take any object literal so long as all of the values in the object are strings:
function getFirstLetters(obj: { [key: string]: string }): string {
return Object.keys(obj).map(key => obj[key][0]);
}
This works well for any indexed type, the problem arises when I try to use non-indexed objects:
interface SomeData {
user: string;
loc: string;
}
const someData: SomeData = {
user: "coolGuy42",
loc: "New York",
};
function getFirstLetters(obj: { [key: string]: string }): string {
return Object.keys(obj).map(key => obj[key][0]);
}
// Argument of type 'SomeData' is not
// assignable to parameter of type
// '{ [key: string]: string; }'.
// Index signature is missing in type 'SomeData'.
getFirstLetters(someData);
The error is straightforward - I have specifically requested that the function validate obj
based on it having an index signature, NOT on the type of its values alone.
Is there any way to make my function work with all objects with a uniform value type without asking anyone who uses it to include an index signature in their interface?
typescript
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a function which can take any object literal so long as all of the values in the object are strings:
function getFirstLetters(obj: { [key: string]: string }): string {
return Object.keys(obj).map(key => obj[key][0]);
}
This works well for any indexed type, the problem arises when I try to use non-indexed objects:
interface SomeData {
user: string;
loc: string;
}
const someData: SomeData = {
user: "coolGuy42",
loc: "New York",
};
function getFirstLetters(obj: { [key: string]: string }): string {
return Object.keys(obj).map(key => obj[key][0]);
}
// Argument of type 'SomeData' is not
// assignable to parameter of type
// '{ [key: string]: string; }'.
// Index signature is missing in type 'SomeData'.
getFirstLetters(someData);
The error is straightforward - I have specifically requested that the function validate obj
based on it having an index signature, NOT on the type of its values alone.
Is there any way to make my function work with all objects with a uniform value type without asking anyone who uses it to include an index signature in their interface?
typescript
I have a function which can take any object literal so long as all of the values in the object are strings:
function getFirstLetters(obj: { [key: string]: string }): string {
return Object.keys(obj).map(key => obj[key][0]);
}
This works well for any indexed type, the problem arises when I try to use non-indexed objects:
interface SomeData {
user: string;
loc: string;
}
const someData: SomeData = {
user: "coolGuy42",
loc: "New York",
};
function getFirstLetters(obj: { [key: string]: string }): string {
return Object.keys(obj).map(key => obj[key][0]);
}
// Argument of type 'SomeData' is not
// assignable to parameter of type
// '{ [key: string]: string; }'.
// Index signature is missing in type 'SomeData'.
getFirstLetters(someData);
The error is straightforward - I have specifically requested that the function validate obj
based on it having an index signature, NOT on the type of its values alone.
Is there any way to make my function work with all objects with a uniform value type without asking anyone who uses it to include an index signature in their interface?
typescript
typescript
asked Nov 9 at 20:54
Sandy Gifford
2,4911639
2,4911639
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You can make the function generic and require the input parameter to be any type whose known properties are all strings:
function getFirstLetters<T extends Record<keyof T, string>>(obj: T): string {
return Object.keys(obj).map(key => obj[key][0]); // error
}
But the compiler (rightly) complains that it doesn't know what obj[key]
might be. After all, the known keys of T
are string
-valued, but types in TypeScript are not exact. A value of type {foo: string}
might have any number of extra properties. We know that its foo
property is a string
, but for all we know it might have a bar
property that's a number
.
