Mapped Type For Optional Arguments
interface Foo{
one: string;
two?: number;
someFunc: (args: string|number) => string
}
So, what i am looking for is a way to declare the type for 'args';
I was wondering if there was a way to specify the type for args.The paramater 'two' is optional. In case the user specifies the parameter 'two', i want someFunc to have args as number and string other wise, is there a way of doing this except using union type.
Thanks.
typescript typescript-typings
asked Nov 19 '18 at 4:19
Amol GuptaAmol Gupta
648
add a comment |
interface Foo{
one: string;
two?: number;
someFunc: (args: string|number) => string
}
So, what i am looking for is a way to declare the type for 'args';
I was wondering if there was a way to specify the type for args.The paramater 'two' is optional. In case the user specifies the parameter 'two', i want someFunc to have args as number and string other wise, is there a way of doing this except using union type.
Thanks.
typescript typescript-typings
asked Nov 19 '18 at 4:19
Amol GuptaAmol Gupta
648
add a comment |
interface Foo{
one: string;
two?: number;
someFunc: (args: string|number) => string
}
So, what i am looking for is a way to declare the type for 'args';
I was wondering if there was a way to specify the type for args.The paramater 'two' is optional. In case the user specifies the parameter 'two', i want someFunc to have args as number and string other wise, is there a way of doing this except using union type.
Thanks.
typescript typescript-typings
asked Nov 19 '18 at 4:19
Amol GuptaAmol Gupta
648
interface Foo{
one: string;
two?: number;
someFunc: (args: string|number) => string
}
So, what i am looking for is a way to declare the type for 'args';
I was wondering if there was a way to specify the type for args.The paramater 'two' is optional. In case the user specifies the parameter 'two', i want someFunc to have args as number and string other wise, is there a way of doing this except using union type.
Thanks.
typescript typescript-typings
typescript typescript-typings
asked Nov 19 '18 at 4:19
Amol GuptaAmol Gupta
648
asked Nov 19 '18 at 4:19
Amol GuptaAmol Gupta
648
asked Nov 19 '18 at 4:19
Amol GuptaAmol Gupta
648
asked Nov 19 '18 at 4:19
Amol GuptaAmol Gupta
648
asked Nov 19 '18 at 4:19
Amol GuptaAmol Gupta
648
648
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You can use a union instead of the interface. One union member will have two required an if type number, and have the appropriate type for args, the other member of the union will have twooptional and of type never to ensure its not compatible with the other union member. You will loose inference for the argument type, but the compiler will validate that the function does have the appropriate parameter type.
type Foo = {
one: string;
two?: never;
someFunc: (args: number) => string
} | {
one: string;
two: number;
someFunc: (args: string) => string
}
let foo: Foo = {
one: '',
someFunc : (a: string) => a
} // err
let foo2: Foo = {
one: '',
someFunc : (a: number) => a.toString()
} // ok
let foo3: Foo = {
one: '',
two:2,
someFunc : (a: string) => a
} // ok
let foo4: Foo = {
one: '',
two:3
someFunc : (a: number) => a.toString()
} // err
answered Nov 19 '18 at 5:28
Titian Cernicova-DragomirTitian Cernicova-Dragomir
61.1k33755
Great Thanks. Works Great.
– Amol Gupta
Nov 19 '18 at 6:08
add a comment |
You want:
type Foo =
| {one: string, two: number, someFunc(args: string): string}
| {one: string, someFunc(args: number): string}
answered Nov 19 '18 at 5:21
bchernybcherny
2,0491524
Thanks For all the help bcherny appreciate it.
– Amol Gupta
Nov 19 '18 at 6:09
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53368260%2fmapped-type-for-optional-arguments%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use a union instead of the interface. One union member will have two required an if type number, and have the appropriate type for args, the other member of the union will have twooptional and of type never to ensure its not compatible with the other union member. You will loose inference for the argument type, but the compiler will validate that the function does have the appropriate parameter type.
type Foo = {
one: string;
two?: never;
someFunc: (args: number) => string
} | {
one: string;
two: number;
someFunc: (args: string) => string
}
let foo: Foo = {
one: '',
someFunc : (a: string) => a
} // err
let foo2: Foo = {
one: '',
someFunc : (a: number) => a.toString()
} // ok
let foo3: Foo = {
one: '',
two:2,
someFunc : (a: string) => a
} // ok
let foo4: Foo = {
one: '',
two:3
someFunc : (a: number) => a.toString()
} // err
answered Nov 19 '18 at 5:28
Titian Cernicova-DragomirTitian Cernicova-Dragomir
61.1k33755
Great Thanks. Works Great.
