Can I construct a slice of a generic type with different type parameters?-CodePudding

In the following code I have a token type that contains a literal value. By using an empty interface I can create a slice of tokens and append tokens that have different types. I don't think it's possible to accomplish the same using generics since we wouldn't be able to infer a type for the Slice of tokens. Is this assumption correct?

type Token struct {
    TokenType string
    Literal interface{}
}

func main(){
    tok1 := &Token{TokenType: "string", Literal: "foo"}
    tok2 := &Token{TokenType: "integer", Literal: 10}
    tokS := []*Token{tok1, tok2}
}

CodePudding user response：

Is this assumption correct?

Yes. Your can have a slice of Token[string] or Token[int].

CodePudding user response：

since we wouldn't be able to infer a type for the Slice of tokens. Is this assumption correct?

Almost. More precisely, the slice of Token wouldn't infer anything because you yourself must construct it with a concrete instantiation of the generic type.

Given a parametrized Token type as:

type Token[T any] struct {
    TokenType string
    Literal   T
}

each instantiation with a different type parameter produces a different (named) type.

So when you construct a slice with a specific instance of Token[T any], different instances are simply not assignable to its element type:

tokS := []*Token[string]{tok1, tok2}
// invalid: cannot use tok2 (variable of type *Token[int]) as type *Token[string] in array or slice literal

The only slice that can hold different types, as Token[string] and Token[int] is []interface{} or []any.

A further note, type inference is used to deduce missing type parameters from those already supplied, or in case of functions, from function arguments. Generic types must be instantiated with a non-empty type parameter list. When the type param is only one, like Token[T any], you must provide that explicitly and there's nothing left to infer.