In the following, would there be a temporary object created before const reference is used to a non-const object?
const int y = 2000;
const int &s = y // ok, const reference to const object.
int x = 1000;
const int &r = x; // any temporary copy here?
If no then, how does this work?
const int z = 3000;
int &t = z // ok, why can't you do this?
Best Answer
No.
A reference is simply an alias for an existing object.
const
is enforced by the compiler; it simply checks that you don't attempt to modify the object through the referencer
.* This doesn't require a copy to be created.Given that
const
is merely an instruction to the compiler to enforce "read-only", then it should be immediately obvious why your final example doesn't compile.const
would be pointless if you could trivially circumvent it by taking a non-const
ref to aconst
object.* Of course, you are still free to modify the object via
x
. Any changes will also be visible viar
, because they refer to the same object.