code: def ssh_accept(cfg, code, key_file, yes, **kwargs):
      
  
      for name, value in kwargs.items():
          setattr(cfg, name, value)
      from . import cmd_ssh
      kind, keyid, pubkey = cmd_ssh.find_public_key(key_file)
      print("Sending public key type='{}' keyid='{}'".format(kind, keyid))
      if yes is not True:
          click.confirm(
              "Really send public key '{}' ?".format(keyid), abort=True)
      cfg.public_key = (kind, keyid, pubkey)
      cfg.code = code
  
      return go(cmd_ssh.accept, cfg)
docs: Send your SSH public-key
  
      In response to a 'wormhole ssh invite' this will send public-key
      you specify (if there's only one in ~/.ssh/* that will be sent).