[ Upstream commit
d6b732666a1bae0df3c3ae06925043bba34502b1 ]
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, "$");
printf("test1: %d\n", is_arm_mapping_symbol(str1));
printf("test2: %d\n", is_arm_mapping_symbol(str2));
Result
------
test1: 0
test2: 1
Signed-off-by: Masahiro Yamada <masahiroy@kernel.org>
Reviewed-by: Nick Desaulniers <ndesaulniers@google.com>
Signed-off-by: Sasha Levin <sashal@kernel.org>
static inline int is_arm_mapping_symbol(const char *str)
{
- return str[0] == '$' && strchr("axtd", str[1])
+ return str[0] == '$' &&
+ (str[1] == 'a' || str[1] == 'd' || str[1] == 't' || str[1] == 'x')
&& (str[2] == '\0' || str[2] == '.');
}