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owipcalc: fix contains not respect default route

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In IPv4 the default route can be written as
 0.0.0.0/0

In IPv6 the default route can be written as
 ::/0

If u try
  owipcalc 0.0.0.0/0 contains 1.1.1.1
or
  owipcalc ::/0 contains ::1
owipcalc will respond with 0 meaning that the "default prefixes" do not
contain the routes.
That is why we check now for 0 prefix.

Furthermore, if the prefix is 0, i will be 16. We will access a negative
array entry in the line:
 uint8_t net1 = x->s6_addr[15-i] & m;

Divide by % 16 to prevent i becoming 16:
 uint8_t i = ((128 - a->prefix) / 8) % 16;

Signed-off-by: Nick Hainke <vincent@systemli.org>
This commit is contained in:
Nick Hainke 2021-01-23 13:48:11 +01:00
parent 64c917b702
commit e974acaac7
2 changed files with 6 additions and 5 deletions
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2
net/owipcalc/Makefile
View file

@ -7,7 +7,7 @@
include $(TOPDIR)/rules.mk
PKG_NAME:=owipcalc
PKG_RELEASE:=5
PKG_RELEASE:=6
PKG_LICENSE:=Apache-2.0
include $(INCLUDE_DIR)/package.mk

9
net/owipcalc/src/owipcalc.c
View file

@ -227,7 +227,7 @@ static bool cidr_contains4(struct cidr *a, struct cidr *b)
if (printed)
qprintf(" ");
if ((b->prefix >= a->prefix) && (net1 == net2))
if ((a->prefix == 0) || ((b->prefix >= a->prefix) && (net1 == net2)))
{
qprintf("1");
return true;
@ -529,7 +529,7 @@ static bool cidr_contains6(struct cidr *a, struct cidr *b)
{
struct in6_addr *x = &a->addr.v6;
struct in6_addr *y = &b->addr.v6;
uint8_t i = (128 - a->prefix) / 8;
uint8_t i = ((128 - a->prefix) / 8) % 16;
uint8_t m = ~((1 << ((128 - a->prefix) % 8)) - 1);
uint8_t net1 = x->s6_addr[15-i] & m;
uint8_t net2 = y->s6_addr[15-i] & m;
@ -537,8 +537,9 @@ static bool cidr_contains6(struct cidr *a, struct cidr *b)
if (printed)
qprintf(" ");
if ((b->prefix >= a->prefix) && (net1 == net2) &&
((i == 15) || !memcmp(&x->s6_addr, &y->s6_addr, 15-i)))
if ((a->prefix == 0) ||
((b->prefix >= a->prefix) && (net1 == net2) &&
((i == 15) || !memcmp(&x->s6_addr, &y->s6_addr, 15-i))))
{
qprintf("1");
return true;