是否可以在不“窃取”输出的情况下捕获已启动过程的输出？

Question

许多其他问题的答案告诉您，您可以使用类似于以下代码的代码捕获流程的输出：

void UnsortedType::SplitLists(UnsortedType list,
                              ItemType item,
                              UnsortedType& list1,
                              UnsortedType& list2){
    ItemType whichItem;

    int numItems = list.GetLength();

    //Loop through all items in the list
    for(int i = 0; i < numItems; i ++){
        whichItem = list.GetNextItem();
        try{
            switch(whichItem.ComparedTo(item)){
                case LESS:
                case EQUAL:
                    if(list1->isFull()){//Error thrown on this line
                        throw std::string("List1 is full.");
                        return;
                    }
                    //add item to list1
                    list1->PutItem(whichItem);//Error thrown on this line
                break;
                case GREATER:
                    if(list2->isFull()){//Error thrown on this line
                        throw std::string("List2 is full.");
                        return;
                    }
                    //add item to list2
                    list2->PutItem(whichItem);//Error thrown on this line
                break;

            }
        }
        catch(std::string e){
            std::cout << e << std::endl;
        }
    }
};

有许多变体或多或少实现相同的功能，但是它们都重定向标准输出/错误-有没有一种方法可以捕获它们，同时仍将它们显示在cmd窗口中？