How can you get a reflect.Type instance of an struct without physically creating the struct?

Question

I want to create a registry of struct types to enable dynamic loading of solutions to 'Project Euler' problems. My current solution however requires the struct to first be created and zeroed before it's type can be registered:

package solution

import (
    "errors"
    "fmt"
    "os"
    "reflect"
)

type Solution interface {
    Load()
    Solve() string
}

type SolutionRegister map[string]reflect.Type

func (sr SolutionRegister) Set(t reflect.Type) {
    fmt.Printf("Registering %s\n", t.Name())
    sr[t.Name()] = t
}

func (sr SolutionRegister) Get(name string) (Solution, error) {
    if typ, ok := sr[name]; ok {
        sol := reflect.New(typ).Interface().(Solution)
        return sol, nil
    }
    return nil, errors.New("Invalid solution: " + name)
}

var solutionsRegistry = make(SolutionRegister)

func Register(sol Solution) {
    solutionsRegistry.Set(reflect.TypeOf(sol).Elem())
}

func Load(s string) Solution {
    sol, err := solutionsRegistry.Get(s)
    if err != nil {
        fmt.Printf("Error loading solution  %s (%s)\n", s, err)
        os.Exit(-1)
    }
    sol.Load()
    return sol
}

type DummySolution struct {
    data [100 * 1024 * 1024 * 1024]uint8
}

func (s *DummySolution) Load() {
}

func (s *DummySolution) Solve() string {
    return ""
}

func Init() {
    Register(&DummySolution{})
}

In this example the type of 'DummySolution struct' is registered inside the Init() function. This structure is purposefully absurdly large to illustrate the problem.

Is there a way I could access the type of DummySolution and other Solutions without having to create an instance of the structure beforehand?

Answer 1

You can use reflect.TypeOf((*DummySolution)(nil)).Elem() . Making a nil pointer doesn't allocate space for the whole struct, and Elem (described under the definition of reflect.Type ) gets you from a pointer (or slice, array, channel, or map) to its element type.