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Typescript is a typed superset of Javascript. For this challenge, we'll consider the following classic basic types:

  • string
  • number
  • boolean
  • undefined
  • null

And two "meta" types:

  • any
  • never

The type any annotates that any type is valid:

const a: any = "foo";
const b: any = 42;
const c: any = true;
const d: any = undefined;

Whereas never annotates that no type is valid:

const e: never = <no valid literal assignment possible>;

We'll also consider two compositions of types: unions and intersections, annotated as | and & respectively.

A union type expects either one or the other type:

const f: string | number = "foo";
const g: string | number = 42;

While an intersection creates a combination of both types:

const h: string & any = "foo";
const i: string & any = 42;
const j: string & number = <no valid literal assignment possible>;

The order of operands within unions and intersections doesn't matter, string | number is equal to number | string.

The challenge

Given one of the above type declarations, return the resulting resolved type. The input will be one of the following:

  • a basic type as listed above or;
  • a meta type as listed above or;
  • a union of two basic and/or meta types or;
  • an intersection of two basic and/or meta types.

Examples

// All basic and meta types simply map to themselves
string → string
number → number
any → any
never → never

// Unions between basic types create unions except with themselves
string | number → string | number
string | undefined → string | undefined
string | string → string

// Intersections between basic types create never except with themselves
string & number → never
string & undefined → never
string & string → string

// Unions with any create any
any | string → any
any | never → any

// Intersections with any result in any except with never
any & string → any
any & number → any
any & never → never

// Unions with never yield the other half
never | string → string
never | any → any
never | never → never

// Intersections with never return never
never & string → never
never & any → never
never & never → never

A good way to get a feel for the types is to try out the Typescript playground.

Any reasonable and consistent input-output format, standard loop holes, .

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  • 2
    \$\begingroup\$ Why does any & something not give something? I would imagine a number is not both any and string \$\endgroup\$
    – hyper-neutrino
    Commented Apr 14, 2021 at 10:38
  • 1
    \$\begingroup\$ @hyper-neutrino It's easier explained with object types, but I omitted them to avoid making the challenge needlessly complex. A simple example is const x: { foo: string } & { bar: number } = { foo: "foo", bar: 42 };. \$\endgroup\$
    – Etheryte
    Commented Apr 14, 2021 at 10:44
  • 2
    \$\begingroup\$ oh, okay. that makes a bit more sense but still quite strange and definitely inconvenient for golfing \$\endgroup\$
    – hyper-neutrino
    Commented Apr 14, 2021 at 10:45
  • 1
    \$\begingroup\$ @user No, all the types you have to deal with are outlined in the challenge section. \$\endgroup\$
    – Etheryte
    Commented Apr 14, 2021 at 13:53
  • 3
    \$\begingroup\$ FYI, this challenge encodes a fundamental misunderstanding of TypeScript. any is not the type that annotates that any type is valid; unknown is. Rather, any is a non-type that’s used for interoperability with untyped JavaScript, behaving contradictorily as both a subtype and supertype of every type. This is why any & string is any, while unknown & string is string as you’d expect. any should not be used in typed code, while unknown is safe. \$\endgroup\$ Commented Apr 16, 2021 at 17:01

7 Answers 7

5
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JavaScript (ES6),  113 110 106  99 bytes

f=b=>(N='never',[x,o,y]=b.split` `,X=x<f?2:x!=N,Y=y<f?2:y!=N,!y|x==y?x:X*Y>1?'any':o>f?X?Y?b:x:y:N)

Try it online!

How?

Although this function is not recursive, it is named explicitly and uses the argument b so that we can test if a string is any (which is the first of all types in lexicographical order) by comparing it with \$f\$. The same trick is used to distinguish between & and |.

f = b => ...

We load the first type in \$x\$, the operator in \$o\$ and the second type in \$y\$. Both \$o\$ and \$y\$ may be undefined if only one type is passed in the input string.

[x, o, y] = s.split` `

The pair of strings \$(x,y)\$ is turned into a pair of integers \$(X,Y)\$ by applying the following rules:

  • if the type is any, map it to \$2\$
  • if the type is never, map it to \$0\$
  • if the type is anything else, map it to \$1\$
N = 'never'
X = x < f ? 2 : x != N
Y = y < f ? 2 : y != N

We then apply the following logic:

!y | x == y ? // if y is undefined or x = y:
  x           //   return x
:             // else:
  X * Y > 1 ? //   if X * Y > 1, which means that one of the types
              //   is 'any' and the other one is *not* 'never':
    'any'     //     return 'any'
  :           //   else:
    o > f ?   //     if this is a union, i.e. o = '|':
      X ?     //       if X is not equal to 0:
        Y ?   //         if Y is not equal to 0:
          b   //           return the original union
        :     //         else (Y = 0 / y = 'never'):
          x   //           return x
      :       //       else (X = 0 / x = 'never'):
        y     //         return y
    :         //     else (intersection):
      N       //       return 'never'
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4
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C++ TMP, 476 466 bytes

I'm surprised I got this far with it. U is union, I is intersection, R is the solution function that defines the solution type t. Online version with tests.

