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Breaking change: Use of 'void' in control flow constructs is now disallowed #26262
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- addedBreaking ChangeWould introduce errors in existing codeWould introduce errors in existing code
on Aug 7, 2018 RyanCavanaugh commented
on Aug 7, 2018 MemberAuthorMore actionsThis is now an error.
Reacted by Geoffrey TestelinHi, I just updated my CI/CD pipeline env to 3.1.1 and I encounter this error while building docker image.
An expression of type 'void' cannot be tested for truthiness
My code is:(err: Error) => (err ? console.log('Upload failed.') || rej(err) : console.log('Upload completed.'))
I think that using the expression, that evaluates to void, like this is perfectly legit, because as typescript docs says "Declaring variables of type void is not useful because you can only assign undefined or null to them". So if it is allowed to check both, null and undefined, in this kind of expression, also void must be permitted.
In addition to that, JS's undefined value is same as void, since JS adds anyway a "return undefined" if a functions hasn't any return value.I think that TS here forgets that it is a JS superset and this makes me sad.
Reacted by Juan Pablo de la Torre, Gilbert, Ben Steward, Qwerty (Vítězslav Ackermann Ferko), Andrej Bartko, alex a, Dovid Gefen, Valen Cheung, Stav Alfi, Zuo Zongyuan and 5 moreRyanCavanaugh commented
on Oct 1, 2018 MemberAuthorMore actionsMaksim Sinik (@fox1t) the problem with assuming a void-returning function is going to return a falsy value is that this code is legal:
var x: () => void = () => true; // somethingImportant does not run if (x() || somethingImportant()) { }
You should use the comma operator if you intend for two sequenced operations to always occur.
I think that TS here forgets that it is a JS superset and this makes me sad.
https://stackoverflow.com/questions/41750390/what-does-all-legal-javascript-is-legal-typescript-mean
Reacted by Juan Pablo de la Torre, Robert Starbuck, Trotyl Yu, BlackGlory, Dan Mercer and snarbles2Reacted by Qwerty (Vítězslav Ackermann Ferko), Andrej Bartko, Kaat, Stav Alfi, Scott Burton, Adam Matheson, Yury Goltsman, Fidanze and jwalkerinterpresReacted by Stav AlfiRyan Cavanaugh (@RyanCavanaugh) I think that there is no need to assuming anything because in JS we have truthy and falsy values and all operators check for that. In addition to that conditional operators are short-circuited and return the value and don't operate any conversion type to boolean.
I assume, because of docs I linked, that also TS would work the same. If we "introduce real" true and false checks (I mean the boolean one and not truthy and falsy), also
const foo = (bar) => bar && bar.baz // or const baz = (name) => name || 'defaultName'
can be considered "illegal" because neither
barornameare not boolean values.The following isn't legal code:
(err: Error) => err ? console.log('Upload failed.'), rej(err) : console.log('Upload completed.')
Can't use comma operator here, but || is legit. I use TS to have a better FP experience in JS, but this type of changes, moreover in minor version, are really annoying.
Anyway, just to point it out, I am here to understand and I am not blaming anyone.
Good point about me being sad about wrongly understand ts is js's "superset". :)
Reacted by Juan Pablo de la Torre, Gilbert, Jan Starzak, Qwerty (Vítězslav Ackermann Ferko), Andrej Bartko, Romario Raffington, Kaat, Valen Cheung, Stav Alfi, Richard Richardson and 1 moreDoes this change mean that Typescript insteads to remove boolean coercion all together? If not, what makes this type different from all other types?
RyanCavanaugh commented
on Oct 4, 2018 MemberAuthorMore actionsIf not, what makes this type different from all other types?
Giving different behavior to different types is why we have types in the first place.
The odds that you meant to write code like the sample in the OP are approximately 0%. Even if you intended to write code like
voidReturningFunc() || someImportantOp(), it's unlikely you fully understood the implication ofvoid- many people believe that avoid-returning function will always return a falsy value, but this is not the case.Moreover, code like
someCall() || someImportantOp()is an extreme code smell because it is not obvious to a reader of the code that the evaluation ofsomeImportantOpis intended to actually be unconditional -- the comma operator should have been used here for clarity and certainty.Reacted by AnyhowStep, S. B. Tam, Trotyl Yu, Dan Mercer and snarbles2Reacted by Andrej Bartko, Stav Alfi, Yang and Adam MathesonGiving different behavior to different types is why we have types in the first place.
