如何转换为typeof（field）？

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 } }

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 
}

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.