NSTimer因访问不良而崩溃

Question

我有以下方法来更新显示简单时间的标签

-(void)updateTimeLabel:(NSTimer *)timer{
    NSInteger secondsSinceStart = (NSInteger)[[NSDate date] timeIntervalSinceDate:_startTime];

    NSInteger seconds = secondsSinceStart % 60;
    NSInteger minutes = (secondsSinceStart / 60) % 60;
    NSInteger hours = secondsSinceStart / (60 * 60);
    NSString *result = nil;
    if (hours > 0) {
        result = [NSString stringWithFormat:@"%02d:%02d:%02d", hours, minutes, seconds];
    }
    else {
    result = [NSString stringWithFormat:@"%02d:%02d", minutes, seconds];
    }

    _totalTime = result;
    _totalTimeLabel.text = result;
}

然后我将其称为按钮操作:

-(IBAction) startTimer{
    _startTime = [NSDate date];
    _walkRouteTimer = [NSTimer scheduledTimerWithTimeInterval:1 target:self selector:@selector(updateTimeLabel:) userInfo:nil repeats:YES];
    [_walkRouteTimer fire];
}

但是,当我运行应用程序时,我得到一个错误的访问错误,应用程序崩溃,任何人都可以帮我这个吗？

提前致谢

Answer 1

你在用ARC吗？如果没有,_startTime = [NSDate date];这一行将导致您的问题.[NSDate date]返回一个autorelease对象,如果你没有使用ARC(或使用ARC但声明_startTime为弱),_startTime将不会保留它.

如果是这样,请尝试为其添加保留

-(IBAction) startTimer{
    //_startTime = [NSDate date]
    _startTime = [[NSDate date] retain];
    _walkRouteTimer = [NSTimer scheduledTimerWithTimeInterval:1 target:self selector:@selector(updateTimeLabel:) userInfo:nil repeats:YES];
    [_walkRouteTimer fire];
}

当你完成计时后,在打电话后[_walkRouteTimer invalidate],打电话[_startTime release].

或者甚至更简单,如果你使用属性startTime并将其声明为保留.只需使用点符号:

-(IBAction) startTimer{
    //_startTime = [NSDate date]
    self.startTime = [NSDate date];
    _walkRouteTimer = [NSTimer scheduledTimerWithTimeInterval:1 target:self selector:@selector(updateTimeLabel:) userInfo:nil repeats:YES];
    [_walkRouteTimer fire];
}
...
//After [_walkRouteTimer invalidate]
self.startTime = nil;