Ah sorry, I thought it was obvious from context that I meant what makes void being coerced to a boolean value invalid typescript, as opposed to a string being coerced to boolean, or float or any other type.
From a purity standpoint, it doesn't make sense to coerce a string to boolean either, right? The only false string is '' and it seems more correct to use === to test that. So if you all are ratcheting up the purity of coercion in TypeScript, I don't think it's unreasonable to ask if you all are planning on making boolean coercion illegal for other types.
Reacted by Qwerty (Vítězslav Ackermann Ferko)Reacted by Juan Pablo de la TorreKit Westneat (@kitwestneat) just get the point I was trying to understand here. In this very moment there are 2 different behaviour of the language for the same operation, and it is wrong from every point of view. In addition can you Ryan Cavanaugh (@RyanCavanaugh) pleas elaborate "many people believe that a void-returning function will always return a falsy value, but this is not the case"?
Also the TS docs say: "Declaring variables of type void is not useful because you can only assign undefined or null to them". So it is legit to expect to use void here, since it is the "same" as undefined and null...Reacted by Qwerty (Vítězslav Ackermann Ferko)RyanCavanaugh commented
on Oct 4, 2018 MemberAuthorMore actionsIt's legal to coerce a string to a boolean because a) this is idiomatic in JS, b) some strings are falsy and some strings are truthy according to rules which are more or less predictable, and c) it's easily conceivable why you might want to do some things with truthy strings and other things with falsy strings
How do you get an expression of type
void? Effectively there are two ways:- Evaluating a non-aliased function call, which means it always returns
undefined- In which case you're writing code that appears to have control flow but doesn't, which is extremely suspect
- Evaluating an aliased function call, which means its return value is truly unknowable
- In which case you have no idea what's going to happen, nor does anyone else reading the code
So it is legit to expect to use void here, since it is the "same" as undefined and null...
I think this has been clearly explained already: #26262 (comment)
Reacted by Kit Westneat, Trotyl Yu and alex a- Evaluating a non-aliased function call, which means it always returns
Ok, thanks for the explanation.
RyanCavanaugh commented
on Oct 4, 2018 MemberAuthorMore actionsLet's look at a longer example:
function myForEach<T>(arr: T[], callback: (arg: T) => void) { let i = 0; // love too sequence with || while (i < arr.length) callback(arr[i]) || i++; } // Yep myForEach([1, 2, 3], n => console.log(n)); // Let's copy this array into another array let dst: number[] = []; // Out of memory error, wtf? myForEach([1, 2, 3], n => dst.push(n));
Reacted by Qwerty (Vítězslav Ackermann Ferko) and mrs-umgReacted by YangOk, this is really a nice example! Love it.
Don't you think that the real problem is before||operation. What is now bothering me is:
n => dst.push(n)has(n: any) => numbertype and the callback parameter is declared ascallback: (arg: T) => void). Isn't this already a problem on its own that I am allowed to pass a callback with wrong signature?Reacted by Qwerty (Vítězslav Ackermann Ferko)6 remaining items
So, wrapping up all of the great comments here, I think we can say that
voidis meant to be used for functions that only have side-effects and that is different both fromanyandundefined, even if
const bar: void = undefinedis valid syntax.
There is still one question left though. As i pointed out before, official TS docs say "Declaring variables of type void is not useful because you can only assign undefined or null to them", but trying to writeconst foo: void = nullgivesType 'null' is not assignable to type 'void'.error. Docs update needed?Reacted by Chris Harvey, Juan Pablo de la Torre and Qwerty (Vítězslav Ackermann Ferko)Maksim Sinik (@fox1t) , I believe that docs are correct.
