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ZMQII-51: Implement O(1) topic matching

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This commit is contained in:
Martin Sustrik
2010-01-31 20:14:30 +01:00
parent 70ea8e9d4b
commit 00b9a5dede
5 changed files with 242 additions and 47 deletions
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44
src/sub.cpp
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@@ -26,7 +26,6 @@
zmq::sub_t::sub_t (class app_thread_t *parent_) :
socket_base_t (parent_),
all_count (0),
has_message (false)
{
options.requires_in = true;
@@ -72,30 +71,14 @@ int zmq::sub_t::xsetsockopt (int option_, const void *optval_,
size_t optvallen_)
{
if (option_ == ZMQ_SUBSCRIBE) {
if (!optvallen_)
all_count++;
else
subscriptions.insert (std::string ((const char*) optval_,
optvallen_));
subscriptions.add ((unsigned char*) optval_, optvallen_);
return 0;
}
if (option_ == ZMQ_UNSUBSCRIBE) {
if (!optvallen_) {
if (!all_count) {
errno = EINVAL;
return -1;
}
all_count--;
}
else {
subscriptions_t::iterator it = subscriptions.find (
std::string ((const char*) optval_, optvallen_));
if (it == subscriptions.end ()) {
errno = EINVAL;
return -1;
}
subscriptions.erase (it);
if (!subscriptions.rm ((unsigned char*) optval_, optvallen_)) {
errno = EINVAL;
return -1;
}
return 0;
}
@@ -181,21 +164,6 @@ bool zmq::sub_t::xhas_out ()
bool zmq::sub_t::match (zmq_msg_t *msg_)
{
// If there is at least one * subscription, the message matches.
if (all_count)
return true;
// Check whether the message matches at least one prefix subscription.
// TODO: Make this efficient - O(log(n)) where n is number of characters in
// the longest subscription string.
for (subscriptions_t::iterator it = subscriptions.begin ();
it != subscriptions.end (); it++) {
size_t msg_size = zmq_msg_size (msg_);
size_t sub_size = it->size ();
if (sub_size <= msg_size &&
memcmp (zmq_msg_data (msg_), it->data (), sub_size) == 0)
return true;
}
return false;
return subscriptions.check ((unsigned char*) zmq_msg_data (msg_),
zmq_msg_size (msg_));
}