How to create a generic struct?

Question

How can I construct a generic struct?

I tried:

type SafeSet[type T] struct {
    Values map[T]bool
}

---

I want to be able to do eg

SafeSet{ Values: make(map[net.Conn]bool) }
SafeSet{ Values: make(map[string]  bool) }
SafeSet{ Values: make(map[int]     bool) }

Answer 1

You can't do that with the current Go version, 1.17. Unfortunately there's nothing else to say.

---

After generics will be added to the language, probably in Go 1.18 (early 2022), the syntax for such a parametrized types, according to the current accepted proposal, will be:

type SafeSet[T comparable] struct {
    Values map[T]bool
}

In particular:

• The type constraint comes after the type name T
• If you want to use T as a map key, you must use the built-in constraint comparable , because map keys must be comparable — ie support == operator.

Then you have to instantiate the parametrized type with an actual type argument:

Example:

To use a generic type, you must supply type arguments. This is called instantiation. The type arguments appear in square brackets, as usual. When we instantiate a type by supplying type arguments for the type parameters, we produce a type in which each use of a type parameter in the type definition is replaced by the corresponding type argument.

    s0 := SafeSet[net.Conn]{Values: make(map[net.Conn]bool)}
    s1 := SafeSet[string]{Values: make(map[string]bool)}
    s2 := SafeSet[int]{Values: make(map[int]bool)}

Since instantiating SafeSet literals looks kinda verbose, you can use a generic constructor func:

func NewSafeSet[T comparable]() SafeSet[T] {
    return SafeSet[T]{Values: make(map[T]bool)}
}

The syntax is, obviously, the same, except that in this case you are explicitly instantiating the function with the type arg:

    s3 := NewSafeSet[uint64]()
    s3.Values[200] = true

Gotip Playground: https://gotipplay.golang.org/p/Qyd6zTLdkRn