枚举类"无法转换为unsigned int"

Question

我有一个像这样的枚举类:

    typedef unsigned int binary_instructions_t;

    enum class BinaryInstructions : binary_instructions_t
    {
        END_INSTRUCTION = 0x0,

        RESET,

        SET_STEP_TIME,
        SET_STOP_TIME,
        START,

        ADD
    };

我试图在这样的switch语句中使用枚举的成员:

const std::string& function(binary_instructions_t arg, bool& error_detect)
{
    switch(arg)
    {
        case (unsigned int)BinaryInstructions::END_INSTRUCTION:
            return "end";
        break;
    }
    translate_error = true;
    return "ERROR";
}

(unsigned int)当底层类型已经是一个时,为什么需要强制转换unsigned int？

Answer 1

这是因为"枚举类"是"强类型",因此不能隐式转换为任何其他类型.http://en.wikipedia.org/wiki/C%2B%2B11#Strongly_typed_enumerations

Answer 2

因为C++ 11 强类型枚举不能通过设计隐式转换为整数类型.底层类型是a的unsigned int事实并不意味着枚举的类型unsigned int.是的BinaryInstructions.

但是你实际上并不需要转换因为arg是无符号的int,你需要一个强制转换,但static_cast为了清晰起见你应该更喜欢:

switch(arg)
{
    case static_cast<unsigned int>(BinaryInstructions::END_INSTRUCTION) :
        return "end";
    break;
}