Problem: In many cases, KDB treats invalid commands as numbers and
instead of printing a usage error, goes ahead and just prints the number
in hex
Example: This can be demonstrated when typing for example "aaazzz", this
confuses KDB into thinking this is the hexadecimal 0xAAA
Solution: Before assuming that the input from the user is a number,
check that it contains only characters that represent numbers.
Also, along the way, transition to using kstrtoul instead of
simple_strtoul (better practice as stated in the definition of the
function)
Signed-off-by: Nir Lichtman <n...@lichtman.org>
---
kernel/debug/kdb/kdb_main.c | 10 +++++-----
1 file changed, 5 insertions(+), 5 deletions(-)
diff --git a/kernel/debug/kdb/kdb_main.c b/kernel/debug/kdb/kdb_main.c
index f5f7d7fb5936..4efdc4d25a59 100644
--- a/kernel/debug/kdb/kdb_main.c
+++ b/kernel/debug/kdb/kdb_main.c
@@ -402,18 +402,18 @@ static void kdb_printenv(void)
*/
int kdbgetularg(const char *arg, unsigned long *value)
{
- char *endp;
unsigned long val;
- val = simple_strtoul(arg, &endp, 0);
+ if ((strpbrk(arg, hex_asc) == NULL)
+ && (strpbrk(arg, hex_asc_upper) == NULL))
+ return KDB_BADINT;
- if (endp == arg) {
+ if (kstrtoul(arg, 0, &val) != 0) {
/*
* Also try base 16, for us folks too lazy to type the
* leading 0x...
*/
- val = simple_strtoul(arg, &endp, 16);
- if (endp == arg)
+ if (kstrtoul(arg, 16, &val) != 0)
return KDB_BADINT;
}
--
2.39.2
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