为什么 const int 适合 char brace init？

Question

我认为大括号初始化不允许缩小。但为什么 int const 允许 char 大括号初始化？

int value1 = 12;
char c1{value1};  // error! no narrowing

const int value2 = 12;
char c2{value2};   // why is this fine?

看到了on Godbolt.

Answer 1

const int value2 = 12;

value2 是编译时常量。编译器可以轻松（并且必须）证明该值为 12，恰好在 char.

可表示的值范围内

int value1 = 12;

value1 不是编译时常量。变量的值可能会在运行时更改。

标准规则的确切措辞（引用最新草案，强调已加）：

[dcl.init.list]/7

A narrowing conversion is an implicit conversion

from an integer type or unscoped enumeration type to an integer type that cannot represent all the values of the original type, except where the source is a constant expression whose value after integral promotions will fit into the target type.

为什么 const int 适合 char brace init？

Why is const int fine for char brace init?

c++

c++11

constants

narrowing