x [0] ==当x是const int []时，C ++ 11中的1个常量表达式？ [英] x[0] == 1 constant expression in C++11 when x is const int[]?
问题描述
以下C ++ 11程序格式错误吗？
Is the following C++11 program ill-formed?
const int x[] = {1,2,3};
static_assert(x[0] == 1, "yay");
int main() {}
gcc和clang似乎这么认为，但为什么 x [0] == 1 不是常数表达式？
gcc and clang seem to think so, but why isn't x[0] == 1 a constant expression?
x[0] == 1
subscript operator
*(x+0) == 1
array-to-pointer conversion (int* p = x)
*(p+0) == 1
pointer addition
*p == 1
indirection (lvalue y = x[0])
y == 1
lvalue-to-rvalue conversion:
一个非易失性glvalue（是，x [0]是整数（是，它具有const int类型）或枚举类型的glvalue和非易失性，该枚举类型是指通过先前的初始化（是yes初始化）的非易失性const对象（是，它具有const int类型） 1），用常量表达式初始化（是1是常量表达式）
a non-volatile glvalue (yes, x[0] is a glvalue and non-volatile) of integral (yes it has type const int) or enumeration type that refers to a non-volatile const object (yes it has type const int) with a preceding initialization (yes initialized with 1), initialized with a constant expression (yes 1 is constant expression)
似乎是真的，<$ c的第一个元素$ c> x 数组满足这些条件。
Seems true, the first element of the x array satisfies these conditions.
1 == 1
？
是这是编译器错误，标准缺陷还是我缺少什么？
Is this a compiler bug, standard defect, or am i missing something?
5.19 [expr.const]的哪一部分表示这不是常量表达式？
What part of 5.19 [expr.const] says this isn't a constant expression?
推荐答案
在5.19中：
A [。 ..]表达式是一个常量表达式，除非它涉及以下[...]之一：
A [...]expression is a constant expression unless it involves one of the following [...]:
-
左值-rvalue转换（4.1），除非将其应用于
an lvalue-to-rvalue conversion (4.1) unless it is applied to
- 整数或枚举类型的glvalue，该值引用具有以下内容的非易失性const对象：先前的初始化，使用常量表达式进行初始化，或者
- 文字类型的glvalue引用用constexpr定义的非易失性对象，或者将
引用到子表达式-这样的对象的对象，或者 - 文字类型的glvalue，它是指用常量
表达式初始化的非易失性临时对象
插入简而言之，如果满足以下条件，则只能在常量表达式中进行左值到右值转换：
Putting it plainly, lvalue-to-rvalue conversion can only be done in constant expressions if:
- 已初始化的常量积分（或枚举）声明带有常量：
const int x = 3;。 - 带有
constexpr ：constexpr int x [] = {1,2,3};。 - 使用常量表达式初始化的临时对象...
- a constant integral (or enum) declaration initialized with a constant:
const int x = 3;. - a declaration with
constexpr:constexpr int x[] = {1,2,3};. - a temporary object initialized with a constant expression...
您的示例确实包含了左值到右值的转换，但没有这些例外，因此 x 不是常量表达式。但是，如果将其更改为：
Your example does include lvalue-to-rvalue conversion, but has none of these exceptions, so x is not a constant expression. If, however, you change it to:
constexpr int x[] = {1,2,3};
static_assert(x[0] == 1, "yay");
int main() {}
然后一切都很好。
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