@@ -122,8 +122,8 @@ static int kdb_getchar(void)
{
#define ESCAPE_UDELAY 1000
#define ESCAPE_DELAY (2*1000000/ESCAPE_UDELAY) /* 2 seconds worth of udelays */
- char escape_data[5]; /* longest vt100 escape sequence is 4 bytes */
- char *ped = escape_data;
+ char buf[4]; /* longest vt100 escape sequence is 4 bytes */
+ char *pbuf = buf;
int escape_delay = 0;
get_char_func *f, *f_escape = NULL;
int key;
@@ -145,27 +145,22 @@ static int kdb_getchar(void)
continue;
}
- if (escape_delay == 0 && key == '\e') {
- escape_delay = ESCAPE_DELAY;
- ped = escape_data;
+ /* Reset state on first character from an input source */
+ if (f_escape != f) {
f_escape = f;
- }
- if (escape_delay) {
- if (f_escape != f)
- return '\e';
-
- *ped++ = key;
- key = kdb_read_handle_escape(escape_data,
- ped - escape_data);
- if (key < 0)
- return '\e';
- if (key == 0)
- continue;
+ pbuf = buf;
+ escape_delay = ESCAPE_DELAY;
}
- break; /* A key to process */
+ *pbuf++ = key;
+ key = kdb_read_handle_escape(buf, pbuf - buf);
+ if (key < 0) /* no escape sequence; return first character */
+ return buf[0];
+ if (key > 0)
+ return key;
}
- return key;
+
+ unreachable();
}
/*
Currently if an escape timer is interrupted by a character from a different input source then the new character is discarded and the function returns '\e' (which will be discarded by the level above). It is hard to see why this would ever be the desired behaviour. Fix this to return the new character rather then the '\e'. This is a bigger refactor that might be expected because the new character needs to go through escape sequence detection. Signed-off-by: Daniel Thompson <daniel.thompson@linaro.org> --- kernel/debug/kdb/kdb_io.c | 33 ++++++++++++++------------------- 1 file changed, 14 insertions(+), 19 deletions(-) -- 2.17.1