– Amol Gupta
Nov 19 '18 at 6:08
add a comment |
You can use a union instead of the interface. One union member will have two required an if type number, and have the appropriate type for args, the other member of the union will have twooptional and of type never to ensure its not compatible with the other union member. You will loose inference for the argument type, but the compiler will validate that the function does have the appropriate parameter type.
type Foo = {
one: string;
two?: never;
someFunc: (args: number) => string
} | {
one: string;
two: number;
someFunc: (args: string) => string
}
let foo: Foo = {
one: '',
someFunc : (a: string) => a
} // err
let foo2: Foo = {
one: '',
someFunc : (a: number) => a.toString()
} // ok
let foo3: Foo = {
one: '',
two:2,
someFunc : (a: string) => a
} // ok
let foo4: Foo = {
one: '',
two:3
someFunc : (a: number) => a.toString()
} // err
answered Nov 19 '18 at 5:28
Titian Cernicova-DragomirTitian Cernicova-Dragomir
61.1k33755
Great Thanks. Works Great.
– Amol Gupta
Nov 19 '18 at 6:08
add a comment |
You can use a union instead of the interface. One union member will have two required an if type number, and have the appropriate type for args, the other member of the union will have twooptional and of type never to ensure its not compatible with the other union member. You will loose inference for the argument type, but the compiler will validate that the function does have the appropriate parameter type.
type Foo = {
one: string;
two?: never;
someFunc: (args: number) => string
} | {
one: string;
two: number;
someFunc: (args: string) => string
}
let foo: Foo = {
one: '',
someFunc : (a: string) => a
} // err
let foo2: Foo = {
one: '',
someFunc : (a: number) => a.toString()
} // ok
let foo3: Foo = {
one: '',
two:2,
someFunc : (a: string) => a
} // ok
let foo4: Foo = {
one: '',
two:3
someFunc : (a: number) => a.toString()
} // err
answered Nov 19 '18 at 5:28
Titian Cernicova-DragomirTitian Cernicova-Dragomir
61.1k33755
You can use a union instead of the interface. One union member will have two required an if type number, and have the appropriate type for args, the other member of the union will have twooptional and of type never to ensure its not compatible with the other union member. You will loose inference for the argument type, but the compiler will validate that the function does have the appropriate parameter type.
type Foo = {
one: string;
two?: never;
someFunc: (args: number) => string
} | {
one: string;
two: number;
someFunc: (args: string) => string
}
let foo: Foo = {
one: '',
someFunc : (a: string) => a
} // err
let foo2: Foo = {
one: '',
someFunc : (a: number) => a.toString()
} // ok
let foo3: Foo = {
one: '',
two:2,
someFunc : (a: string) => a
} // ok
let foo4: Foo = {
one: '',
two:3
someFunc : (a: number) => a.toString()
} // err
answered Nov 19 '18 at 5:28
Titian Cernicova-DragomirTitian Cernicova-Dragomir
61.1k33755
answered Nov 19 '18 at 5:28
Titian Cernicova-DragomirTitian Cernicova-Dragomir
61.1k33755
answered Nov 19 '18 at 5:28
Titian Cernicova-DragomirTitian Cernicova-Dragomir
61.1k33755
answered Nov 19 '18 at 5:28
Titian Cernicova-DragomirTitian Cernicova-Dragomir
61.1k33755
61.1k33755
Great Thanks. Works Great.
– Amol Gupta
Nov 19 '18 at 6:08
add a comment |
Great Thanks. Works Great.
– Amol Gupta
Nov 19 '18 at 6:08
Great Thanks. Works Great.
– Amol Gupta
Nov 19 '18 at 6:08
Great Thanks. Works Great.
– Amol Gupta
Nov 19 '18 at 6:08
add a comment |
You want:
type Foo =
| {one: string, two: number, someFunc(args: string): string}
| {one: string, someFunc(args: number): string}
answered Nov 19 '18 at 5:21
bchernybcherny
2,0491524
Thanks For all the help bcherny appreciate it.
– Amol Gupta
Nov 19 '18 at 6:09
add a comment |
You want:
type Foo =
| {one: string, two: number, someFunc(args: string): string}
| {one: string, someFunc(args: number): string}
answered Nov 19 '18 at 5:21
bchernybcherny
2,0491524
Thanks For all the help bcherny appreciate it.
– Amol Gupta
Nov 19 '18 at 6:09
add a comment |
You want:
type Foo =
| {one: string, two: number, someFunc(args: string): string}
| {one: string, someFunc(args: number): string}
answered Nov 19 '18 at 5:21
bchernybcherny
2,0491524
You want:
type Foo =
| {one: string, two: number, someFunc(args: string): string}
| {one: string, someFunc(args: number): string}
answered Nov 19 '18 at 5:21
bchernybcherny
2,0491524
answered Nov 19 '18 at 5:21
bchernybcherny
2,0491524
answered Nov 19 '18 at 5:21
bchernybcherny
2,0491524
answered Nov 19 '18 at 5:21
bchernybcherny
2,0491524
2,0491524
Thanks For all the help bcherny appreciate it.
– Amol Gupta
Nov 19 '18 at 6:09
add a comment |
Thanks For all the help bcherny appreciate it.
– Amol Gupta
Nov 19 '18 at 6:09
Thanks For all the help bcherny appreciate it.
– Amol Gupta
Nov 19 '18 at 6:09
Thanks For all the help bcherny appreciate it.
– Amol Gupta
Nov 19 '18 at 6:09
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53368260%2fmapped-type-for-optional-arguments%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
