[Asterisk-Dev] ast_waitfor*() question

[Luigi Rizzo]

functions ast_waitfor_n_fd() and ast_waitfor_nandfds() have a performance
issue that I would like to fix.
 
ast_waitfor_n_fd() has an O(N^2) loop to figure out the 'winner' fd 
because ast_fdisset() itself is O(N):
   
    for (x=0;x<n;x++) {
        if (fds[x] > -1) {
            if ((res = ast_fdisset(pfds, fds[x], y, &spoint))) {
                ...
            }
        }
    }
 
ast_waitfor_nandfds() is even worse - the loop is O(N^2 * AST_MAX_FDS)
 
    for (x=0;x<n;x++) {
        for (y=0;y<AST_MAX_FDS;y++) {
            if (c[x]->fds[y] > -1) {
                if ((res = ast_fdisset(pfds, c[x]->fds[y], max, &spoint))) {
                    ....
                }
            }
        }
    }
 
 
Both can be reduced to O(N), trivially for ast_waitfor_n_fd() :

    for (x = 0; x < y; x++) {
        if ((res = pfds[x].revents) != 0) {
            winner = pfds[x].fd;
            if (exception)
                *exception = (res & POLLPRI) ? -1 : 0;
        }
    }


and only slightly harder for ast_waitfor_nandfds() by keeping track,
while building the pfds[] array, which channel is associated to each entry.
 
I have only one question though - if multiple channels are ready,
ast_waitfor_nandfds() seems to privilege the highest numbered one.
Do I have to preserve this behaviour or it does not matter who
the winner is ? I do not see it specified elsewhere.
 
        cheers
        luigi