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I have a function that returns an rvalue reference.

auto function() -> int&& {
    int x{10};
    std::cout << x << std::endl; // to check the value of x
    return std::move(x);
}

Now, when I use the following code:

std::cout << function() << std::endl;
std::vector<int> v;
v.push_back(function());
std::cout << v[0] << std::endl;

This results in the following output

10
0

It looks like even though function() returns an rvalue-reference, vector is pushing back a default constructed int.

Interestingly, if I have a code like this:

auto x = function();
v.push_back(std::move(x));

This works perfectly.

It appears that if I had simply returned a local variable from the function then RVO would have done a copy elision anyway. But since I am doing an explicity std::move() I am tripping off RVO resulting in a temporary variable being returned.

Now, in v.push_back(function()) I am calling the vector::push_back(T&&) function, results in referring to a reference to a temp variable. In this case it always tend to be 0 rather than a garbage (I am guessing that this is because of optimisation turned on).

However, if I try to capture the return value of function() in a local variable it works, presumably because the value of the temp variable created in the process of return value is copied over to this local variable. This does not look like an I got lucky with an undefined behaviour.

abhijit
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1 Answers1

6

Your function returns a reference to a function-local variable, which is destroyed before anyone can use it. (This is just as invalid using rvalue references as using lvalue references.)

aschepler
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  • This is a bit confusing to understand. When I use something like `auto x = function()` I can see that x actually hold a value of 10 and not a default constructed one or garbage for that matter. I expected that std::move would transfer the ownership of the local object and hence would not automatically get destroyed. – abhijit Apr 19 '17 at 00:11
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    Nope, the lifetime of `x` ends with the function. If it seems to work sometimes, that's just Undefined Behavior for you. – aschepler Apr 19 '17 at 00:38