c++ - 带位的长枚举

Question

在超过第 32 个标志后，我的 enum 一直面临问题:

enum ConditionType_t {
    CONDITION_NONE                          = 0,
    CONDITION_LIGHT                         = 1 << 0,
    CONDITION_INFIGHT                       = 1 << 1,
    CONDITION_MUTED                         = 1 << 2,
    ...
    CONDITION_LUCKY                         = 1 << 32,
}

知道 enums 基本上是 8bit，CONDITION_LUCKY 将等于 CONDITION_NONE。所以我实现了 C++11 的 enum classes:

enum class ConditionType_t : uint64_t {
    CONDITION_NONE                          = 0,
    CONDITION_LIGHT                         = 1 << 0,
    CONDITION_INFIGHT                       = 1 << 1,
    CONDITION_MUTED                         = 1 << 2,
    ...
    CONDITION_LUCKY                         = 1 << 32,
}

现在我收到数百万条警告，例如:

warning C4293: '<<' : shift count negative or too big, undefined behavior

错误如:

error C2065: 'CONDITION_NONE' : undeclared identifier

显然，位移位与枚举类不相容。

有什么想法吗？

Answer 1

表达式1 << 32是未定义的行为。来自 [expr.shift]:

The behavior is undefined if the right operand is negative, or greater than or equal to the length in bits of the promoted left operand.

自 1是一个 int ，它只有 32 位。即使这不是未定义的，你也不会得到你想要的值，因为 1 << 32 的类型是int ，无论如何都不能保存 33 位值。

您需要使用 uint64_t从头开始:

enum class ConditionType_t : uint64_t {
    CONDITION_NONE                          = 0,
    CONDITION_LIGHT                         = 1ULL << 0,
    CONDITION_INFIGHT                       = 1ULL << 1,
    CONDITION_MUTED                         = 1ULL << 2,
    ...
    CONDITION_LUCKY                         = 1ULL << 32,
};