如何从一个向量在C++

Question

如何在C++ 矢量删除节点删除节点此代码甚至不编译

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()); 

编译器抱怨查询分配。

Answer 1

我觉得你这样做是很难的。您可以使用带有lambda表达式的标准库中的std::remove_if来代替手写循环遍历容器。

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; 
} 

上述功能将删除满足条件所有元素。如果您只想删除第一个匹配元素，则这是一个简单的更改。

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; 
}