程序在运行时收到信号SIGABRT

时间:2011-12-06 15:50:16

标签: xcode4

当我运行该程序时,它总是显示程序接收信号SIGABRT。下面是我修改的代码,尤其是AVAudioPlayer。我的代码有什么问题吗?问题是什么?这是断点:int retVal = UIApplicationMain(argc, argv, nil, nil);

-(IBAction) start {
        {
         self.playBgMusic.enabled = YES;
        [self.player play];
        }   
    }

    -(IBAction) stop {
        {
            self.playBgMusic.enabled = NO;
            [self.player stop];
        }   
    }

    -(void) audioPlayerDidFinishPlaying:(AVAudioPlayer *)player successfully:(BOOL)completed 
    {
        if(completed == YES){
            self.playBgMusic.enabled = YES;
        }
    }


    - (void)viewDidLoad
    {

        NSString *path = [[NSBundle mainBundle] pathForResource:@"music" ofType:@"mp3"];
        self.player=[[AVAudioPlayer alloc] initWithContentsOfURL:[NSURL fileURLWithPath:path] error:NULL]; 
        player.delegate = self;
        [player play];
        player.numberOfLoops = -1;

        [super viewDidLoad];
    }


    - (void)viewDidUnload
    {  
        [player stop];
        [super viewDidUnload];

    }

1 个答案:

答案 0 :(得分:0)

您可能需要保留self.player,但您的代码中并不清楚这一点。如果您已将player(正确)定义为:

@property (nonatomic, retain) AVAudioPlayer *player;

...然后你有一个不同的问题。