Storage reuse Undefined behaviour

Question

Reading in the cppreference for lifetime, there is the following statement for storage reuse

A program is not required to call the destructor of an object to end its lifetime if the object is trivially-destructible or if the program does not rely on the side effects of the destructor. However, if a program ends the lifetime of an non-trivially destructible object that is a variable explicitly, it must ensure that a new object of the same type is constructed in-place (e.g. via placement new) before the destructor may be called implicitly, i.e. due to scope exit or exception for automatic objects, due to thread exit for thread-local objects, or due to program exit for static objects; otherwise the behavior is undefined.

There is the following example

class T {}; // trivial
 
struct B
{
    ~B() {} // non-trivial
};
 
void x()
{
    long long n; // automatic, trivial
    new (&n) double(3.14); // reuse with a different type okay
} // okay
 
void h()
{
    B b; // automatic non-trivially destructible
    b.~B(); // end lifetime (not required, since no side-effects)
    new (&b) T; // wrong type: okay until the destructor is called
} // destructor is called: undefined behavior

Why when the destructor of T is called is undefined behavior?

In the above example, if std::launder is used , is the UB issue solved? Look at the following demo