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290 | // Copyright (C) 2020-2024 Internet Systems Consortium, Inc. ("ISC")
//
// This Source Code Form is subject to the terms of the Mozilla Public
// License, v. 2.0. If a copy of the MPL was not distributed with this
// file, You can obtain one at http://mozilla.org/MPL/2.0/.
#include <config.h>
#include <dhcpsrv/ip_range_permutation.h>
#include <gtest/gtest.h><--- Include file: not found. Please note: Cppcheck does not need standard library headers to get proper results.
#include <set><--- Include file: not found. Please note: Cppcheck does not need standard library headers to get proper results.
#include <vector><--- Include file: not found. Please note: Cppcheck does not need standard library headers to get proper results.
using namespace isc;
using namespace isc::asiolink;
using namespace isc::dhcp;
namespace {
// This test verifies that the object can be successfully constructed for
// both IPv4 and IPv6 address range.
TEST(IPRangePermutationTest, constructor) {
ASSERT_NO_THROW({
AddressRange range(IOAddress("192.0.2.10"), IOAddress("192.0.2.100"));
IPRangePermutation perm(range);
});
ASSERT_NO_THROW({
AddressRange range(IOAddress("3000::"), IOAddress("3000::10"));
IPRangePermutation perm(range);
});
}
// This test verifies that a permutation of IPv4 address range can
// be generated and each time a different permutation is generated.
TEST(IPRangePermutationTest, ipv4) {<--- syntax error
// Create address range with 91 addresses.
AddressRange range(IOAddress("192.0.2.10"), IOAddress("192.0.2.100"));
std::vector<std::vector<IOAddress>> iterations;
for (auto i = 0; i < 2; ++i) {
IPRangePermutation perm(range);
// This set will record unique IP addresses generated.
std::set<IOAddress> addrs;
// This vector will record the addresses assignment order.
std::vector<IOAddress> ordered_addrs;
bool done = false;
// Call the next() function 95 times. The first 91 calls should return non-zero
// IP addresses.
for (auto j = 0; j < 95; ++j) {
auto next = perm.next(done);
if (!next.isV4Zero()) {
// Make sure the returned address is within the range.
EXPECT_LE(range.start_, next);
EXPECT_LE(next, range.end_);
}
// If we went over all addresses in the range, the flags indicating that
// the permutation is exhausted should be set to true.
if (j >= 90) {
EXPECT_TRUE(done);
EXPECT_TRUE(perm.exhausted());
} else {
// We're not done yet, so these flag should still be false.
EXPECT_FALSE(done);
EXPECT_FALSE(perm.exhausted());
}
// Insert the address returned to the set and vector.
addrs.insert(next);
ordered_addrs.push_back(next);
}
// We should have recorded 92 unique addresses, including the zero address.
EXPECT_EQ(92, addrs.size());
EXPECT_TRUE(addrs.begin()->isV4Zero());
iterations.push_back(ordered_addrs);
}
// We want to make sure that each new permutation instance produces a different
// sequence of addresses. It checks whether or not the random device has been
// initialized properly. If the random device uses the same seed for each
// new permutation, the output sequence is always the same. The test below
// checks that the sequences are different by comparing the respective addresses
// for two different permutations. It is ok if some of them are equal because it
// is statistically probable. The threshold of 20% should guard against some
// of them being equal without a risk of sporadic test failures.
int overlaps = 0;
for (auto i = 0; i < iterations[0].size(); ++i) {
if (iterations[0][i] == iterations[1][i]) {
++overlaps;
}
}
EXPECT_LE(overlaps, iterations[0].size()/5)
<< "The number of overlapping random address between the test two iterations"
<< " is greater than 20% of all allocated addresses in each iteration."
<< " It means that the permutation mechanism does not sufficiently randomize"
<< " addresses. Perhaps the randomization device is not properly initialized?";
}
// This test verifies that a permutation of IPv6 address range can
// be generated and each time a different permutation is generated.
TEST(IPRangePermutationTest, ipv6) {
AddressRange range(IOAddress("2001:db8:1::1:fea0"),
IOAddress("2001:db8:1::2:abcd"));
std::vector<std::vector<IOAddress>> iterations;
for (auto i = 0; i < 2; ++i) {
IPRangePermutation perm(range);
std::set<IOAddress> addrs;
std::vector<IOAddress> ordered_addrs;
bool done = false;
for (auto j = 0; j < 44335; ++j) {
auto next = perm.next(done);
if (!next.isV6Zero()) {
// Make sure that the address is within the range.
EXPECT_LE(range.start_, next);
EXPECT_LE(next, range.end_);
}
// If we went over all addresses in the range, the flags indicating that
// the permutation is exhausted should be set to true.
if (j >= 44333) {
EXPECT_TRUE(done);
EXPECT_TRUE(perm.exhausted());
} else {
// We're not done yet, so these flag should still be false.
EXPECT_FALSE(done);
EXPECT_FALSE(perm.exhausted());
}
// Insert the address returned to the set and vector.
addrs.insert(next);
ordered_addrs.push_back(next);
}
// We should have recorded 44335 unique addresses, including the zero address.
