Edit: It looks like this has more to do with being an interface method than an embedded type.
https://play.golang.org/p/I5XPdWR_O0 Is there a generic way to get the caller safely, or should I just check for <autogenerated> and increment? On Sun, Aug 14, 2016 at 3:02 PM, Tim Hockin <thoc...@google.com> wrote: > It is, in fact, a method on an embedded type. That means I can document it > and it ceases to be a magic number! Thanks. > > > On Aug 14, 2016 12:27 PM, "Ian Lance Taylor" <i...@golang.org> wrote: >> >> On Sun, Aug 14, 2016 at 11:10 AM, 'Tim Hockin' via golang-nuts >> <golang-nuts@googlegroups.com> wrote: >> > I was playing with a method that wants to print the file:line of the >> > caller. `runtime.Caller(1)` seems like it should do the trick, but I >> > get '<autogenerated>:2'. If I call `runtime.Caller(2)` instead it >> > seems to work, but I don't know why, so I am reluctant to depend on >> > that. >> > >> > Can anyone enlighten me? How can I know how many frames to crawl back >> > to my real caller? >> > >> > go1.6.1 on linux/amd64 >> >> An `autogenerated` frame means that you are calling a method that is >> defined on some type T1 that is embedded in some other type T2. The >> `autogenerated` code is the little wrapper that takes a value of type >> T2 and invokes the method on the embedded field of type T1. As far as >> I know you will never see autogenerated code called by autogenerated >> code. So you only need to worry about this in a method of a type that >> your program will embed into some other type. If you don't know at >> run time whether the type is embedded or not, call runtime.Caller(1), >> and if you see autogenerated call runtime.Caller(2). >> >> Ian -- You received this message because you are subscribed to the Google Groups "golang-nuts" group. To unsubscribe from this group and stop receiving emails from it, send an email to golang-nuts+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