#define Z template
#define C class
#define K Z<C A>C R
#define T(x){public:using t=x;};
C string;C number;C boolean;C undefined;C null;C any;C never;using X=any;using Y=never;Z<C...>C U;Z<C...>C I;K T(A)K<U<A,A>>T(A)K<I<A,A>>T(A)Z<C A,C B>C R<I<A,B>>T(Y)K<U<A,X>>T(X)K<U<X,A>>T(X)K<I<X,A>>T(X)K<I<A,X>>T(X)Z<>C R<I<Y,X>>T(Y)Z<>C R<I<X,Y>>T(Y)K<U<A,Y>>T(A)K<U<Y,A>>T(A)Z<>C R<U<Y,X>>T(X)Z<>C R<U<X,Y>>T(X)Z<>C R<U<Y,Y>>T(Y)K<I<Y,A>>T(Y)K<I<A,Y>>T(Y)Z<>C R<I<Y,Y>>T(Y)

I'm certain this could be compacted further, maybe even to around 350 bytes. If anyone wants to give it a try, here is the version without preprocessor directives:

struct string;
struct number;
struct boolean;
struct undefined;
struct null;
struct any;
struct never;
using X=any;
using Y=never;
template<class A>struct T{using t=A;};
template<class...>struct U;
template<class...>struct I;
template<class A>struct R:T<A>{};
template<class A>struct R<U<A,A>>:T<A>{};
template<class A>struct R<I<A,A>>:T<A>{};
template<class A,class B>struct R<I<A,B>>:T<Y>{};
template<class A>struct R<U<A,X>>:T<X>{};
template<class A>struct R<U<X,A>>:T<X>{};
template<class A>struct R<I<X,A>>:T<X>{};
template<class A>struct R<I<A,X>>:T<X>{};
template<>struct R<I<Y,X>>:T<Y>{};
template<>struct R<I<X,Y>>:T<Y>{};
template<class A>struct R<U<A,Y>>:T<A>{};
template<class A>struct R<U<Y,A>>:T<A>{};
template<>struct R<U<Y,X>>:T<X>{};
template<>struct R<U<X,Y>>:T<X>{};
template<>struct R<U<Y,Y>>:T<Y>{};
template<class A>struct R<I<Y,A>>:T<Y>{};
template<class A>struct R<I<A,Y>>:T<Y>{};
template<>struct R<I<Y,Y>>:T<Y>{};
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  • \$\begingroup\$ Holy moly, C++ templates was not a solution I expected to see for this, but I'm very glad I did. Nice work, very impressive. \$\endgroup\$
    – Etheryte
    Commented Apr 16, 2021 at 15:55
  • \$\begingroup\$ This is nightmare fuel. \$\endgroup\$ Commented Aug 4, 2022 at 12:24
3
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Python 3, 261 bytes

lambda x,o="&",y="":("any"in x+y)*(o<"|")*-~-("never"in x+y)*"any"or(lambda r:r==31and"any"or"|".join("ubsnnnotuudormleliblfene iagr nn   e    d"[i::5]for i in range(5)if r&2**i)or"never")(eval(o.join(str(min(31,2**"dormlyv".find(k[2])%64))for k in[x,y or x])))

Try it online!

-1 byet thanks to ovs

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2
  • \$\begingroup\$ (1-("never"in x+y)) can be shortened to -~-("never"in x+y) or 0**("never"in x+y). \$\endgroup\$
    – ovs
    Commented Apr 14, 2021 at 12:24
  • \$\begingroup\$ @ovs 0** is very clever. also i did not realize using complement would actually make it short enough; should've tried more combinations \$\endgroup\$
    – hyper-neutrino
    Commented Apr 14, 2021 at 13:22
2
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Perl 5 -apl, 73 bytes

@F=sort@F;$_=$F[1]eq$F[2]?pop@F:/y/*(/,/+!/v/)?any:/&/?never:/v/?$F[2]:$_

Try it online!

Uses , instead of | so that the sort works without a customized function.

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2
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Python 3, 177 bytes

import enum;
def f(x):
	class T(enum.Flag):never=0;string=2;number=1;boolean=4;undefined=8;null=16;any=31
	return[str(eval(x,{},T))[2:],"any"][("&"in x)*("y"in x)*("v"not in x)]

Try it online!

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1
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Retina 0.8.2, 120 bytes

never \| (\w+)|(\w+) (\| never|[|&] \1)
$1$2
any & never|never & any
never
any [|&] \w+|\w+ [|&] any
any
\w+ & \w+
never

Try it online! Link includes test cases. Explanation:

never \| (\w+)|(\w+) (\| never|[|&] \1)
$1$2

Simplify unions with never and self-unions and intersections.

any & never|never & any
never

Simplify intersections between any and never.

any [|&] \w+|\w+ [|&] any
any

Simplify any other intersections of any.

\w+ & \w+
never

Simplify any other intersections.

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1
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TypeScript's type system, 38 bytes

type F<A,B=A,C=A>=B extends"&"?A&C:A|C

Try it at the TS playground

Takes input like F<string> or F<never,"|",number>.

This is admittedly a very cheaty I/O method, here's a more normal one:

TypeScript's type system, 163 bytes

//@ts-ignore
type F<A,B=A,C=A,D={string:string;number:number;boolean:boolean;undefined:undefined;null:null;any:any;never:never},X=D[A],Y=D[C]>=B extends"&"?X&Y:X|Y

Try it at the TS playground

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