Different behavior between "int &&" and "auto &&"

Question

In C ++ 11, if I try to do this:

int x = 5;

int && y = x;

It will not be able to compile with an error indicating that the r-value reference cannot be bound to an lvalue.

However, if I do this:

int x = 5;

auto && y = x;

It compiles without errors. Why is this happening? I tried to get the type y, but typeid()removes the reference attributes. auto &&automatically collapses in &or &&depending on what is assigned?

Answer 1

In the first case, the int && yvariable ycan only be associated with an rvalue, for which it xdoes not.

auto && y y , y — - — :

auto && y = x;

x lvalue, auto int&, :

int& && y = x;

:

int & y = x;

.

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, :

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Answer 2

int&& r-value int. auto&& .

Answer 3

In the second example auto(c auto&&) there will be everything that would make the code correct. In this case, it int& &&is correct, since it int& &&collapses into int&- thus, everything works well.