ios STPopup通过在后台点击来解雇

Question

我已经使用STPopup library在iOS应用中使用Objective-c创建了一个弹出窗口，所以一切都很好。

现在我想通过点击后台来解除弹出，但我的代码不起作用并收到错误backgroundViewDidTap]: unrecognized selector sent to instance 0x15780b7b0

ViewController.m

- (void)showPopupDaudit{
    STPopupController *popupController = [[STPopupController alloc] initWithRootViewController:[POPViewController new]];
    [STPopupNavigationBar appearance].tintColor = MP_HEX_RGB(MAINCOLOR);;

    popupController.containerView.layer.cornerRadius = 4;
    if (NSClassFromString(@"UIBlurEffect")) {
        UIBlurEffect *blurEffect = [UIBlurEffect effectWithStyle:UIBlurEffectStyleDark];
        popupController.backgroundView = [[UIVisualEffectView alloc] initWithEffect:blurEffect];
    }
    [popupController.backgroundView addGestureRecognizer:[[UITapGestureRecognizer alloc] initWithTarget:popupController
                                                                                                 action:@selector(backgroundViewDidTap)]];
    [popupController presentInViewController:self];
}

POPUPController.m

- (IBAction)backgroundViewDidTap
{
    [self.popupController dismiss];
}

有什么想法吗？