如何從一個向量在C++

Question

如何在C++ 矢量刪除節點刪除節點此代碼甚至不編譯

typedef vector<simple_node> container; 
//====================================== 
// remove an item from the queue that matches what we arelady have 
container parser::removeFromQueue(container local_container, simple_node *node_to_remove) 
{ 
    for (auto i = local_container.begin(); i != local_container.end(); i++) 
    { 
     if ((i->toy == node_to_remove->toy) && 
      (i->type == node_to_remove->type)) 
      { 
       local_container.erase(remove(local_container.begin(), local_container.end(), i), local_container.end()); 
       break; 
      } 
    } 
    return local_container; 

} 

它未能就行了

local_container.erase(remove(local_container.begin(), local_container.end(), i), local_container.end()); 

編譯器抱怨查詢分配。

Answer 1

我覺得你這樣做是很難的。您可以使用帶有lambda表達式的標準庫中的std::remove_if來代替手寫循環遍歷容器。

container parser::removeFromQueue(container local_container, simple_node *node_to_remove) 
{ 
    auto pred = [node_to_remove](const simple_node &node) { 
     return node.toy == node_to_remove->toy && node.type == node_to_remove->type; 
    }; 
    local_container.erase(std::remove_if(local_container.begin(), local_container.end(), pred), local_container.end()); 
    return local_container; 
} 

上述功能將刪除滿足條件所有元素。如果您只想刪除第一個匹配元素，則這是一個簡單的更改。

container parser::removeFromQueue(container local_container, simple_node *node_to_remove) 
{ 
    auto pred = [node_to_remove](const simple_node &node) { 
     return node.toy == node_to_remove->toy && node.type == node_to_remove->type; 
    }; 
    auto itr = std::find_if(local_container.begin(), local_container.end(), pred); 
    if (itr != local_container.end()) { 
     local_container.erase(itr); 
    } 
    return local_container; 
}