请求发送获取错误：[CFString release]：消息发送到已释放的实例 0x6a83e00

Question

我想检查苹果的应用程序版本，因此我

- (void)connectToCheckVersion{
NSString *url = @"http://itunes.apple.com/lookup?id=466424846";
TTURLRequest *_request = [TTURLRequest requestWithURL:url delegate:self];
_request.httpMethod = @"GET";
_request.cachePolicy = TTURLRequestCachePolicyNone;
_request.shouldHandleCookies = NO;
TTURLJSONResponse* response = [[TTURLJSONResponse alloc] init];
_request.response = response;
TT_RELEASE_SAFELY(response); 
[_request send];
}


- (void)requestDidFinishLoad:(TTURLRequest*)request {
    TTURLJSONResponse* response = request.response;
    NSDictionary* json = response.rootObject;

    NSArray *results = [json objectForKey:@"results"];
    NSString *version;
    for (NSDictionary *rawResult in results) {

        version = [rawResult objectForKey:@"version"];
    }
    NSString *currentVersion = [[NSBundle mainBundle] objectForInfoDictionaryKey:@"CFBundleShortVersionString"];
    if (version != nil && currentVersion != nil && ![version isEqualToString:currentVersion]) {

        UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"info" 
                                                        message:@"newer version" 
                                                       delegate:self 
                                              cancelButtonTitle:@"ok"
                                              otherButtonTitles:nil, nil];
        [alert show];
        [alert release];

    }

}

在 [_request send] 之后发送如下请求；将得到 [CFString release]：发送到已释放实例 0x6a83e00 的消息。我检查了此方法中的所有字符串似乎都没有问题，并且我仍然可以从远程获得正确的响应。

如果我注释掉这个 connectToCheckVersion 方法，那么就没有任何问题。 有死吗？

Answer 1

我认为您应该保留 _request 变量并将其保存为成员。
因为函数返回后会自动释放。

请求成功或失败后都必须释放它。

谢谢。