Question

如何从C ++中的向量中删除节点这段代码甚至没有编译

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;

}

行失败

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

编译器抱怨不平等分配。

Answer 1

我认为你很难做到这一点。您可以使用带有lambda表达式的标准库中的std::remove_if来代替手写循环来迭代容器。

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;
}

上述功能将删除符合条件的所有元素。如果您只想删除第一个匹配元素，那么这只是一个简单的更改。

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;
}

如何从C ++中的向量中删除节点

1 个答案: