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fix sscanf %n match at end of input string

I was playing with some code that sometimes got a string where a %n
match should have been done where the input string ended, for example
like this:

  sscanf("abc123", "abc%d%n", &a, &n);  /* doesn't work */
  sscanf("abc123a", "abc%d%n", &a, &n); /* works */

However, the scanf function in the kernel doesn't convert the %n in that
case because it has already matched the complete input after %d and just
completely stops matching then. This patch fixes that.

[akpm@linux-foundation.org: cleanups]
Signed-off-by: Johannes Berg <johannes@sipsolutions.net>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
Johannes Berg 18 жил өмнө
parent
commit
c6b40d16d1
1 өөрчлөгдсөн 11 нэмэгдсэн , 0 устгасан
  1. 11 0
      lib/vsprintf.c

+ 11 - 0
lib/vsprintf.c

@@ -825,6 +825,17 @@ int vsscanf(const char * buf, const char * fmt, va_list args)
 			break;
 		str = next;
 	}
+
+	/*
+	 * Now we've come all the way through so either the input string or the
+	 * format ended. In the former case, there can be a %n at the current
+	 * position in the format that needs to be filled.
+	 */
+	if (*fmt == '%' && *(fmt + 1) == 'n') {
+		int *p = (int *)va_arg(args, int *);
+		*p = str - buf;
+	}
+
 	return num;
 }