EXPECT_EQ(44335, addrs.size());
EXPECT_TRUE(addrs.begin()->isV6Zero());
iterations.push_back(ordered_addrs);
}
// We want to make sure that each new permutation instance produces a different
// sequence of addresses. It checks whether or not the random device has been
// initialized properly. If the random device uses the same seed for each
// new permutation, the output sequence is always the same. The test below
// checks that the sequences are different by comparing the respective addresses
// for two different permutations. It is ok if some of them are equal because it
// is statistically probable. The threshold of 20% should guard against some
// of them being equal without a risk of sporadic test failures.
int overlaps = 0;
for (auto i = 0; i < iterations[0].size(); ++i) {
if (iterations[0][i] == iterations[1][i]) {
++overlaps;
}
}
EXPECT_LE(overlaps, iterations[0].size()/5)
<< "The number of overlapping random address between the test two iterations"
<< " is greater than 20% of all allocated addresses in each iteration."
<< " It means that the permutation mechanism does not sufficiently randomize"
<< " addresses. Perhaps the randomization device is not properly initialized?";
}
// This test verifies that a permutation of delegated prefixes can be
// generated and each time a different permutation is generated.
TEST(IPRangePermutationTest, pd) {
PrefixRange range(IOAddress("3000::"), 112, 120);
std::vector<std::vector<IOAddress>> iterations;
for (auto i = 0; i < 2; ++i) {
IPRangePermutation perm(range);
std::set<IOAddress> addrs;
std::vector<IOAddress> ordered_addrs;
bool done = false;
for (auto j = 0; j < 257; ++j) {
auto next = perm.next(done);
if (!next.isV6Zero()) {
// Make sure the prefix is within the range.
EXPECT_LE(range.start_, next);
EXPECT_LE(next, range.end_);
}
// If we went over all delegated prefixes in the range, the flags indicating
// that the permutation is exhausted should be set to true.
if (j >= 255) {
EXPECT_TRUE(done);
EXPECT_TRUE(perm.exhausted());
} else {
// We're not done yet, so these flag should still be false.
EXPECT_FALSE(done);
EXPECT_FALSE(perm.exhausted());
}
// Insert the prefix returned to the set and vector.
addrs.insert(next);
ordered_addrs.push_back(next);
}
// We should have recorded 257 unique addresses, including the zero address.
EXPECT_EQ(257, addrs.size());
EXPECT_TRUE(addrs.begin()->isV6Zero());
iterations.push_back(ordered_addrs);
}
ASSERT_EQ(2, iterations.size());
// We want to make sure that each new permutation instance produces a different
// sequence of prefixes. It checks whether or not the random device has been
// initialized properly. If the random device uses the same seed for each
// new permutation, the output sequence is always the same. The test below
// checks that the sequences are different by comparing the respective prefixes
// for two different permutations. It is ok if some of them are equal because it
// is statistically probable. The threshold of 20% should guard against some
// of them being equal without a risk of sporadic test failures.
int overlaps = 0;
for (auto i = 0; i < iterations[0].size(); ++i) {
if (iterations[0][i] == iterations[1][i]) {
++overlaps;
}
}
EXPECT_LE(overlaps, iterations[0].size()/5)
<< "The number of overlapping random prefixes between the test two iterations"
<< " is greater than 20% of all allocated addresses in each iteration."
<< " It means that the permutation mechanism does not sufficiently randomize"
<< " prefixes. Perhaps the randomization device is not properly initialized?";
}
// This test verifies that a permutation of delegated prefixes is
// generated from the prefix range specified using first and last
// address.
TEST(IPRangePermutationTest, pdStartEnd) {
PrefixRange range(IOAddress("3000::"), IOAddress("3000::ffff"), 120);
IPRangePermutation perm(range);
std::set<IOAddress> addrs;
bool done = false;
for (auto i = 0; i < 257; ++i) {
auto next = perm.next(done);
if (!next.isV6Zero()) {
// Make sure the prefix is within the range.
EXPECT_LE(range.start_, next);
EXPECT_LE(next, range.end_);
}
// If we went over all delegated prefixes in the range, the flags indicating
// that the permutation is exhausted should be set to true.
if (i >= 255) {
EXPECT_TRUE(done);
EXPECT_TRUE(perm.exhausted());
} else {
// We're not done yet, so these flag should still be false.
EXPECT_FALSE(done);
EXPECT_FALSE(perm.exhausted());
}
// Insert the prefix returned to the set.
addrs.insert(next);
}
// We should have recorded 257 unique addresses, including the zero address.
EXPECT_EQ(257, addrs.size());
EXPECT_TRUE(addrs.begin()->isV6Zero());
}
// This test verifies that it is possible to reset the permutation state.
TEST(IPRangePermutationTest, reset) {
// Create address range with 11 addresses.
AddressRange range(IOAddress("192.0.2.10"), IOAddress("192.0.2.20"));
IPRangePermutation perm(range);
// This set will record unique IP addresses generated.
std::set<IOAddress> addrs;
bool done = false;
// Call the next() function several times to consume several addresses.
for (auto i = 0; i < 5; ++i) {
auto next = perm.next(done);
EXPECT_FALSE(next.isV4Zero());
addrs.insert(next);
}
EXPECT_EQ(5, addrs.size());
// Reset the permutation. We should be able to get all addresses again.
perm.reset();
for (auto i = 0; i < 11; ++i) {
auto next = perm.next(done);
EXPECT_FALSE(next.isV4Zero());
addrs.insert(next);
}
EXPECT_EQ(11, addrs.size());
}
} // end of anonymous namespace
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