[PATCH 4.19 207/213] kdb: Merge identical case statements in kdb_read()

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4.19-stable review patch.  If anyone has any objections, please let me know.

------------------

From: Daniel Thompson <daniel.thompson@xxxxxxxxxx>

commit 6244917f377bf64719551b58592a02a0336a7439 upstream.

The code that handles case 14 (down) and case 16 (up) has been copy and
pasted despite being byte-for-byte identical. Combine them.

Cc: stable@xxxxxxxxxxxxxxx # Not a bug fix but it is needed for later bug fixes
Reviewed-by: Douglas Anderson <dianders@xxxxxxxxxxxx>
Tested-by: Justin Stitt <justinstitt@xxxxxxxxxx>
Link: https://lore.kernel.org/r/20240424-kgdb_read_refactor-v3-4-f236dbe9828d@xxxxxxxxxx
Signed-off-by: Daniel Thompson <daniel.thompson@xxxxxxxxxx>
Signed-off-by: Greg Kroah-Hartman <gregkh@xxxxxxxxxxxxxxxxxxx>
---
 kernel/debug/kdb/kdb_io.c |   10 +---------
 1 file changed, 1 insertion(+), 9 deletions(-)

--- a/kernel/debug/kdb/kdb_io.c
+++ b/kernel/debug/kdb/kdb_io.c
@@ -314,6 +314,7 @@ poll_again:
 		}
 		break;
 	case 14: /* Down */
+	case 16: /* Up */
 		memset(tmpbuffer, ' ',
 		       strlen(kdb_prompt_str) + (lastchar-buffer));
 		*(tmpbuffer+strlen(kdb_prompt_str) +
@@ -328,15 +329,6 @@ poll_again:
 			++cp;
 		}
 		break;
-	case 16: /* Up */
-		memset(tmpbuffer, ' ',
-		       strlen(kdb_prompt_str) + (lastchar-buffer));
-		*(tmpbuffer+strlen(kdb_prompt_str) +
-		  (lastchar-buffer)) = '\0';
-		kdb_printf("\r%s\r", tmpbuffer);
-		*lastchar = (char)key;
-		*(lastchar+1) = '\0';
-		return lastchar;
 	case 9: /* Tab */
 		if (tab < 2)
 			++tab;






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