Concatenate empty vector with two other vectors

[Ganado]

I am trying to concatenate two std::vectors together.
I'm using http://stackoverflow.com/a/201729 as a basis for my code.

Foo operator+(const Foo& lhs, const Foo& rhs)
{
    Foo ret; // creates object initally with empty m_vector.
    
    // Concatenate m_vector
    ret.m_vector.insert(ret.m_vector.end(),
                        lhs.m_vector.begin(),
                        lhs.m_vector.end() );
    ret.m_vector.insert(ret.m_vector.end(),
                        rhs.m_vector.begin(),
                        rhs.m_vector.end() );

Is there a way to do this efficiently with just one insert call since I know that ret initially has an empty vector? I should probably learn more about iterators, so sorry if this is an obvious question.

Or would it be better to have a constructor that copies lhs's m_vector into ret's m_vector, and then call insert() to insert the rhs vector?

[anup30]

#include <iostream>
#include <vector>

template<typename T>
std::vector<T> addVec(const std::vector<T>& v1, const std::vector<T>& v2)
{
	std::vector<T> vec;
	vec = v1;
	vec.insert(vec.end(), v2.begin(), v2.end());
    return vec;
}


int main(){	

	std::vector<int> v1{1,2,3,4,5};
	std::vector<int> v2{6,7,8};
	std::vector<int> v3;
	
	v3 = addVec(v1, v2);
	
	for(auto x:v3){
		std::cout << x << " ";
	}
	  
return 0;
}

[anup30]

#include <iostream>
#include <vector>

template<typename T>
std::vector<T> operator+(const std::vector<T>& v1, const std::vector<T>& v2)
{
	std::vector<T> vec;
	vec = v1;
	vec.insert(vec.end(), v2.begin(), v2.end());
    return vec;
}

int main(){	

	std::vector<int> v1{1,2,3,4,5};
	std::vector<int> v2{6,7,8};
	std::vector<int> v3;
	
	v3 = v1+v2;
	
	for(auto x:v3){
		std::cout << x << " ";
	}
	  
return 0;
}

EDIT:

template<typename T>
std::vector<T> operator+( std::vector<T> v1, const std::vector<T>& v2)
{
	v1.insert(v1.end(), v2.begin(), v2.end());
	return v1;
}

[Ganado]

Ah right... you just copy construct it and then insert the right-hand side. Thanks for the example.