modpost: fix undefined behavior of is_arm_mapping_symbol()

The return value of is_arm_mapping_symbol() is unpredictable when "$"
is passed in.

strchr(3) says:
  The strchr() and strrchr() functions return a pointer to the matched
  character or NULL if the character is not found. The terminating null
  byte is considered part of the string, so that if c is specified as
  '\0', these functions return a pointer to the terminator.

When str[1] is '\0', strchr("axtd", str[1]) is not NULL, and str[2] is
referenced (i.e. buffer overrun).

Test code
---------

  char str1[] = "abc";
  char str2[] = "ab";

  strcpy(str1, "$");
  strcpy(str2, "$");

  printf("test1: %d\n", is_arm_mapping_symbol(str1));
  printf("test2: %d\n", is_arm_mapping_symbol(str2));

Result
------

  test1: 0
  test2: 1

Signed-off-by: Masahiro Yamada <masahiroy@kernel.org>
Reviewed-by: Nick Desaulniers <ndesaulniers@google.com>
This commit is contained in:
Masahiro Yamada 2022-05-24 01:46:22 +09:00
Родитель b5beffa20d
Коммит d6b732666a
1 изменённых файлов: 2 добавлений и 1 удалений

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@ -1180,7 +1180,8 @@ static int secref_whitelist(const struct sectioncheck *mismatch,
static inline int is_arm_mapping_symbol(const char *str)
{
return str[0] == '$' && strchr("axtd", str[1])
return str[0] == '$' &&
(str[1] == 'a' || str[1] == 'd' || str[1] == 't' || str[1] == 'x')
&& (str[2] == '\0' || str[2] == '.');
}