I have a C++ template class С and its member function foo(). I also have tags file generated with Exuberant Ctags.

I want to access this function via the :tag command, something like this:

:tag C::foo

However it doesn't seem to work for me. Inspecting tags file for understanding what should I have passed as :tag argument didn't help me.

Is it possible? How should I specify template class?

P.S. :tag foo works fine. The problem is that there a too many of them in different classes so it's hard to find appropriate tag in the taglist. It turned out the simplest way is to use :tag C and then finding /foo which is still a bit awkward.

2 Answers 2


There are (universal?) tags options that can request functions names to be prefixed by full scope (namespace + class). Unfortunately, I can't remember which one it is among:

--c++-kinds=+f --fields=+imaSftn --extra=+q

Any way, I highly recommend universal ctags instead of exuberant ctags for C++. And BTW, in lh-tags (that I'm maintaining), you'll find a more ergonomic way to select which overload you wish to jump to.


For what it's worth, you can use ts:

:ts foo
 # pri kind tag               file
 1 F C f    foo               b.cpp
               void foo(){
 2 F C f    foo               c.cpp
               void foo(){
Type number and <Enter> (empty cancels):

You can see all matching elements, and select them. But I did not see any option to specify directly the parent element.

  • 1
    Please see my PS -- I have about a hundred of foo's and it's a pain to look through all the pages of ts to find my match
    – DimG
    Apr 15, 2016 at 17:03

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