How to pass type to function argument in Go

Question

ERROR: type CustomStruct is not an expression.

type CustomStruct struct { }  func getTypeName(t interface{}) string {     rt := reflect.TypeOf(t).Elem()     return rt.Name() }  getTypeName(CustomStruct) 

How can I pass struct type to function without type instance?

This will work

getTypeName((*CustomStruct)(nil)) 

But I wonder if there is more simple version..

Answer 1

You can't. You can only pass a value, and CustomStruct is not a value but a type. Using a type identifier is a compile-time error.

Usually when a "type" is to be passed, you pass a reflect.Type value which describes the type. This is what you "create" inside your getTypeName(), but then the getTypeName() will have little left to do:

func getTypeName(t reflect.Type) string {     return t.Name() }  // Calling it: getTypeName(reflect.TypeOf(CustomStruct{})) 

(Also don't forget that this returns an empty string for anonymous types such as []int.)

Another way is to pass a "typed" nil pointer value as you did, but again, you can just as well use a typed nil value to create the reflect.Type too, without creating a value of the type in question, like this:

t := reflect.TypeOf((*CustomStruct)(nil)).Elem() fmt.Println(t.Name()) // Prints CustomStruct