将指针转换为迭代器 [英] convert pointer to iterator
问题描述
我有两个结构,指向对方
struct Person {
string name;
string born;
int age
Id * p_id;
};
struct Id {
string id_number;
人* p_person;
};
这些结构体存储在两个向量的ponters向量结构中,称为vec_id和vec_person。
我需要在vec_person中找到Person的函数,然后删除向量vec_id中的匹配ID。
我的问题是将p_id转换为指针。
我的代码示例:
<
std :: vector< Person *> vec_person;
std :: vector< Id *> vec_id;
vector< Person *> :: iterator lowerb = std :: lower_bound(vec_person.begin(),vec_person.end(),Peter,gt);
// gt是在其他地方定义的匹配函数
// peter是struct Person的现有实例
// lowerb是迭代器,工作正常。
vec_id.erase((* lowerb) - > p_id);
//给出错误:没有匹配的函数调用'std :: vector< Person *> :: erase(Person *&)'|
//如果我可以转换指针(*低) - > pnumber到迭代器,它会被解决(我猜)。
Thx帮助人
你不能只是将一个值(在这种情况下是指针)转换为一个迭代器。你必须搜索向量中的值并删除它。您可以使用std :: remove_if算法从范围中删除某些值。如果每个Person都链接到一个id,或者使用不同的容器,如地图,你也可以考虑不保留两个向量。
I have 2 structs, that point to each other
struct Person{
string name;
string born;
int age;
Id* p_id;
};
struct Id{
string id_number;
Person* p_person;
};
those structs are stored in two vectors of ponters to those structures called vec_id and vec_person. I need function which finds Person in vec_person, and then deletes matching Id in vector vec_id. My problem is converting p_id to pointer.
example of my code :
std::vector<Person*> vec_person;
std::vector<Id*> vec_id;
vector <Person*>::iterator lowerb=std::lower_bound (vec_person.begin(), vec_person.end(), Peter, gt);
//gt is matching function which is defined elsewhere
//peter is existing instance of struct Person
// lowerb is iterator, that works fine.
vec_id.erase((*lowerb)->p_id);
//gives error: no matching function for call to ‘std::vector<Person*>::erase(Person*&)’|
//if i can convert pointer (*low)->pnumber to iterator, it would be solved(i guess).
Thx for help guys
You can't just 'convert' from a value (pointer in this case) to an iterator. You'd have to search for the value inside the vector and remove it. You could use the std::remove_if algorithm to remove certain values from a range. You might also consider not keeping two vectors if each Person is linked to an id, or maybe using a different container, such as a map.
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