Avoiding copy of objects with the "return" statement

I have a very basic question in C++. How to avoid copy when returning an object ?

Here is an example :

std::vector<unsigned int> test(const unsigned int n)
{
    std::vector<unsigned int> x;
    for (unsigned int i = 0; i < n; ++i) {
        x.push_back(i);
    }
    return x;
}

As I understand how C++ works, this function will create 2 vectors : the local one (x), and the copy of x which will be returned. Is there a way to avoid the copy ? (and I don't want to return a pointer to an object, but the object itself)

What would be the syntax of that function using "move semantics" (which was stated in the comments)?

Solution 1:

There seems to be some confusion as to how the RVO (Return Value Optimization) works.

A simple example:

#include <iostream>

struct A {
    int a;
    int b;
    int c;
    int d;
};

A create(int i) {
    A a = {i, i+1, i+2, i+3 };
    std::cout << &a << "\n";
    return a;
}

int main(int argc, char*[]) {
    A a = create(argc);
    std::cout << &a << "\n";
}

And its output at ideone:

0xbf928684
0xbf928684

Surprising ?

Actually, that is the effect of RVO: the object to be returned is constructed directly in place in the caller.

How ?

Traditionally, the caller (main here) will reserve some space on the stack for the return value: the return slot; the callee (create here) is passed (somehow) the address of the return slot to copy its return value into. The callee then allocate its own space for the local variable in which it builds the result, like for any other local variable, and then copies it into the return slot upon the return statement.

RVO is triggered when the compiler deduces from the code that the variable can be constructed directly into the return slot with equivalent semantics (the as-if rule).

Note that this is such a common optimization that it is explicitly white-listed by the Standard and the compiler does not have to worry about possible side-effects of the copy (or move) constructor.

When ?

The compiler is most likely to use simple rules, such as:

// 1. works
A unnamed() { return {1, 2, 3, 4}; }

// 2. works
A unique_named() {
    A a = {1, 2, 3, 4};
    return a;
}

// 3. works
A mixed_unnamed_named(bool b) {
    if (b) { return {1, 2, 3, 4}; }

    A a = {1, 2, 3, 4};
    return a;
}

// 4. does not work
A mixed_named_unnamed(bool b) {
    A a = {1, 2, 3, 4};

    if (b) { return {4, 3, 2, 1}; }

    return a;
}

In the latter case (4), the optimization cannot be applied when A is returned because the compiler cannot build a in the return slot, as it may need it for something else (depending on the boolean condition b).

A simple rule of thumb is thus that:

RVO should be applied if no other candidate for the return slot has been declared prior to the return statement.

Solution 2:

This program can take advantage of named return value optimization (NRVO). See here: http://en.wikipedia.org/wiki/Copy_elision

In C++11 there are move constructors and assignment which are also cheap. You can read a tutorial here: http://thbecker.net/articles/rvalue_references/section_01.html

Solution 3:

Named Return Value Optimization will do the job for you since the compiler tries to eliminate redundant Copy constructor and Destructor calls while using it.

std::vector<unsigned int> test(const unsigned int n){
    std::vector<unsigned int> x;
    return x;
}
...
std::vector<unsigned int> y;
y = test(10);

with return value optimization:

1. y is created
2. x is created
3. x is assigned into y
4. x is destructed

(in case you want to try it yourself for deeper understanding, look at this example of mine)

or even better, just like Matthieu M. pointed out, if you call test within the same line where y is declared, you can also avoid construction of redundant object and redundant assignment as well (x will be constructed within memory where y will be stored):

std::vector<unsigned int> y = test(10);

check his answer for better understanding of that situation (you will also find out that this kind of optimization can not always be applied).

OR you could modify your code to pass the reference of vector to your function, which would be semantically more correct while avoiding copying:

void test(std::vector<unsigned int>& x){
    // use x.size() instead of n
    // do something with x...
}
...
std::vector<unsigned int> y;
test(y);