Can references cause memory leaks? [duplicate]

Question

Consider the following C++ code.

struct foo { std::string value; }

inline foo bar() { return { "42" }; }

Now imagine I have a function that uses bar() in the following way.

std::string my_func()
{
    const auto &x = bar();
    return x.value;
}

Does this leak memory Because my_func only holds a reference to x? Or does x still get cleaned up after my_func terminates?

I know this is not how references are supposed to be used. But I just realized this compiles fine and wondered what the semantics of it are.

Answer 1

But I just realized this compiles fine

Code provided should not compile, as trying to assign temporary to lvalue reference.

error: invalid initialization of non-const reference of type ‘foo&’ from an rvalue of type ‘foo’

If you fix code, by

std::string my_func()
{
    const auto &x = bar();
    return x.value;
}

then it would be fine, as const reference extends lifetime of temporary for lifetime of const reference.

Answer 2

Short answer: no.

Longer answer: In that case, the compiler will make sure, that the referenced temporary object will live to the end of the current scope. bar() returns n object by value. That will be copied into a temporary, anonymous object and the reference will then reference that temporary object.

There are other situations similar to that, where the standard has this explicit requirement: temporaries that are bound to references live until the end of the current scope is reached.