C++11: does "decltype(1+2)" declare an xvalue or a prvalue?

Question

Does decltype(1+2) declare an xvalue or or a prvalue?

cppreference says, decltype(expression) will declare: 1. T&& if expression is an xvalue 2. T if expression is an prvalue 3. T& if expression is lvalue

But my question is: how to generate an expression that's an xvalue? I suppose that return value and temp objects should be xvalue, but actually they seem to be xvalue, in my experiment:

struct S{};
S f();
int main()
{
    int i=2;
    decltype(i+1) j=i;
    ++j;
    printf("i=%d\n",i);
    S obj;
    decltype(f()) k=obj;

    return 0;
}

This program compiles: I could judge that

decltype(i+1) declares (i+1) as a prvalue

because if it's an xvalue, then decltype gets T&&, which cannot bind to a left value variable of "i". It's also weird that decltype(f()) is also giving me f() as a prvalue?

So my question is: how to write an expression so that decltype(expression) gives me an xvalue? Thanks.

Answer 1

Decltype resolves to a type, not an expression - you can't say that it "declares a prvalue" or anything like that.

i+1 is a prvalue, and not an id-expression. So decltype yields a non-reference type: decltype(i+1) j = i; means int j = i;.

The second case is similar; f() is a prvalue, so decltype(f()) is S.

To get decltype(expression) to resolve to an rvalue reference type, the expression must be an xvalue. For example decltype( std::move(f()) ) is S&&.