[c++] Why is it OK to return a 'vector' from a function?

Please consider this code. I have seen this type of code several times. words is a local vector. How is it possible to return it from a function?

Can we guarantee it will not die?

 std::vector<std::string> read_file(const std::string& path)
 {
    std::ifstream file("E:\\names.txt");

    if (!file.is_open())
    {
        std::cerr << "Unable to open file" << "\n";
        std::exit(-1);
    }

    std::vector<string> words;//this vector will be returned
    std::string token;

    while (std::getline(file, token, ','))
    {
        words.push_back(token);
    }

    return words;
}

I think you are referring to the problem in C (and C++) that returning an array from a function isn't allowed (or at least won't work as expected) - this is because the array return will (if you write it in the simple form) return a pointer to the actual array on the stack, which is then promptly removed when the function returns.

But in this case, it works, because the std::vector is a class, and classes, like structs, can (and will) be copied to the callers context. [Actually, most compilers will optimise out this particular type of copy using something called "Return Value Optimisation", specifically introduced to avoid copying large objects when they are returned from a function, but that's an optimisation, and from a programmers perspective, it will behave as if the assignment constructor was called for the object]

As long as you don't return a pointer or a reference to something that is within the function returning, you are fine.

I do not agree and do not recommend to return a vector:

vector <double> vectorial(vector <double> a, vector <double> b)
{
    vector <double> c{ a[1] * b[2] - b[1] * a[2], -a[0] * b[2] + b[0] * a[2], a[0] * b[1] - b[0] * a[1] };
    return c;
}

This is much faster:

void vectorial(vector <double> a, vector <double> b, vector <double> &c)
{
    c[0] = a[1] * b[2] - b[1] * a[2]; c[1] = -a[0] * b[2] + b[0] * a[2]; c[2] = a[0] * b[1] - b[0] * a[1];
}

I tested on Visual Studio 2017 with the following results in release mode:

8.01 MOPs by reference
5.09 MOPs returning vector

In debug mode, things are much worse:

0.053 MOPS by reference
0.034 MOPs by return vector

To well understand the behaviour, you can run this code:

#include <iostream>

class MyClass
{
  public:
    MyClass() { std::cout << "run constructor MyClass::MyClass()" << std::endl; }
    ~MyClass() { std::cout << "run destructor MyClass::~MyClass()" << std::endl; }
    MyClass(const MyClass& x) { std::cout << "run copy constructor MyClass::MyClass(const MyClass&)" << std::endl; }
    MyClass& operator = (const MyClass& x) { std::cout << "run assignation MyClass::operator=(const MyClass&)" << std::endl; }
};

MyClass my_function()
{
  std::cout << "run my_function()" << std::endl;
  MyClass a;
  std::cout << "my_function is going to return a..." << std::endl;
  return a;
}

int main(int argc, char** argv)
{
  MyClass b = my_function();

  MyClass c;
  c = my_function();

  return 0;
}

The output is the following:

run my_function()
run constructor MyClass::MyClass()
my_function is going to return a...
run constructor MyClass::MyClass()
run my_function()
run constructor MyClass::MyClass()
my_function is going to return a...
run assignation MyClass::operator=(const MyClass&)
run destructor MyClass::~MyClass()
run destructor MyClass::~MyClass()
run destructor MyClass::~MyClass()

Note that this example was provided in C++03 context, it could be improved for C++ >= 11

This is actually a failure of design. You shouldn't be using a return value for anything not a primitive for anything that is not relatively trivial.

The ideal solution should be implemented through a return parameter with a decision on reference/pointer and the proper use of a "const\'y\'ness" as a descriptor.

On top of this, you should realise that the label on an array in C and C++ is effectively a pointer and its subscription are effectively an offset or an addition symbol.

So the label or ptr array_ptr === array label thus returning foo[offset] is really saying return element at memory pointer location foo + offset of type return type.

Pre C++11:

The function will not return the local variable, but rather a copy of it. Your compiler might however perform an optimization where no actual copy action is made.

See this question & answer for further details.

C++11:

The function will move the value. See this answer for further details.