Without registering calls (no password for acc_add) calls can be repeated
without restarting the stack
With registering after adding the account a call must be made immediately,
then a following call
has no sound.. somehow the rtp fails to make it to the sound device
see all the options tried , ha ha
t,z will repeat t,z t,z and call out
t,r will register ,then y will fail.. (all ok but no sound from rtp to the
sound device)
ha ha
however without registering (no password) for add_acc
repeated call hangup .. no need to reload the stack..
the high level api seems stuck.. somehow
and all the samples found seem to bear that out.
for (;;) {
char option[10];
puts("Press 'h' to hangup all calls, 'q' to quit");
if (fgets(option, sizeof(option), stdin) == NULL) {
puts("EOF while reading stdin, will quit now..");
break;
}
if (option[0] == 'q')
break;
if (option[0] == 'h')
{ pjsua_call_hangup_all();
}
if(option[0] == 'x')
{ pjsua_acc_del(acc_id);
}
if(option[0] == 'y')
{
// status = pjsua_acc_add(&cfg1, PJ_TRUE, &acc_id);
status = pjsua_call_make_call(acc_id, &uri, 0, NULL, NULL, NULL);}
if(option[0] == 'r')
status = pjsua_acc_add(&cfg1, PJ_TRUE, &acc_id);
// startpj();
if(option[0] == 's')
{ startpj();
status = pjsua_acc_add(&cfg1, PJ_TRUE, &acc_id);
}
if(option[0] == 't')
{ startpj();
// status = pjsua_acc_add(&cfg1, PJ_TRUE, &acc_id);
}
if(option[0] == 'u')
{ startpj();
status = pjsua_acc_add(&cfg1, PJ_TRUE, &acc_id);
status = pjsua_call_make_call(acc_id, &uri, 0, NULL, NULL, NULL);
}
if(option[0] == 'v')
{
status = pjsua_acc_add(&cfg1, PJ_TRUE, &acc_id);
status = pjsua_call_make_call(acc_id, &uri, 0, NULL, NULL, NULL);}
if(option[0] == 'w')
{ startpj();
status = pjsua_acc_add(&cfg1, PJ_TRUE, &acc_id);
}
if(option[0] == 'z')
{
status = pjsua_acc_add(&cfg1, PJ_TRUE, &acc_id);
status = pjsua_call_make_call(acc_id, &uri, 0, NULL, NULL, NULL);
}
--
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