How C++ reference works

Question

After working 15 years in C++ I found that I don\'t understand references completely...

class TestClass
{
public:
    TestClass() : m_nData(0)
    {
    }

    T         


        

Answer 1

Your code suffers from multiple problems and ultimately won't make sense. However, let's hack through it.

1) You can only bind a temporary to a const reference, thus extending its lifetime:

const TestClass & c = TestClass();

2) Now we can't use dump, because you didn't declare it const:

void Dump() const

3) Saying c = TestClass() is an assignment. However, c is now a reference-to-const, which cannot be assigned to, since assignment is non-constant (for obvious reasons). Let's hack around this:

const_cast(c) = TestClass(10);

Now we've assigned a new value to the temporary-but-extended object c, and all is as it should be:

main started
data = 0  ptr = 0x0xbfa8219c
destructor
data = 10  ptr = 0x0xbfa8219c
main ended
destructor

The pointers are the same because there's only one object, namely the (temporary) one referenced by c. Assigning to it is a hack that's undefined behaviour in general, but we get away with it for the purpose of this demonstration.

The intermediate destructor is that of the second temporary TestClass(10).