std::launder has a precondition that all bytes reachable from the would-be-returned pointer are reachable through the passed pointer.
My understanding is that this is meant to allow compiler optimizations, so that e.g.
struct A {
int a[2];
int b;
};
void f(int&);
int g() {
A a{{0,0},2};
f(a.a[0]);
return a.b;
}
can be optimized to always return 2. (see Pointer interconvertibility vs having the same address and Can std::launder be used to convert an object pointer to its enclosing array pointer?)
Does this mean the reachability precondition should also apply to placement-new? Otherwise is there anything preventing me from writing f as follows?:
void f(int& x) {
new(&x) A{{0,0},0};
}
The address of a.a[0] is the same as that of a and the new A object is transparently replaceable with the old A object, so that a.b in g should now be 0.
