如何转换为typeof（field）？

Question

Given something like this:

type Foo struct {
  x int
}

type FooFoo struct {
  foo *Foo
}

type Bar struct {
  x int
}

Where Foo is hidden (in my case due to vendoring), how can I create a FooFoo struct with a valid foo entry?

If Foo were available, I could do

foofoo := &FooFoo{ foo: &Foo{5} }

or even

foofoo := &FooFoo{ foo: (*Foo)&Bar{ 5 } }

But I can't find a way to do it without mentioning Foo .

I think I would need something like:

foofoo := &FooFoo{ foo: (*typeof(FooFoo.foo)) &Bar{ 5 } }

Answer 1

You shouldn't set the private method from another library as per this answer . However, the library should have an appropriate constructor. The library should have a method that looks like

func FooFooGenerator(appropriateInfo int) FooFoo {
 ... Library does the work 
}

Answer 2

You just need to export a "constructor" function for FooFoo and keep your foo unexported.

func NewFooFoo() *FooFoo {
    f := &foo{ /* initialize foo */ }
    return &FooFoo{foo:f}
}

Then clients of you package will have access to NewFooFoo , and FooFoo , but not foo .

And as for casting Foo to Bar , not sure why you would want that, but you can do it, if you are on Go1.8+ , this way (*Foo)(&Bar{ 5 }) playground .