C++ 编程 : Error when assigning the iterator to vector

Question

    vector<vector<string> > vvs;
    vector<string> vs;

    vs.push_back("r1-c1");
    vs.push_back("r1-c2");
    vs.push_back("r1-c3");

    vvs.push_back(vs);

    for (vector<vector<string> >::iterator vvsi = vvs.begin(); vvsi != vvs.end(); vvsi++) {
        vector<string> vec_str = *vvsi;
        for (vector<string>::iterator vsi = vec_str.begin(); vsi != vec_str.end(); vsi++) {
            cout << *vsi << ", ";
        }
        cout << "\n";
    }

在上面的 C++ 代码中，为了避免 vector(vector vec_str = *vvsi) 的复制，我尝试了下面的代码

vector<string> *vec_str = vvsi.base(); //Working. which (returns a const pointer&)     
vector<string> *vec_str = &(*vvsi); //Working. Assigning the address

但是

vector<string> *vec_str = vvsi; //Error. Not able to assign

错误

(build error : cannot convert 'std::vector<std::vector<std::basic_string<char> > >::iterator to 'std::vector<std::basic_string<char> >*' in initialization)

以整数为例

    int a=10;
    int *b = &a; //working. Assigning address
    int *c = &(*b); //working. Assigning address
    int *d = b; //working. Assigning address
    *c=11;
    std::cout << a<<"\n";
    *d=12;
    std::cout << a<<"\n";

在 vector 的情况下，为什么在赋值时构建错误(无法从 C++ 迭代器文档中理解)？

Answer 1

迭代器不是指针。它的接口(interface)被用作指针，但它肯定不是指针。

但是，如果您只是想遍历父 vector 中每个 vector 的元素，则无需将其分配给临时 vector 或指针，迭代器本身就足够了:

for (vector<vector<string> >::iterator vvsi = vvs.begin(); vvsi != vvs.end(); ++vvsi) {
    for (vector<string>::iterator vsi = vvsi->begin(); vsi != vvsi->end(); ++vsi) {
        cout << *vsi << ", ";
    }
    cout << "\n";
}

此外，由于您将 c++11 作为问题中的标签，我认为您可能对更现代的表达方式感兴趣:

for(auto& vst : vvs){
    for(auto& st : vst){
        cout << st << ", ";
    }
    cout << endl;
}