CODEWITHSUNDEEP
c++referenceobject-lifetime
Richard Dally
Richard Dally

Reputation: 1450

Lifetime of const references bound to destroyed stack variable

I'm wondering if it's by chance a pointer to const reference bound to a destroyed stack variable can be working.
I read const reference lifetime is extended on rvalues, so this is "normal" const reference works but at the end of the ctor of Storage ref should be destroyed, isn't it ?
Does const reference lifetime is also extended because I retrieved it's address in a pointer or is this pure luck ?

Live example

#include <iostream>

class Storage
{
public:
    Storage(const int& ref)
    {
        p = &ref;
    }

    const int* Get() const
    {
        return p;
    }

private:    
    const int* p;
};

int main()
{
    Storage* s = nullptr;

    {
        int someValue = 42;
        std::cout << &someValue << std::endl;
        s = new Storage(someValue);
    }

    const int* p = s->Get();
    std::cout << p << std::endl;
    std::cout << *p << std::endl;
}

It's also working with a struct Live example

#include <iostream>

struct Dummy
{   
    int value;
};

class Storage
{
public:
    Storage(const Dummy& ref)
    {
        p = &ref; // Get address of const reference
    }

    const Dummy* Get() const
    {
        return p;
    }

private:    
    const Dummy* p;
};

int main()
{
    Storage* s = nullptr;

    {
        Dummy dummy;
        dummy.value = 42;
        std::cout << &dummy << std::endl;
        s = new Storage(dummy);
    }

    const Dummy* p = s->Get();
    std::cout << p << std::endl;
    std::cout << p->value << std::endl;
}

Upvotes: 4

Views: 153

Answers (2)

Dave Emberton
Dave Emberton

Reputation: 86

The compiler probably assigns the stack space for all the local variables at the start, so someValue is still in the same place in memory even after it goes out of scope. Of course a different compiler might do something different, and it could get overwritten by a subsequent local variable.

For your second example if you wrote a ~Dummy(){ value=0; }, then it wouldn't work, proving that the lifetime of the dummy object hasn't been extended.

Upvotes: 0

Cory Kramer
Cory Kramer

Reputation: 117876

Your s variable indeed has a dangling pointer as someValue has fallen out of scope. Your code therefore exhibits undefined behavior.

Your comment about "const reference lifetime is extended on rvalues" is true in some circumstances, but someValue is an lvalue.

Upvotes: 9

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