ios - 如何在 iOS 中创建应由库调用的回调

Question

我正在开发一个使用 PJSIP 库 的 VOIP_App，它是用 C 语言 编写的，该库中编写的大部分方法都是根据情况自动调用的。

有一个名为 on_incoming_call 的方法会自动调用并且用户会收到电话，我想添加一些用户交互来接听电话，我需要创建一些回调，应该在这个方法中调用，并且方法定义应该用Objective-C来写。

代码片段如下:

/* Callback called by the library upon receiving incoming call */
static void on_incoming_call(pjsua_acc_id acc_id, pjsua_call_id call_id,
                         pjsip_rx_data *rdata)
{
    pjsua_call_info ci;

    PJ_UNUSED_ARG(acc_id);
    PJ_UNUSED_ARG(rdata);

    pjsua_call_get_info(call_id, &ci);



    PJ_LOG(3,(THIS_FILE, "....\n\n\n Incoming call from %.*s!!  \n\n\n",
          (int)ci.remote_info.slen,
          ci.remote_info.ptr));

    /* Automatically answer incoming calls with 200/OK */
    pjsua_call_answer(call_id, 200, NULL, NULL);
}

Answer 1

您可以通过从 PJSIP 库中的回调发布通知来实现您的目标

static void on_incoming_call(pjsua_acc_id acc_id, pjsua_call_id call_id,
                     pjsip_rx_data *rdata){
   pjsua_call_info ci;
   NSString *callerNumber = [NSString stringWithUTF8String:ci.remote_info.ptr];

   [[NSNotificationCenter defaultCenter] postNotificationName:
     @"IncomingCall" object:callerNumber];

pjsua_call_answer(call_id, 200, NULL, NULL); //Comment out this line and instead call this in your UIViewController to accept the call on click of some button or whatever UI action you desire 

}

现在您可以在 AppDelegate 中观察到此通知

- (void)applicationDidFinishLaunching:(UIApplication *)application
{
 [[NSNotificationCenter defaultCenter] addObserver:self selector:@selector(launchIncomingVC:) name:@"IncomingCall" object:nil];
}

然后这个函数显示你的 UIViewController 接受或拒绝来电

-(void)launchIncomingVC:(NSNotification *) notif
{
    NSLog(@"LAUNCH INCOMING CALL VC");

    YourViewController *rvc = [[YourViewController alloc] init];
    UIStoryboard *storyboard = [UIStoryboard storyboardWithName:@"YourStoryBoardName" bundle:nil];
    rvc = [storyboard instantiateViewControllerWithIdentifier:@"YourViewControllerIdentifier"];

    rvc.paramCallerNumber = noti.object; //pass the caller number to show it on the incoming view

        dispatch_async(dispatch_get_main_queue(), ^{

            [self.window makeKeyAndVisible];

            UIViewController *topRootViewController = [UIApplication sharedApplication].keyWindow.rootViewController;
            while (topRootViewController.presentedViewController)
            {
                topRootViewController = topRootViewController.presentedViewController;
            }

            [topRootViewController presentViewController:incomingVC animated:YES completion:nil];
        });

}