typ := r.doTyp()
assert(typ != nil)
+ // For recursive type declarations involving interfaces and aliases,
+ // above r.doTyp() call may have already set pr.typs[idx], so just
+ // double check and return the type.
+ //
+ // Example:
+ //
+ // type F = func(I)
+ //
+ // type I interface {
+ // m(F)
+ // }
+ //
+ // The writer writes data types in following index order:
+ //
+ // 0: func(I)
+ // 1: I
+ // 2: interface{m(func(I))}
+ //
+ // The reader resolves it in following index order:
+ //
+ // 0 -> 1 -> 2 -> 0 -> 1
+ //
+ // and can divide in logically 2 steps:
+ //
+ // - 0 -> 1 : first time the reader reach type I,
+ // it creates new named type with symbol I.
+ //
+ // - 2 -> 0 -> 1: the reader ends up reaching symbol I again,
+ // now the symbol I was setup in above step, so
+ // the reader just return the named type.
+ //
+ // Now, the functions called return, the pr.typs looks like below:
+ //
+ // - 0 -> 1 -> 2 -> 0 : [<T> I <T>]
+ // - 0 -> 1 -> 2 : [func(I) I <T>]
+ // - 0 -> 1 : [func(I) I interface { "".m(func("".I)) }]
+ //
+ // The idx 1, corresponding with type I was resolved successfully
+ // after r.doTyp() call.
if typ := pr.typs[idx]; typ != nil {
- // This happens in fixedbugs/issue27232.go.
- // TODO(mdempsky): Explain why/how this happens.
return typ
}