I know that the following should be undefined behavior
since the original variable is const. How is it possible, however,
that the two addresses &x and &rx are the same, yet the values
which they print (I don't say hold, since it simply cannot be),
aren't. Thanks!
const int x=10;
int& rx = const_cast<int&>(x);
rx++;
cout << x << " and " << rx << endl;
cout << "is &x == &xr: " << (&x==&rx) << endl;
the output from G++ 4.9 is
10 and 11
is &x == &xr: 1
The compiler (well, clang++ 3.7.0 as of last week) does indeed optimise the "intention" of the code, regardless of the legality of it:
movl $_ZSt4cout, %edi
movl $10, %esi
callq _ZNSolsEi
movq %rax, %rbx
movl $.L.str, %esi
movl $5, %edx
movq %rbx, %rdi
callq _ZSt16__ostream_insertIcSt11char_traitsIcEERSt13basic_ostreamIT_T0_ES6_PKS3_l
movl $11, %esi
movq %rbx, %rdi
callq _ZNSolsEi
As always, it's worth noting that the behaviour of undefined behaviour does cover "does what you think it will do", as well as "doesn't do what you think it will do", and this CERTAINLY applies here.
const_cast of a value that was originally const is undefined behaviour, and at that point you have given up all rights to "sane behaviour" from the compiler. What happens now is whatever the compiler writer thinks is the right thing - and if that means the value actually got placed in a read-only bit of memory, then your code will not succeed in updating the value. But in this case, it simply optimises x to the constant 10, and rx becomes 11 - since you don't actually "do" anything else with x and rx, that's "fine" by the compilers standards.