You have this error because you have
strictNullChecks: true(orstrict: truefor that matter) in yourtsconfig.json. By defaultstrictNullChecksisfalse, andvoidis assignable fromnull.Btw, I've got a nice type compatibility table. Paste it into any editor capable of highlighting TypeScript errors and you will see which types are compatible and which are not (depending on the active
tsconfig.json).// Type compatibility table function fx(): never { throw Error("Never returns"); } function fv(): void { } var x: never = fx(); var u: undefined = undefined; var v: void = fv(); var n: null = null; var t: {} = {}; // random type var a: any = {}; /* |x | --+-----+ y |x = y| (typeof X "assignable from" typeof Y) --+-----+ */ /* | never | undefined | void | null | T | any | -----------+--------------+--------------+--------------+--------------+--------------+--------------+ never |*/ x = x; /*|*/ u = x; /*|*/ v = x; /*|*/ n = x; /*|*/ t = x; /*|*/ a = x; /*| undefined |*/ x = u; /*|*/ u = u; /*|*/ v = u; /*|*/ n = u; /*|*/ t = u; /*|*/ a = u; /*| void |*/ x = v; /*|*/ u = v; /*|*/ v = v; /*|*/ n = v; /*|*/ t = v; /*|*/ a = v; /*| null |*/ x = n; /*|*/ u = n; /*|*/ v = n; /*|*/ n = n; /*|*/ t = n; /*|*/ a = n; /*| T |*/ x = t; /*|*/ u = t; /*|*/ v = t; /*|*/ n = t; /*|*/ t = t; /*|*/ a = t; /*| any |*/ x = a; /*|*/ u = a; /*|*/ v = a; /*|*/ n = a; /*|*/ t = a; /*|*/ a = a; /*| -----------+--------------+--------------+--------------+--------------+--------------+--------------+ */ // Observations (under `strict: true`): // // 1. `never` assignable-from only `never` // 2. `never` assignable-to any type // 3. `any` assignable-from any type // 4. `any` assignable-to any type (except for `never`, see p.1) // 5. `undefined` assignable-to `void` // 6. `undefined`, `null`, `T` are invariant //
Demo screenshot:
Update: Published this table in a dedicated repo:
https://github.com/earshinov/typescript-type-compatibility-tableReacted by Ryan Cavanaugh, Paul Irish, Brendan Kenny, Connor Clark, Ilie Ciotir, Donald Pipowitch, Steven, Jordi Oliveras Rovira, Martin Hochel, OliverGrack and 13 moreReacted by Ryan Cavanaugh, Mohsen Azimi, Giovanni Gonzaga, Jordi Oliveras Rovira, Kitson Kelly, Rudi Kovač, Harryson Palheta, Qwerty (Vítězslav Ackermann Ferko), Joey Wunderlich, Conor Dockry and 2 moreRyanCavanaugh commented
on Oct 19, 2018 MemberAuthorMore actionsearshinov that is gorgeous 😍
connorjclark commented
on Oct 19, 2018 ContributorMore actionsAdding unknown to above :)
// Type compatibility table function fx(): never { throw Error("Never returns"); } function fv(): void { } var x: never = fx(); var u: undefined = undefined; var v: void = fv(); var n: null = null; var t: {} = {}; // random type var a: any = {}; var k: unknown = {}; /* |x | --+-----+ y |x = y| (typeof X "assignable from" typeof Y) --+-----+ */ /* | never | undefined | void | null | T | any | unknown | -----------+--------------+--------------+--------------+--------------+--------------+--------------+--------------+ never |*/ x = x; /*|*/ u = x; /*|*/ v = x; /*|*/ n = x; /*|*/ t = x; /*|*/ a = x; /*|*/ k = x; /*| undefined |*/ x = u; /*|*/ u = u; /*|*/ v = u; /*|*/ n = u; /*|*/ t = u; /*|*/ a = u; /*|*/ k = u; /*| void |*/ x = v; /*|*/ u = v; /*|*/ v = v; /*|*/ n = v; /*|*/ t = v; /*|*/ a = v; /*|*/ k = v; /*| null |*/ x = n; /*|*/ u = n; /*|*/ v = n; /*|*/ n = n; /*|*/ t = n; /*|*/ a = n; /*|*/ k = n; /*| T |*/ x = t; /*|*/ u = t; /*|*/ v = t; /*|*/ n = t; /*|*/ t = t; /*|*/ a = t; /*|*/ k = t; /*| any |*/ x = a; /*|*/ u = a; /*|*/ v = a; /*|*/ n = a; /*|*/ t = a; /*|*/ a = a; /*|*/ k = a; /*| k |*/ x = k; /*|*/ u = k; /*|*/ v = k; /*|*/ n = k; /*|*/ t = k; /*|*/ a = k; /*|*/ k = k; /*| -----------+--------------+--------------+--------------+--------------+--------------+--------------+--------------+ */ // Observations (under `strict: true`): // // 1. `never` assignable-from only `never` // 2. `never` assignable-to any type // 3. `any` assignable-from any type // 4. `any` assignable-to any type (except for `never`, see p.1) // 5. `undefined` assignable-to `void` // 6. `undefined`, `null`, `T` are invariant //Reacted by earshinov, Martynas Žilinskas, Terry, Jordi Oliveras Rovira, Tiago Moraes, Qwerty (Vítězslav Ackermann Ferko) and Drew ShowalterGuys, is there a solution to this issue ?
