Why is not deleting an object that has a destructor with a side effect undefined behavior in C++11?

Question

This answer quotes C++11 Standard 3.8:

if there is no explicit call to the destructor or if a delete-expression (5.3.5) is not used to release the sto

Answer 1

I believe this is put into the standard to allow for garbage collection. And to point out that C++ behaves differently that some other languages, by not calling destructors while collecting.

It specifically says that storage can be reused without calling the destructors of the objects in that memory area. If the program depends on the destructors being run, it will not work as expected.

If it doesn't depend on the destructors, all is fine.