CODEWITHSUNDEEP
c++vector
Ahmed
Ahmed

Reputation: 3458

How to return a 2 dimensional vector?

I have a function that creates a 2D vector

void generate(int n)
{
   vector< vector<int> > V (n, vector<int>(1 << n , 0 ));
   .......

}//n is used to determine the size of vector

Now, I need to return the created vector to use it in another function .If I did 

return V ;

it will be wrong because V is a local variable but I can't define V outside the function because this functions defines the size of V . What should I do ?

Upvotes: 2

Views: 23500

Answers (5)

Smit Bhatt
Smit Bhatt

Reputation: 11

Just like we do for other,we can do that here also.See my code for better explanation I have given my code for matrix multiplication in which i have made a function for it and returned a matrix as a 2D vector. `

#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> mul(vector<vector<int>> a,vector<vector<int>> b)
{
    int n=a.size();
    int k=a[0].size();
    int m=b[0].size();
    vector<vector<int>> c(n,vector<int> (m,0));
    
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            c[i][j]=0;
            for(int m=0;m<k;m++)
            {
                c[i][j]+=(a[i][m]*b[m][j]);
            }
        }
    }
    
    return c;
}
int main() {
    vector<vector<int>> l={{1,2}};
    vector<vector<int>> r={{1,2,3},{4,5,6}};
     vector<vector<int>> m;
    m=mul(l,r);
    
    for(int i=0;i<m.size();i++)
    {
        for(int j=0;j<m[i].size();j++)
        {
            cout<<m[i][j]<<" ";
        }
        cout<<endl;
    }
    
    return 0;
}

`

Upvotes: 1

cybevnm
cybevnm

Reputation: 2566

As alternative to other answers you can return pointer to heap allocated vector. For securing from memory leaks you can use "move constructor" idiom. 

typedef auto_ptr<vector<vector<int> > > vector_ptr;
vector_ptr generate(int n)
{
   vector_ptr V(new vector<vector<int> >(n, vector<int>(1 << n , 0 )));
   return V;
}

// ...

vector_ptr V(generate(some_number));

This idiom can be used for any "heavy" object. If you whant to prolonge lifetime of returned object you can assign it to shared_ptr (as example boost::shared_ptr). But I think it is better return copy whenever it is possible, for minimizing complexity of your code.

Upvotes: 0

mtvec
mtvec

Reputation: 18316

vector<vector<int> > generate(int n)
{
    vector<vector<int> > v(n, vector<int>(1 << n, 0));
    //...
    return v;
}

The return value is a copy of the local variable v so there is no problem at all.

If you're concerned about copying the vector, maybe you could do something like this:

void generate(int n, vector<vector<int> >& v)
{
    v.clear(); //not necessary if you're sure it's empty
    v.resize(n, vector<int>(1 << n, 0));
    //...
}

Upvotes: 6

user229044
user229044

Reputation: 239270

You can return V with no issues - it will return a copy of the local variable. Issues only arise when you return a reference or pointer to a variable with local scope; when the function ends, the local variable falls out of scope and is destroyed and the reference/pointer is no longer valid.

Alternatively, you can accept a reference to a vector as your argument, write to it and return void:

void generate(int n, std::vector< std::vector<int> >& vec) {
    vec.resize(n, std::vector<int>(1 << n, 0));
}

int main() {
    std::vector< std::vector<int> > v;
    generate(10, v);
}

This is faster than returning a copy of the local member, which can be expensive for large objects such as multi-dimensional vectors.

Upvotes: 11

Dustin Getz
Dustin Getz

Reputation: 21801

if the vector is small, just return it by value, it will be copied. if the vector is large, the caller can pass the dest vector by reference.

Upvotes: 0

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