If you are sure that only exact-like types will be passed to getFirstLetters
, then you can use a type assertion to convince the compiler that you are doing something safe:
function getFirstLetters<T extends Record<keyof T, string>>(obj: T): string {
// no error now
return (Object.keys(obj) as (Array<keyof T>)).map(key => obj[key][0]);
}
And it should work as you expect when you call it:
getFirstLetters(someData); // no error
And it will reject values with known properties whose values are not string
s:
getFirstLetters({a: "a", b: 23}); // error on b, not a string
But again, keep in mind that you can pass it some things with unknown non-string properties that will cause problems at runtime:
const whoopsie: SomeData = Object.assign({}, someData, { oops: null });
// whoopsie is a SomeData with an extra "oops" property that the
// compiler has explicitly forgotten about
getFirstLetters(whoopsie); // no compiler error
// but calls null[0] at runtime and explodes!! 💥
It's up to you if you care about those edge cases and how to deal with them if so. Anyway, hope that helps. Good luck!
huh... so this certainly solves the above problem, but not my actual problem (I simplified it a bunch with a function for brevity). I'll have to rephrase in a separate question.
– Sandy Gifford
Nov 12 at 15:31
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can make the function generic and require the input parameter to be any type whose known properties are all strings:
function getFirstLetters<T extends Record<keyof T, string>>(obj: T): string {
return Object.keys(obj).map(key => obj[key][0]); // error
}
But the compiler (rightly) complains that it doesn't know what obj[key]
might be. After all, the known keys of T
are string
-valued, but types in TypeScript are not exact. A value of type {foo: string}
might have any number of extra properties. We know that its foo
property is a string
, but for all we know it might have a bar
property that's a number
.
If you are sure that only exact-like types will be passed to getFirstLetters
, then you can use a type assertion to convince the compiler that you are doing something safe:
function getFirstLetters<T extends Record<keyof T, string>>(obj: T): string {
// no error now
return (Object.keys(obj) as (Array<keyof T>)).map(key => obj[key][0]);
}
And it should work as you expect when you call it:
getFirstLetters(someData); // no error
And it will reject values with known properties whose values are not string
s:
getFirstLetters({a: "a", b: 23}); // error on b, not a string
But again, keep in mind that you can pass it some things with unknown non-string properties that will cause problems at runtime:
const whoopsie: SomeData = Object.assign({}, someData, { oops: null });
// whoopsie is a SomeData with an extra "oops" property that the
// compiler has explicitly forgotten about
getFirstLetters(whoopsie); // no compiler error
// but calls null[0] at runtime and explodes!! 💥
It's up to you if you care about those edge cases and how to deal with them if so. Anyway, hope that helps. Good luck!
huh... so this certainly solves the above problem, but not my actual problem (I simplified it a bunch with a function for brevity). I'll have to rephrase in a separate question.
– Sandy Gifford
Nov 12 at 15:31
add a comment |
up vote
1
down vote
accepted
You can make the function generic and require the input parameter to be any type whose known properties are all strings:
function getFirstLetters<T extends Record<keyof T, string>>(obj: T): string {
return Object.keys(obj).map(key => obj[key][0]); // error
}
But the compiler (rightly) complains that it doesn't know what obj[key]
might be. After all, the known keys of T
are string
-valued, but types in TypeScript are not exact. A value of type {foo: string}
might have any number of extra properties. We know that its foo
property is a string
, but for all we know it might have a bar
property that's a number
.
If you are sure that only exact-like types will be passed to getFirstLetters
, then you can use a type assertion to convince the compiler that you are doing something safe:
function getFirstLetters<T extends Record<keyof T, string>>(obj: T): string {
// no error now
return (Object.keys(obj) as (Array<keyof T>)).map(key => obj[key][0]);
}
And it should work as you expect when you call it:
getFirstLetters(someData); // no error
And it will reject values with known properties whose values are not string
s:
getFirstLetters({a: "a", b: 23}); // error on b, not a string
But again, keep in mind that you can pass it some things with unknown non-string properties that will cause problems at runtime:
const whoopsie: SomeData = Object.assign({}, someData, { oops: null });
// whoopsie is a SomeData with an extra "oops" property that the
// compiler has explicitly forgotten about
getFirstLetters(whoopsie); // no compiler error
// but calls null[0] at runtime and explodes!! 💥
It's up to you if you care about those edge cases and how to deal with them if so. Anyway, hope that helps. Good luck!
huh... so this certainly solves the above problem, but not my actual problem (I simplified it a bunch with a function for brevity). I'll have to rephrase in a separate question.
