c++ 是`f().a [0]`一个x值？

struct S{
    int a[3] = {1,2,3};
};

S&& f(){return S();}

&f().a;       //[Error] taking address of xvalue (rvalue reference)
&f().a[0];    //ok in GCC 5.1.0 and Clang 3.6.0

S s;
&static_cast<S&&>(s).a;     //[Error] taking address of xvalue (rvalue reference)
&static_cast<S&&>(s).a[0];  //ok in GCC 5.1.0 and Clang 3.6.0

那么,是f().a [0] xvalue？

我认为f().a [0]应该是一个x值.

[EDIT1]

忽略& f().和& f().a [0];因为12.2 [class.temporary] p5.2

The lifetime of a temporary bound to the returned value in a function return statement (6.6.3) is not  extended; the temporary is destroyed at the end of the full-expression in the return statement

static_cast<S&&>(s).a是一个x值(7.2和7.3).

“除了在数组操作数的情况下,如果该操作数是一个左值,另一个值则是一个左值.

所以我认为static_cast<S&&>(s).a [0]应该是一个x值,但是

&安培;的static_cast< S&安培;&安培;&将(S).A [0]; //在GCC 5.1.0和Clang 3.6.0中

追问：

我错了吗？如果我错了,给我一个例子,下标数组会产生一个x值.

这说：

Because the subscripting operation is defined as indirection through a pointer value, the result of a subscript operator applied to an xvalue array is an lvalue, not an xvalue. This could be surprising to some.

而这一改变的部分5.2.1 [expr.sub]如下：

A postfix expression followed by an expression in square brackets is a

postfix expression. One of the expressions shall have the type

“array of T” or“pointer to T” and the other shall have unscoped enumeration or integral type. The result is an lvalue of type

“T.” The type “T” shall be a completely-defined object type.62 The

expression E1[E2] is identical (by definition) to *((E1)+(E2)) [Note:

see 5.3 [expr.unary] and 5.7 [expr.add] for details of * and + and

那么确实是f()的结果a [0];而static_cast<S&&>(s).a [0]应为x值.

这个缺陷在2012年12月之前没有提出决议, clangs defect report support 列出了该缺陷报告的支持是未知的,所以很可能实施者还没有修复这个缺陷.

更新

提出cl ang bug report: Subscript operator applied to an temporary array results in an lvalue .