c++ - 枚举警告 : outside the range

Question

我有以下代码。有两个枚举。一个有 16 个值 (A_) 和另一个有 15 个值 (A1_)。

#include <iostream>

typedef enum
{   
    A_0 = 0,
    A_1,
    A_2,
    A_3,
    A_4,
    A_5,
    A_6,
    A_7,
    A_8,
    A_9,
    A_10,
    A_11,
    A_12,
    A_13,
    A_14,
    A_15,
    A_FIRST = A_0,
    A_LAST = A_15
} a_t;

typedef enum
{   
    A1_0 = 0,
    A1_1,
    A1_2,
    A1_3,
    A1_4,
    A1_5,
    A1_6,
    A1_7,
    A1_8,
    A1_9,
    A1_10,
    A1_11,
    A1_12,
    A1_13,
    A1_14,
    A1_FIRST = A1_0,
    A1_LAST = A1_14
} a1_t;

int 
main()
{   
    a_t m = static_cast<a_t> (static_cast<int> (A_LAST) + 1); // WARNING
    std::cout << m;
    a1_t m1 = static_cast<a1_t> (static_cast<int> (A1_LAST) + 1); // NO WARNING
    std::cout << m1;
    return 0;
}

g++编译:

g++ -Wconversion enum_testing.cpp

我收到一条警告消息:

enum_testing.cpp: In function ‘int main()’:
enum_testing.cpp:49:64: warning: the result of the conversion is     unspecified because ‘16’ is outside the range of type ‘a_t’ [-Wconversion]
     a_t m = static_cast<a_t> (static_cast<int> (A_LAST) + 1);

问题:为什么我在第一次使用枚举 A_ 声明时收到警告消息，而不是使用枚举 A1_ 的消息。

Answer 1

enum 的范围是:

[...]If the underlying type is not fixed, the range is all values possible for the smallest bit field large enough to hold all enumerators of the target enumeration[...]

所以 15 适合这个范围，而 16 不适合。这是 C++17 之前的未指定行为，之后将是未定义的行为。更改是由于 DR 1766: Values outside the range of the values of an enumeration 其中说:

Although issue 1094 clarified that the value of an expression of enumeration type might not be within the range of the values of the enumeration after a conversion to the enumeration type (see 5.2.9 [expr.static.cast] paragraph 10), the result is simply an unspecified value. This should probably be strengthened to produce undefined behavior, in light of the fact that undefined behavior makes an expression non-constant. See also 9.6 [class.bit] paragraph 4.

解决方案是使其成为未定义的行为。