iOS-APP崩溃,APP不退出应用怎么实现

简介

当APP遇到崩溃时，APP会闪退，那要怎么实现不让他闪退，比如弹出一个提示框

1231665759176_.pic.jpg

这就要用到runloop技术了

强行让runloop继续运行

上代码

@implementation ViewController

- (void)viewDidLoad {
    [super viewDidLoad];
    [self installUncaughtSignalExceptionHandler];
}

/// 获取崩溃
- (void)installUncaughtSignalExceptionHandler{
    NSSetUncaughtExceptionHandler(&LGExceptionHandlers);
}

void LGExceptionHandlers(NSException *exception) {
    UIAlertController *alert = [UIAlertController alertControllerWithTitle:@"Crash" message:@"The App has crashed and will attempt to send a crash report" preferredStyle:UIAlertControllerStyleAlert];
    UIAlertAction *sure = [UIAlertAction actionWithTitle:@"OK" style:UIAlertActionStyleDestructive handler:nil];
    [alert addAction:sure];
    [[UIViewController hll_topVC] presentViewController:alert animated:YES completion:nil];
    
    // 本次异常处理， 保证弹框能正常弹出
    CFRunLoopRef runloop = CFRunLoopGetCurrent();
    CFArrayRef   allMode = CFRunLoopCopyAllModes(runloop);
    while (YES) {
    for (NSString *mode in (__bridge NSArray *)allMode) {
            CFRunLoopRunInMode((CFStringRef)mode, 0.0001, false);
        }
    }
    CFRelease(allMode);
}

- (void)touchesBegan:(NSSet *)touches withEvent:(UIEvent *)event {
    /// 崩溃代码
    @[@""][2];
}