Why do un-named C++ objects destruct before the scope block ends?

Question

The following code prints one,two,three. Is that desired and true for all C++ compilers?

class Foo
{
      const char* m_name;
public:
      Foo(const char         


        

Answer 1

Yes, it is desired.

Foo foo("three") creates a normal object which will be destroyed when the scope ends.

Foo("one") creates a temporary object, which is destroyed at the end of the instruction[1]. Why? Because there is no way you can access it after the instruction has ended.

[1] Deliberate simplification: I should have said sequence point.