gdb/split-name.c

   1 /* Split a symbol name.
   2
   3    Copyright (C) 2022-2024 Free Software Foundation, Inc.
   4
   5    This file is part of GDB.
   6
   7    This program is free software; you can redistribute it and/or modify
   8    it under the terms of the GNU General Public License as published by
   9    the Free Software Foundation; either version 3 of the License, or
  10    (at your option) any later version.
  11
  12    This program is distributed in the hope that it will be useful,
  13    but WITHOUT ANY WARRANTY; without even the implied warranty of
  14    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
  15    GNU General Public License for more details.
  16
  17    You should have received a copy of the GNU General Public License
  18    along with this program.  If not, see <http://www.gnu.org/licenses/>.  */
  19
  20 #include "defs.h"
  21 #include "split-name.h"
  22 #include "cp-support.h"
  23
  24 /* See split-name.h.  */
  25
  26 std::vector<std::string_view>
  27 split_name (const char *name, split_style style)
  28 {
  29   std::vector<std::string_view> result;
  30   unsigned int previous_len = 0;
  31
  32   switch (style)
  33     {
  34     case split_style::CXX:
  35       for (unsigned int current_len = cp_find_first_component (name);
  36            name[current_len] != '\0';
  37            current_len += cp_find_first_component (name + current_len))
  38         {
  39           gdb_assert (name[current_len] == ':');
  40           result.emplace_back (&name[previous_len],
  41                                current_len - previous_len);
  42           /* Skip the '::'.  */
  43           current_len += 2;
  44           previous_len = current_len;
  45         }
  46       break;
  47
  48     case split_style::DOT_STYLE:
  49       /* D and Go-style names.  */
  50       for (const char *iter = strchr (name, '.');
  51            iter != nullptr;
  52            iter = strchr (iter, '.'))
  53         {
  54           result.emplace_back (&name[previous_len],
  55                                iter - &name[previous_len]);
  56           ++iter;
  57           previous_len = iter - name;
  58         }
  59       break;
  60
  61     default:
  62       break;
  63     }
  64
  65   result.emplace_back (&name[previous_len]);
  66   return result;
  67 }
  68