How to create a generic struct?

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 e.g.

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

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

<hr>

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 — i.e. 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