kdb: remove redundant check for scancode 0xe0

ColinIanKing · daniel-thompson · commit fdbdd0ccb30a · 2025-09-20T21:19:09.000+01:00
The check for scancode 0xe0 is always false because earlier on
the scan code is masked with 0x7f so there are never going to
be values greater than 0x7f. Remove the redundant check.

Signed-off-by: Colin Ian King &lt;colin.i.king@gmail.com&gt;
Reviewed-by: Douglas Anderson &lt;dianders@chromium.org&gt;
Signed-off-by: Daniel Thompson (RISCstar) &lt;danielt@kernel.org&gt;
diff --git a/kernel/debug/kdb/kdb_keyboard.c b/kernel/debug/kdb/kdb_keyboard.c
@@ -145,9 +145,6 @@ int kdb_get_kbd_char(void)
 		return CTRL('F');
 	}
 
-	if (scancode == 0xe0)
-		return -1;
-
 	/*
 	 * For Japanese 86/106 keyboards
 	 * 	See comment in drivers/char/pc_keyb.c.

Original file line number	Diff line number	Diff line change
`@@ -145,9 +145,6 @@ int kdb_get_kbd_char(void)`
`145`	`145`	`return CTRL('F');`
`146`	`146`	`}`
`147`	`147`
`148`		`- if (scancode == 0xe0)`
`149`		`- return -1;`
`150`		`-`
`151`	`148`	`/*`
`152`	`149`	`* For Japanese 86/106 keyboards`
`153`	`150`	`* See comment in drivers/char/pc_keyb.c.`