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Breaking change: Use of 'void' in control flow constructs is now disallowed #26262

Description

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:

function existsAndIsCool(fileName: string) {
  fs.exists(fileName, exists => {
    return exists && isCool(fileName);
  });
}

if (existsAndIsCool("myFile")) {
  console.log("Cool!");
} else {
  console.log("Not cool :(");
}

Expected behavior: The "Cool!" branch is actually unreachable because existsAndIsCool returns void, not boolean. 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

Activity

  1. RyanCavanaugh commented on Aug 7, 2018

    @RyanCavanaugh
    MemberAuthor

    This is now an error.

  2. fox1t commented on Oct 1, 2018

    @fox1t

    Hi, 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.

  3. RyanCavanaugh commented on Oct 1, 2018

    @RyanCavanaugh
    MemberAuthor

    Maksim 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

  4. fox1t commented on Oct 1, 2018

    @fox1t

    Ryan 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 bar or name are 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". :)

  5. kitwestneat commented on Oct 4, 2018

    @kitwestneat

    Does this change mean that Typescript insteads to remove boolean coercion all together? If not, what makes this type different from all other types?

  6. RyanCavanaugh commented on Oct 4, 2018

    @RyanCavanaugh
    MemberAuthor

    If 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 of void - many people believe that a void-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 of someImportantOp is intended to actually be unconditional -- the comma operator should have been used here for clarity and certainty.

  7. kitwestneat commented on Oct 4, 2018

    @kitwestneat

    Giving 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.

  8. fox1t commented on Oct 4, 2018

    @fox1t

    Kit 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...

  9. RyanCavanaugh commented on Oct 4, 2018

    @RyanCavanaugh
    MemberAuthor

    It'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)

  10. kitwestneat commented on Oct 4, 2018

    @kitwestneat

    Ok, thanks for the explanation.

  11. RyanCavanaugh commented on Oct 4, 2018

    @RyanCavanaugh
    MemberAuthor

    Let'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));
  12. fox1t commented on Oct 4, 2018

    @fox1t

    Ok, 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) => number type and the callback parameter is declared as callback: (arg: T) => void). Isn't this already a problem on its own that I am allowed to pass a callback with wrong signature?

  13. 6 remaining items

  14. fox1t commented on Oct 5, 2018

    @fox1t

    So, wrapping up all of the great comments here, I think we can say that void is meant to be used for functions that only have side-effects and that is different both from any and undefined, even if
    const bar: void = undefined is 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 write const foo: void = null gives Type 'null' is not assignable to type 'void'. error. Docs update needed?

  15. earshinov commented on Oct 19, 2018

    @earshinov

    Maksim Sinik (@fox1t) , I believe that docs are correct.

    You have this error because you have strictNullChecks: true (or strict: true for that matter) in your tsconfig.json. By default strictNullChecks is false, and void is assignable from null.

    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:

    type compat table demo

    Update: Published this table in a dedicated repo:
    https://github.com/earshinov/typescript-type-compatibility-table

  16. RyanCavanaugh commented on Oct 19, 2018

    @RyanCavanaugh
    MemberAuthor

    earshinov that is gorgeous 😍

  17. connorjclark commented on Oct 19, 2018

    @connorjclark
    Contributor

    Adding 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
    //
    

    image

  18. igpeev commented on Dec 9, 2018

    @igpeev

    Guys, 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

  19. ackvf commented on Feb 20, 2019

    @ackvf

    I 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) ) // now
    

    Can 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;
      }
    }
    
  20. thscott commented on Feb 24, 2019

    @thscott

    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.

  21. sharno commented on Nov 21, 2019

    @sharno

    Why is void allowed 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?

  22. weswigham commented on Nov 21, 2019

    @weswigham
    Member

    void is a bad type that predates unknown and undefined and null and how it works is by and large an artifact of how it used to work (ie, we generally don't try to change how void behaves - we change how other things behave in the presence of void instead). If you're writing new type definitions today, it's almost always better to declare things of type unknown or undefined, depending on your intent, instead of void (which sort of acts like either, depending on context, which is why it's so inconsistent).

  23. ackvf commented on Nov 26, 2019

    @ackvf

    This 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'
    }
    
  24. stavalfi commented on Dec 2, 2019

    @stavalfi

    Ryan Cavanaugh (@RyanCavanaugh)

    .then(x=> console.log(x) || x) is a type error.

    How can we turn off this rule?

  25. ackvf commented on Dec 30, 2019

    @ackvf

    Stav Alfi (@stavalfi)

    just 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;
      }
    }
    
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