Here's how it affects my project. I use swagger plugin (swagger-codegen-maven-plugin) in JAVA Spring project to generate the Angular/TS API (i.e. front-end client services and interfaces/modules). Now there's a manual step of fixing the autogenerated code, as the Angular project doesn't build due to this breaking change in TS (Use of 'void' in control flow constructs is now disallowed Error).
So, is TS going to fix this (I tested in Chrome Console JS works just fine with void evaluating to falsy) OR it is going to be a deviation from JS from now on in respect to this particular functionality ?P.S. I also tested manual generators available, but all work probably with the same template and cannot escape the error in the generated code unless manual fixing it ... Thanks if s.o. pays attn to this
Reacted by Jeremy MoritzReacted by Jeremy Moritz and Qwerty (Vítězslav Ackermann Ferko)ackvf commented
on Feb 20, 2019 More actionsI was using the
console.debug(props) ||for quickly debugging just about everything, I even have a snippet for that. With the comma separator, it is necessary to also enclose the code within brackets.const setFilter = (filter: FilterType) => setTheFilter(filter) // code to debug const setFilter = (filter: FilterType) => console.log(filger) || setTheFilter(filter) // before const setFilter = (filter: FilterType) => ( console.log(filger), setTheFilter(filter) ) // nowCan I suppress the error or can I augment the console declaration to look like it's returning a falsy value?
//EDIT
Okay, this makes the job done.
declare global { interface Console { debug(message?: any, ...optionalParams: any[]): undefined; } }Looks like you can still coerce a void function to a number, and then do a truthiness check, so
+console.log("log") || aFn()works to avoid the outer brackets needed with the comma operator.Reacted by Stefan Mirea and Conor DockryReacted by Qwerty (Vítězslav Ackermann Ferko)Why is
voidallowed to be checked for truthiness when it's a discriminated union with another type?(a: A | void) => { a ? a.b : undefined // works };
Do you think I should file a bug for this?
Reacted by Vladimir Lyzovoidis a bad type that predatesunknownandundefinedandnulland how it works is by and large an artifact of how it used to work (ie, we generally don't try to change howvoidbehaves - we change how other things behave in the presence ofvoidinstead). If you're writing new type definitions today, it's almost always better to declare things of typeunknownorundefined, depending on your intent, instead ofvoid(which sort of acts like either, depending on context, which is why it's so inconsistent).ackvf commented
on Nov 26, 2019 More actionsThis still bothers me a bit though
const f: () => void = () => true; // no errors function f(): void { return true; // Error: 'true' is not assignable to type 'void' }Reacted by Vladimir Lyzo, jwalkerinterpres and Qwerty (Vítězslav Ackermann Ferko)Ryan Cavanaugh (@RyanCavanaugh)
.then(x=> console.log(x) || x)is a type error.How can we turn off this rule?
Reacted by jwalkerinterpresackvf commented
on Dec 30, 2019 More actionsjust coerce the console.log to a not a number
.then(x=> +console.log(x) || x)OR augment the type
declare global { interface Console { log(message?: any, ...optionalParams: any[]): undefined; } }- added a commit that references this issue
on Jul 16, 2021


TypeScript Version: 3.1
Search Terms: boolean void truthy falsy if while do for not
An expression of type 'void' cannot be tested for truthiness
Code
Consider this code:
Expected behavior: The "Cool!" branch is actually unreachable because
existsAndIsCoolreturnsvoid, notboolean. I should have been warned about this at some point.Actual behavior: No error, never cool 😢
Playground Link: Link
Related Issues:
PR #26234
#7256
#10942
#7306