What are the type deduction rules of decltype(rvalue expr) ?

Question

I have watched a video about type deduction rules of auto and decltype explained by Scott Meyers ... He explained the following

// decltype(lvalue expr) => reference to the type of the expression
// decltype(lvalue name) => type of the name

I understand these rules ... but he didnt explain the following

// decltype(rvlaue expr) => ???

So I tried to understand it by practicing so I did the following

int x = 8;
int func();  // calling this function is rvlaue expr ...

decltype(32) t1 = 128;    // Ok   t1 is int
decltype(64) t2 = x;      // Ok   t2 is int 
decltype(func()) t3 = x;  // Ok   t3 is int ... obviously

Now the magic

decltype(std::move(x)) t4 = x;  // Error t4 is int&& ... compiler says

isn't std::move(x) a rvalue expression ? why decltype deducing t4 to int&& not just int like the examples above? What are the rules of decltype type deductions for rvalue expressions?

Answer 1

decltype behaves differently based on the type it is used on

if the value category of expression is xvalue, then decltype yields T&&;

if the value category of expression is lvalue, then decltype yields T&;

if the value category of expression is prvalue, then decltype yields T.

As you can see it has two different behaviors for rvalues. If the rvalue is an xvalue then we get T&& otherwise it is a prvalue and we get T .

Now if we look at std::move we see that it returns an xvalue as the return is T&& and not T . So std::move(x) is an xvalue and is correctly deduced to int&&