最後一個模塊返回時的觸發方法

Question

當我確信兩個模塊代碼都已返回時，我該如何觸發一種方法？是這樣的...

// Retrieve Messages Array from Parse 
[ParseManager retrieveAllMessagesForShredderUser:(ShredderUser *)[PFUser currentUser] withCompletionBlock:^(BOOL success, NSError *error, NSArray *objects){ 
     self.messagesArray = objects; 
    }]; 

// Retrieve MessagesPermissions Array from Parse 
[ParseManager retrieveAllMessagePermissionsForShredderUser:(ShredderUser *)[PFUser currentUser] withCompletionBlock:^(BOOL success, NSError *error, NSArray *objects){ 
     self.messagePermissionsArray = objects; 
    }]; 

-(void)methodToRunWhenBothBlocksHaveReturned{ 
} 

Answer 1

如果你能保證塊將在同一個線程（即UI線程）上執行，那麼選擇很簡單，使用__block變量。

-(void)yourMethod { 
    __block int count = 0; 
    [ParseManager retrieveAllMessagesForShredderUser:(ShredderUser *)[PFUser currentUser] withCompletionBlock:^(BOOL success, NSError *error, NSArray *objects){ 
     self.messagesArray = objects; 
     count++; 
     if (count == 2) { 
      [self methodToRunWhenBothBlocksHaveReturned]; 
     } 
    }]; 

    [ParseManager retrieveAllMessagePermissionsForShredderUser:(ShredderUser *)[PFUser currentUser] withCompletionBlock:^(BOOL success, NSError *error, NSArray *objects){ 
     self.messagePermissionsArray = objects; 
     count++; 
     if (count == 2) { 
      [self methodToRunWhenBothBlocksHaveReturned]; 
     } 
    }]; 
} 

-(void)methodToRunWhenBothBlocksHaveReturned{ 
} 

如果不具有相同的線程保證，你可以使用一個lock，以確保該變量（和比較2）的增量將是原子。