Why is my socket's open port not listed by netstat?

Question

If you run this example, you'll see the port is never listed by netstat. Why? And how do I make it so?

#include <WinSock.h>
#include <io.h>
#include <stdio.h>

#pragma comment(lib, "WS2_32")

int main() {
    WORD wVers = MAKEWORD(2, 2);
    WSADATA wsa;
    WSAStartup(wVers, &wsa);
    SOCKET sock = socket(AF_INET, SOCK_STREAM, 6);
    if (sock != INVALID_SOCKET) {
        struct sockaddr_in addr = { 0 };
        addr.sin_family = AF_INET;
        addr.sin_addr.s_addr = htonl(INADDR_LOOPBACK);
        int addrlen = sizeof(addr);
        bind(sock, (struct sockaddr *)&addr, addrlen);
        if (getsockname(sock, (struct sockaddr *)&addr, &addrlen) == 0) {
            fprintf(stdout, "HANDLE = %d, port = %d\n", sock, addr.sin_port);
            fflush(stdout);
            system("netstat -a -n");
        }
        closesocket(sock);
    }
    WSACleanup();
}

Solution

netstat -a only lists connected sockets and listening socket.

  -a            Displays all connections and listening ports.

Neither connect nor listen was called on your socket, so it falls outside the purview of netstat -a.

However, since Windows 10, you can use netstat -q.

  -q            Displays all connections, listening ports, and bound

Comments

[nos]

Actually - at least on windows 10 netstat has this flag " -q Displays all connections, listening ports, and bound" This shows these incomplete sockets with the word "BOUND" in the State column

[ikegami]

@nos, Thanks. Added to my answer.

[user541686]

Now that's a complete answer. Thanks!

[Eli Finkel]

The -q flag seems not to work in conjunction with the -b (shows binary/exe name), but it does works with -o (shows owning pid), which is good enough. (or at least this is my experience on Win10 1903)