关于ios中块的混淆

问题描述：

例如说我有完成BOOL块：

-(void) addNewSubGoalToLocalDatabaseAndParse:(CompletionBlock)cb 
{ 
    SubGoal* subGoalToAdd = [SubGoal new]; 
    subGoalToAdd.name = subGoalName; 
    subGoalToAdd.parentGoal = goal; 
    Score* scoreToAdd = [Score new]; 
    scoreToAdd.score = 0; 
    scoreToAdd.subgoal = subGoalToAdd; 
    [subGoalToAdd pinInBackgroundWithBlock:^(BOOL succeeded, NSError *error) { 
    if (succeeded) { 
     [scoreToAdd pinInBackgroundWithBlock:^(BOOL succeeded, NSError *error){ 
      if (succeeded) { 
       NSLog(@"NEW SUBGOALS AND SCORES ADDED"); 
       [subGoalToAdd saveInBackgroundWithBlock:^(BOOL succeeded, NSError *error) { 
        if (succeeded) { 
         [scoreToAdd saveInBackgroundWithBlock:^(BOOL succeeded, NSError *error) { 
          if (succeeded) { 
           NSLog(@"NEW SUBGOALS AND SCORES ADDED"); 
           cb(true); 
          } 
         }]; 
        } 
       }]; 
      } 
     }]; 
    } 
}]; 
}

现在，这里我发送真正到了块时，所有的操作都completed.Say如果我第一个成功发送后，真正到了块它会退出整个块还是继续异步运行代码？

这看起来相当复杂。您应该重新设计代码，以在后台线程中顺序执行这4件事，然后调用完成处理程序。 – Droppy

答

在你的函数，即回调，不return.You方法将返回first.Because有在你的代码所以很多异步代码，如果

-(void) addNewSubGoalToLocalDatabaseAndParse:(CompletionBlock)cb 
{ 
SubGoal* subGoalToAdd = [SubGoal new]; 
subGoalToAdd.name = subGoalName; 
subGoalToAdd.parentGoal = goal; 
Score* scoreToAdd = [Score new]; 
scoreToAdd.score = 0; 
scoreToAdd.subgoal = subGoalToAdd; 
[subGoalToAdd pinInBackgroundWithBlock:^(BOOL succeeded, NSError *error) { 
if (succeeded) { 
    cb(true); 
    [scoreToAdd pinInBackgroundWithBlock:^(BOOL succeeded, NSError *error){ 
     if (succeeded) { 
      NSLog(@"NEW SUBGOALS AND SCORES ADDED"); 
      [subGoalToAdd saveInBackgroundWithBlock:^(BOOL succeeded, NSError *error) { 
       if (succeeded) { 
        [scoreToAdd saveInBackgroundWithBlock:^(BOOL succeeded, NSError *error) { 
         if (succeeded) { 
          NSLog(@"NEW SUBGOALS AND SCORES ADDED"); 
          cb(true); 
         } 
        }]; 
       } 
      }]; 
     } 
    }]; 
} 
}]; 
}

您将有两个回调。

我与那些代码

-(void)testFunction:(CALLBACK)callback{ 
dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{ 

    sleep(2); 
    callback(@"1"); 
    dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{ 
     sleep(2); 
     callback(@"2"); 
    }); 

}); 
}

测试然后调用

[self testFunction:^(NSString *str) { 
    NSLog(@"%@",str); 
}];

这将输出

2015-05-29 15:57:24.945 OCTest[5009:261291] 1 
2015-05-29 15:57:26.950 OCTest[5009:261291] 2

非常感谢很多人认真的解释..这清除了我的概念.. – soldiershin

答

是的，如果您在第一个成功块中发送cb(true);，它将退出整个块。

没有代码返回，直到它到达其作用域的末尾或使用'return'语句（忽略异常）。为什么块有什么不同？ – Droppy

关于ios中块的混淆

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