Reflection in Go: create a stack[T]

Do you know what can Go's package reflect do?

Whatever you had use it or not. Understanding it is not a bad thing.

A well known thing is Go don't have generic, I'm not going to tell you we have generic, I'm going to tell you some basic trick to have the result like generic.

Real world example: elz-lang/collection/stack

Elz is a language I'm developing, but that's not the point. Point is this collection/stack using the trick I'm going to talk about.

Take a look on the type Stack

type Stack struct {
    stack  []interface{}
    limitT *reflect.Type

limitT is a *reflect.Type, the reason that it's a pointer to reflect.Type rather than reflect.Type is because of we may do not spec it.

We add the Stack<T> by invoke WithT.

func (s *Stack) WithT(v interface{}) *Stack {
    t := reflect.TypeOf(v).Elem()
    s.limitT = &t
    return s

Why is reflect.TypeOf(v).Elem()? Because we can't really get an instance that type is an interface! Instead of that, we can get a type is pointer to an interface!

We have a common idiom is using (*SomeInterface)(nil) to get pointer to interface instance.

Now we know that user code can be

type AST interface {
    Gen() llvm.Value

// main
s := stack.New().WithT((*AST)(nil))

After we do that, user can't push a value do not implement AST.

So, how we do that? We do a check at Push

func (s *Stack) Push(element interface{}) {
    if s.limitT != nil {
	if !reflect.TypeOf(element).Implements(*s.limitT) {
	    panic(fmt.Sprintf("element must implement type: %s", *s.limitT))
    s.stack = append(s.stack, element)

If limitT is not nil, means we do not limit the type, just keep going on.

But if we limit the type, we check that element implements limitT or not.

If not, we panic the process.

Now we have a stack can promise type safe at runtime.

Date: 2018-07-22 Sun 00:00

Author: Lîm Tsú-thuàn