– Sandy Gifford
Nov 12 at 15:31
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can make the function generic and require the input parameter to be any type whose known properties are all strings:
function getFirstLetters<T extends Record<keyof T, string>>(obj: T): string {
return Object.keys(obj).map(key => obj[key][0]); // error
}
But the compiler (rightly) complains that it doesn't know what obj[key]
might be. After all, the known keys of T
are string
-valued, but types in TypeScript are not exact. A value of type {foo: string}
might have any number of extra properties. We know that its foo
property is a string
, but for all we know it might have a bar
property that's a number
.
If you are sure that only exact-like types will be passed to getFirstLetters
, then you can use a type assertion to convince the compiler that you are doing something safe:
function getFirstLetters<T extends Record<keyof T, string>>(obj: T): string {
// no error now
return (Object.keys(obj) as (Array<keyof T>)).map(key => obj[key][0]);
}
And it should work as you expect when you call it:
getFirstLetters(someData); // no error
And it will reject values with known properties whose values are not string
s:
getFirstLetters({a: "a", b: 23}); // error on b, not a string
But again, keep in mind that you can pass it some things with unknown non-string properties that will cause problems at runtime:
const whoopsie: SomeData = Object.assign({}, someData, { oops: null });
// whoopsie is a SomeData with an extra "oops" property that the
// compiler has explicitly forgotten about
getFirstLetters(whoopsie); // no compiler error
// but calls null[0] at runtime and explodes!! 💥
It's up to you if you care about those edge cases and how to deal with them if so. Anyway, hope that helps. Good luck!
You can make the function generic and require the input parameter to be any type whose known properties are all strings:
function getFirstLetters<T extends Record<keyof T, string>>(obj: T): string {
return Object.keys(obj).map(key => obj[key][0]); // error
}
But the compiler (rightly) complains that it doesn't know what obj[key]
might be. After all, the known keys of T
are string
-valued, but types in TypeScript are not exact. A value of type {foo: string}
might have any number of extra properties. We know that its foo
property is a string
, but for all we know it might have a bar
property that's a number
.
If you are sure that only exact-like types will be passed to getFirstLetters
, then you can use a type assertion to convince the compiler that you are doing something safe:
function getFirstLetters<T extends Record<keyof T, string>>(obj: T): string {
// no error now
return (Object.keys(obj) as (Array<keyof T>)).map(key => obj[key][0]);
}
And it should work as you expect when you call it:
getFirstLetters(someData); // no error
And it will reject values with known properties whose values are not string
s:
getFirstLetters({a: "a", b: 23}); // error on b, not a string
But again, keep in mind that you can pass it some things with unknown non-string properties that will cause problems at runtime:
const whoopsie: SomeData = Object.assign({}, someData, { oops: null });
// whoopsie is a SomeData with an extra "oops" property that the
// compiler has explicitly forgotten about
getFirstLetters(whoopsie); // no compiler error
// but calls null[0] at runtime and explodes!! 💥
It's up to you if you care about those edge cases and how to deal with them if so. Anyway, hope that helps. Good luck!
answered Nov 10 at 0:58
jcalz
20.6k21637
20.6k21637
huh... so this certainly solves the above problem, but not my actual problem (I simplified it a bunch with a function for brevity). I'll have to rephrase in a separate question.
– Sandy Gifford
Nov 12 at 15:31
add a comment |
huh... so this certainly solves the above problem, but not my actual problem (I simplified it a bunch with a function for brevity). I'll have to rephrase in a separate question.
– Sandy Gifford
Nov 12 at 15:31
huh... so this certainly solves the above problem, but not my actual problem (I simplified it a bunch with a function for brevity). I'll have to rephrase in a separate question.
– Sandy Gifford
Nov 12 at 15:31
huh... so this certainly solves the above problem, but not my actual problem (I simplified it a bunch with a function for brevity). I'll have to rephrase in a separate question.
– Sandy Gifford
Nov 12 at 15:31
add a comment |
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