Fix unefficient call to memset before memcpu in nla_strlcpy.

Before this commit, nla_strlcpy first memseted dst to 0 then wrote src into it.
This is inefficient because bytes whom number is less than src length are written
twice.

This patch solves this issue by first writing src into dst then fill dst with
0's.
Note that, in the case where src length is higher than dst, only 0 is written.
Otherwise there are as many 0's written to fill dst.

For example, if src is "foo\0" and dst is 5 bytes long, the result will be:
1. "fooGG" after memcpy (G means garbage).
2. "foo\0\0" after memset.

Signed-off-by: Francis Laniel <laniel_francis@privacyrequired.com>
Reviewed-by: Kees Cook <keescook@chromium.org>
Signed-off-by: Jakub Kicinski <kuba@kernel.org>

diff --git a/lib/nlattr.c b/lib/nlattr.c
index 74019c8ebf6beec7de2d8fb183992eae6a152c17..07156e5819978187d92f55c52be49c67d70dac4d 100644 (file)
--- a/lib/nlattr.c
+++ b/lib/nlattr.c
@@ -731,8 +731,9 @@ size_t nla_strlcpy(char *dst, const struct nlattr *nla, size_t dstsize)
        if (dstsize > 0) {
                size_t len = (srclen >= dstsize) ? dstsize - 1 : srclen;
 
-               memset(dst, 0, dstsize);
                memcpy(dst, src, len);
+               /* Zero pad end of dst. */
+               memset(dst + len, 0, dstsize - len);
        }
 
        return srclen;