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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, "$");
printf("test1: %d\n", is_arm_mapping_symbol(str1));
printf("test2: %d\n", is_arm_mapping_symbol(str2));
Result
------
test1: 0
test2: 1
Signed-off-by: Masahiro Yamada <masahiroy@kernel.org>
Reviewed-by: Nick Desaulniers <ndesaulniers@google.com>
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1 changed files with 2 additions and 1 deletions
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@ -1180,7 +1180,8 @@ static int secref_whitelist(const struct sectioncheck *mismatch,
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static inline int is_arm_mapping_symbol(const char *str)
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static inline int is_arm_mapping_symbol(const char *str)
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{
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{
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return str[0] == '$' && strchr("axtd", str[1])
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return str[0] == '$' &&
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(str[1] == 'a' || str[1] == 'd' || str[1] == 't' || str[1] == 'x')
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&& (str[2] == '\0' || str[2] == '.');
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&& (str[2] == '\0' || str[2] == '.');
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}
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}
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