Recursive Type Constraint with Predefined Type
In Go generics, it is possible to specify type constraints on generic types using interfaces. However, it can be challenging to define a constraint that requires the implementation of a method with an argument of the generic type itself.
Current Draft Limitations
The current draft of Go generics does not allow recursive type constraints, making it impossible to achieve the desired behavior. For example, consider the following interface:
type Lesser interface { Less(rhs Lesser) bool }
We can implement this interface for a custom type, such as Int:
type Int int func (lhs Int) Less(rhs Int) bool { return lhs <p>However, the following code will fail to compile because Int does not satisfy the Lesser interface:</p><pre class="brush:php;toolbar:false">func IsLess[T Lesser](lhs, rhs T) bool { return lhs.Less(rhs) } func main() { IsLess[Int](Int(10), Int(20)) }
Solution Using Predefined Interface
A workaround for this limitation is to define a generic interface that references itself as a type parameter:
type Lesser[T any] interface { Less(T) bool }
This interface specifies that any type that implements the Lesser interface must have a Less method that takes an argument of the same type.
We can then define the IsLess function using this generic interface:
func IsLess[T Lesser[T]](x, y T) bool { return x.Less(y) }
Now, we can implement custom types that fulfill the Lesser constraint:
type Apple int func (a Apple) Less(other Apple) bool { return a <p>Using the IsLess function with these custom types will result in correct behavior:</p><pre class="brush:php;toolbar:false">func main() { fmt.Println(IsLess(Apple(10), Apple(20))) // true fmt.Println(IsLess(Orange(30), Orange(15))) // false }
This solution effectively enforces the recursive type constraint by requiring the implementation of a method with an argument of the generic type.
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