CODEWITHSUNDEEP
c++pass-by-referencefunctor
netleap tom
netleap tom

Reputation: 163

c++ calling a template functor but not passing an argument

I'm trying to understand two things in c++, functors and passing reference. Below is example code for the library libtins available here libtins example. The function snifferloop() takes the template functor doo().

The declaration for the functor is bool doo(PDU &some_pdu) so the argument some_pdu of type PDU is passed by reference to the function.

However, doo() is called as sniffer.sniff_loop(doo); and no argument appears to be passed to doo().

Could somebody explain what is happening here please? Apologies if it is a basic.

bool doo(PDU &some_pdu) {
    // Search for it. If there is no IP PDU in the packet, 
    // the loop goes on
    const IP &ip = some_pdu.rfind_pdu<IP>(); // non-const works as well
    std::cout << "Destination address: " << ip->dst_addr() << std::endl;
    // Just one packet please
    return false;
}

void test() {
    SnifferConfiguration config;
    config.set_promisc_mode(true);
    config.set_filter("ip src 192.168.0.100");
    Sniffer sniffer("eth0", config);
    sniffer.sniff_loop(doo);
}

Upvotes: 0

Views: 89

Answers (1)

Some programmer dude
Some programmer dude

Reputation: 409136

sniffer.sniff_loop(doo) doesn't call doo. It calls sniffer.sniff_loop, passing a pointer to doo as argument to sniffer.sniff_loop. Then sniffer.sniff_loop will use that argument to call doo, passing the argument to doo.

Upvotes: 3

Related Questions