问题描述
下面的C ++ 11程序是否错误？
const int x [] = {1,2， 3};
static_assert（x [0] == 1，yay）;
int main（）{}
gcc和clang似乎认为，但是为什么 x [0] == 1 是一个常量表达式？
x [0] == 1
下标运算符
*（x + 0）== 1
数组到指针的转换（int * p = x）
*（p + 0）== 1
指针添加
* p == 1
间接（l值y = x [0]）
y == 1
左值到右值转换：
似乎为真，<$ c的第一个元素$ c> x 数组符合这些条件。
1 == 1
？
这是一个编译器错误，标准缺陷我缺少一些东西？
5.19 [expr.const]的哪一部分说这不是一个常量表达式？
解决方案在5.19中：
很明显，左值到右值转换只能在常量表达式中进行，如果：
- 用常量初始化的常量整数（或枚举）声明：
const int x = 3; li>
- 一个包含
constexpr 的声明： constexpr int x [] = {1,2,3}; 。
- 用常量表达式初始化的临时对象...
您的示例包括lvalue-to-rvalue转换，但没有这些异常，因此 x 不是常量表达式。然而，如果你把它改成：
constexpr int x [] = {1,2,3}
static_assert（x [0] == 1，yay）;
int main（）{}
p>
Is the following C++11 program ill-formed?
const int x[] = {1,2,3};
static_assert(x[0] == 1, "yay");
int main() {}
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:
Seems true, the first element of the x array satisfies these conditions.
1 == 1
?
Is this a compiler bug, standard defect, or am i missing something?
What part of 5.19 [expr.const] says this isn't a constant expression?
解决方案 In 5.19:
Putting it plainly, lvalue-to-rvalue conversion can only be done in constant expressions if:
- 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...
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() {}
Then all is well.
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