It's confusing both pointers and arrays are printed as *. Let's print
array types with [] so that we can identify them easily. Although it's
interchangable, sometimes it can cause confusion with size like in the
below example.
Note that it is not the same with C syntax where it goes to the variable
names, but we want to have it in the type names (like in Go language).
Before:
mov [20] 0x68(reg5) -> reg0 type='struct page**' size=0x80 (die:0x4e61d32)
After:
mov [20] 0x68(reg5) -> reg0 type='struct page*[]' size=0x80 (die:0x4e61d32)
Signed-off-by: Namhyung Kim <namhyung@kernel.org>
Acked-by: Masami Hiramatsu <mhiramat@kernel.org>
Cc: Adrian Hunter <adrian.hunter@intel.com>
Cc: Ian Rogers <irogers@google.com>
Cc: Ingo Molnar <mingo@kernel.org>
Cc: Jiri Olsa <jolsa@kernel.org>
Cc: Kan Liang <kan.liang@linux.intel.com>
Cc: Peter Zijlstra <peterz@infradead.org>
Link: https://lore.kernel.org/r/20240507041338.2081775-1-namhyung@kernel.org
Signed-off-by: Arnaldo Carvalho de Melo <acme@redhat.com>
const char *tmp = "";
tag = dwarf_tag(type_die);
- if (tag == DW_TAG_array_type || tag == DW_TAG_pointer_type)
+ if (tag == DW_TAG_pointer_type)
tmp = "*";
+ else if (tag == DW_TAG_array_type)
+ tmp = "[]";
else if (tag == DW_TAG_subroutine_type) {
/* Function pointer */
return strbuf_add(buf, "(function_type)", 15);