Can a Descriptor change type in Typescript?
1
I'm trying to make descriptor which can change return type in typescript but I don't know how to do this.
Here is the code and what I tried:
function changeReturnType()
symbol,
descriptor: TypedPropertyDescriptor<T>
) =>
const oldValue = descriptor.value;
return
...descriptor,
value(...argv)
// @ts-ignore
return String(oldValue.call(target, ...argv))
;
class Foo
@changeReturnType()
square ()
return 1;
let val = new Foo().square(); // I hope ts know here is string
console.dir(
val,
type: typeof val,
);
javascript node.js typescript types descriptor
edited Nov 11 '18 at 7:15
Roy Scheffers
2,189101826
asked Nov 11 '18 at 6:06
blueloversbluelovers
1608
add a comment |
1
I'm trying to make descriptor which can change return type in typescript but I don't know how to do this.
Here is the code and what I tried:
function changeReturnType()
symbol,
descriptor: TypedPropertyDescriptor<T>
) =>
const oldValue = descriptor.value;
return
...descriptor,
value(...argv)
// @ts-ignore
return String(oldValue.call(target, ...argv))
;
class Foo
@changeReturnType()
square ()
return 1;
let val = new Foo().square(); // I hope ts know here is string
console.dir(
val,
type: typeof val,
);
javascript node.js typescript types descriptor
edited Nov 11 '18 at 7:15
Roy Scheffers
2,189101826
asked Nov 11 '18 at 6:06
blueloversbluelovers
1608
add a comment |
1
1
1
1
I'm trying to make descriptor which can change return type in typescript but I don't know how to do this.
Here is the code and what I tried:
function changeReturnType()
symbol,
descriptor: TypedPropertyDescriptor<T>
) =>
const oldValue = descriptor.value;
return
...descriptor,
value(...argv)
// @ts-ignore
return String(oldValue.call(target, ...argv))
;
class Foo
@changeReturnType()
square ()
return 1;
let val = new Foo().square(); // I hope ts know here is string
console.dir(
val,
type: typeof val,
);
javascript node.js typescript types descriptor
edited Nov 11 '18 at 7:15
Roy Scheffers
2,189101826
asked Nov 11 '18 at 6:06
blueloversbluelovers
1608
I'm trying to make descriptor which can change return type in typescript but I don't know how to do this.
Here is the code and what I tried:
function changeReturnType()
symbol,
descriptor: TypedPropertyDescriptor<T>
) =>
const oldValue = descriptor.value;
return
...descriptor,
value(...argv)
// @ts-ignore
return String(oldValue.call(target, ...argv))
;
class Foo
@changeReturnType()
square ()
return 1;
let val = new Foo().square(); // I hope ts know here is string
console.dir(
val,
type: typeof val,
);
javascript node.js typescript types descriptor
javascript node.js typescript types descriptor
edited Nov 11 '18 at 7:15
Roy Scheffers
2,189101826
asked Nov 11 '18 at 6:06
blueloversbluelovers
1608
edited Nov 11 '18 at 7:15
Roy Scheffers
2,189101826
asked Nov 11 '18 at 6:06
blueloversbluelovers
1608
edited Nov 11 '18 at 7:15
Roy Scheffers
2,189101826
edited Nov 11 '18 at 7:15
Roy Scheffers
2,189101826
edited Nov 11 '18 at 7:15
Roy Scheffers
2,189101826
2,189101826
asked Nov 11 '18 at 6:06
blueloversbluelovers
1608
asked Nov 11 '18 at 6:06
blueloversbluelovers
1608
asked Nov 11 '18 at 6:06
blueloversbluelovers
1608
1608
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
0
This is currently not possible in TypeScript but you may look at this comment and modify hack described there to your needs (i.e. for method decorator usage).
answered Nov 11 '18 at 10:10
syntagmasyntagma
12.6k1249105
The situation is different in linked comment. Returned type is ignored in class decorator, while in method decorator it's taken into account, so decorated method type is incompatible with original method.
– estus
Nov 11 '18 at 15:32
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
);
);
draft saved
draft discarded
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%2f53246289%2fcan-a-descriptor-change-type-in-typescript%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
This is currently not possible in TypeScript but you may look at this comment and modify hack described there to your needs (i.e. for method decorator usage).
answered Nov 11 '18 at 10:10
syntagmasyntagma
12.6k1249105
The situation is different in linked comment. Returned type is ignored in class decorator, while in method decorator it's taken into account, so decorated method type is incompatible with original method.
– estus
Nov 11 '18 at 15:32
add a comment |
0
This is currently not possible in TypeScript but you may look at this comment and modify hack described there to your needs (i.e. for method decorator usage).
answered Nov 11 '18 at 10:10
syntagmasyntagma
12.6k1249105
The situation is different in linked comment. Returned type is ignored in class decorator, while in method decorator it's taken into account, so decorated method type is incompatible with original method.
– estus
Nov 11 '18 at 15:32
add a comment |
0
0
0
This is currently not possible in TypeScript but you may look at this comment and modify hack described there to your needs (i.e. for method decorator usage).
answered Nov 11 '18 at 10:10
syntagmasyntagma
12.6k1249105
This is currently not possible in TypeScript but you may look at this comment and modify hack described there to your needs (i.e. for method decorator usage).
answered Nov 11 '18 at 10:10
syntagmasyntagma
12.6k1249105
answered Nov 11 '18 at 10:10
syntagmasyntagma
12.6k1249105
answered Nov 11 '18 at 10:10
syntagmasyntagma
12.6k1249105
answered Nov 11 '18 at 10:10
syntagmasyntagma
12.6k1249105
12.6k1249105
The situation is different in linked comment. Returned type is ignored in class decorator, while in method decorator it's taken into account, so decorated method type is incompatible with original method.
– estus
Nov 11 '18 at 15:32
add a comment |
The situation is different in linked comment. Returned type is ignored in class decorator, while in method decorator it's taken into account, so decorated method type is incompatible with original method.
– estus
Nov 11 '18 at 15:32
The situation is different in linked comment. Returned type is ignored in class decorator, while in method decorator it's taken into account, so decorated method type is incompatible with original method.
– estus
Nov 11 '18 at 15:32
The situation is different in linked comment. Returned type is ignored in class decorator, while in method decorator it's taken into account, so decorated method type is incompatible with original method.
– estus
Nov 11 '18 at 15:32
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f53246289%2fcan-a-descriptor-change-type-in-typescript%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
