使stringWithContentsOfURL异步 - 它安全吗？

Question

我试图通过从后台线程同步运行它来使[NSString stringWithContentsOfURL:encoding:error:]异步:

__block NSString *result;
dispatch_queue_t currentQueue = dispatch_get_current_queue();

void (^doneBlock)(void) = ^{
    printf("done! %s",[result UTF8String]);
};

dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT,
                                         (unsigned long)NULL), ^(void) {
    result = [NSString stringWithContentsOfURL:[NSURL URLWithString:@"http://www.google.com/"] encoding:NSUTF8StringEncoding error:nil];
    dispatch_sync(currentQueue, ^{
        doneBlock();
    });
});

它的工作正常,最重要的是,它的异步.

我的问题是这样做是否安全,或者是否存在任何线程问题等？

提前致谢 :)

Answer 1

这应该是安全的,但为什么重新发明轮子？

NSURLRequest *req = [NSURLRequest requestWithURL:[NSURL URLWithString:@"http://www.google.com"]];
[NSURLConnection sendAsynchronousRequest:req queue:[NSOperationQueue mainQueue] completionHandler:^(NSURLResponse *response, NSData *data, NSError *error) {
    NSString *result = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
    // etc
}];