是否有将选项传递给Go函数的规范方法？

Question

I am writing a package that exposes this function:

func Marshal(input interface{}) ([]byte, error)

This is fine for the majority of situations, but I also want to prove another function if there are extra options:

type MarshalOptions struct {
    OnlyStdClass bool
}

My first thought is to create another function:

func MarshalWithOptions(input interface{}, options MarshalOptions) ([]byte, error)

Is this the recommended way of doing this? Is there a standard naming convention for function that also provide a more specific version with options?

Answer 1

You can take a *MarshalOptions. The caller can then pass nil if they want the default behavior.

eg. func Marshal(input interface{}, options *MarshalOptions) ([]byte, error)