向量上的push_back问题 <vector<int> &gt;

Question

Dear All,
I used a vector of vector, say vector<vector<int> > no_1_2 to store elements of two vector<int> containers, say no1 & no2. so for eg i would say, no1={2,5,7,10,3} and no2={21,34,15}.

I wanted to store these 2 vectors in one container called no_1_2 and I used no_1_2[0].push_back(no1.at(i)) and no_1_2[1].push_back(no2.at(j)) within two 'for' loops. But, i got error. is there anyway to solve this?

Can't i push_back vector of vector? any help please.. thanks

sorry here is part of my code, //some classes and some codes are here

vector< vector<int> > final_list(int code1,int code2){
   vector<vector<int> > no_1_2;
   vector<int> in;
   vector<int> out;
   for (int foo=0;foo<size();foo++){
       int a_point=foo;
       if (at(a_point).Code()==code1){
           vector<int> closer_points;
           closer_points=at(foo).Closers();
           for (int fee=0;fee<closer_points.size();fee++){
               int a_neb_point=closer_points.at(fee);
               if (at(a_neb_point).Code()==code2){
                    in.push_back(a_point);  
                    out.push_back(a_neb_point);  
                    }
               }
           }
       }
   /remove duplicates // above in and out vectors containing some values repeatedly, so, i remove the duplicates here
   for(vector<int>::iterator i=in.begin();i!=in.end();i++){
     sort(in.begin(),in.end());
     in.erase(unique(in.begin(),in.end()),in.end()); 
     no_1_2[0].push_back(*i);
     }
   for(vector<int>::iterator o=out.begin();o!=out.end();o++){
     sort(out.begin(),out.end());
     out.erase(unique(out.begin(),out.end()),out.end());
     no_1_2[1].push_back(*o);
     }
   return no_1_2;
   }

int main (){
// some code

vector< vector <int> > in_out=mylist.final_list(34,1); //here i just tried for code    values for (34, 1), like that i have many sets
}

Answer 1

Without seeing your actual code and errors, let me guess:

You didn't call no_1_2.resize(2) before you accessed no_1_2[0] or no_1_2[1]

EDIT: After seeing code, the above's no longer guesswork:

You cannot do no_1_2[0].push_back(*i); - no_1_2 is empty and you're trying to access no_1_2[0] . You will need to resize it before the loop.

EDIT2:

Move your sort/unique/erase outside the loops, and use assign or operator=:

sort(in.begin(),in.end());
in.erase(unique(in.begin(),in.end()),in.end()); 
no_1_2[0].assign(in.begin(), in.end());
// OR: no_1_2[0] = in;

Answer 2

std::vector::at returns a reference to an element at position n in the vector.

std::vector<int> vec(2);
vec[0] = 0;
vec[1] = 1;

cout << vec.at(0) ;  // returns 0 but not the vector.

So, in order to return a vector, value at the position must be a vector, which is not the case either in vectors no1 , no2 you mentioned.

std::vector< std::vector<int> > no_1_2;
std::vector<int> no1(5), no2(3) ;
// Assigning values to vector of integers no1, no2

no_1_2.push_back(no1);
no_1_2.push_back(no2);

no_1_2.at(0) ; // This returns std::vector<int>

---

no_1_2[0].push_back(*i); , no_1_2[1].push_back(*o); are the errors. You need to resize the vector size because at the point of no_1_2, in,out declarations, size of the container is not mentioned.

vector<vector<int>> no_1_2 ;
vector<int> in;
vector<int> out;

And since size of no_1_2 is not mentioned, it an error to use an [] on it. However, once after the push_back ,it is valid though. Example -

vector<int> temp;
temp[0] = 10;    // Error : What is temp[0] ? size of temp is 0 and where would the `[0]` take to.
temp.push_back(10);  // Increased the size of vector
temp[0] = 30 ;       // And this is now valid.

Answer 3

I'm assuming that you haven't properly allocated the vector vectors.

vector< vector<int> > vectors; 
  vectors.resize(2); 

FWIW, you're not doing a push_back of a vector of a vector, you're doing a push_back of an int from vector no2 to vector no_1_2[1].