seq_buf: Make seq_buf_puts() null-terminate the buffer

Currently seq_buf_puts() will happily create a non null-terminated
string for you in the buffer. This is particularly dangerous if the
buffer is on the stack.

For example:

  char buf[8];
  char secret = "secret";
  struct seq_buf s;

  seq_buf_init(&s, buf, sizeof(buf));
  seq_buf_puts(&s, "foo");
  printk("Message is %s\n", buf);

Can result in:

  Message is fooªªªªªsecret

We could require all users to memset() their buffer to zero before
use. But that seems likely to be forgotten and lead to bugs.

Instead we can change seq_buf_puts() to always leave the buffer in a
null-terminated state.

The only downside is that this makes the buffer 1 character smaller
for seq_buf_puts(), but that seems like a good trade off.

Link: http://lkml.kernel.org/r/20181019042109.8064-1-mpe@ellerman.id.auAcked-by: Kees Cook <keescook@chromium.org>
Signed-off-by: Michael Ellerman <mpe@ellerman.id.au>
Signed-off-by: Steven Rostedt (VMware) <rostedt@goodmis.org>

lib/seq_buf.c

@@ -144,9 +144,13 @@ int seq_buf_puts(struct seq_buf *s, const char *str)
 
 	WARN_ON(s->size == 0);
 
+	/* Add 1 to len for the trailing null byte which must be there */
+	len += 1;
+
 	if (seq_buf_can_fit(s, len)) {
 		memcpy(s->buffer + s->len, str, len);
-		s->len += len;
+		/* Don't count the trailing null byte against the capacity */
+		s->len += len - 1;
 		return 0;
 	}
 	seq_buf_set_overflow(s);