如何获取枚举元素的数量？ [英] How to get the number of elements of an enum?

问题描述

您好，

我创建了一个结构，例如，

Hello,
I have created one struture like,

typedef enum
{
   NAME = 0,
   ADDRESS,
   BIRTHDATE,
}INFORMATION

我想获得此结构中的元素数量。有没有函数？

And I want to get Number of elements in this structure. Is there any function?

推荐答案

考虑到你给出的结构是：

Considering the structure that is given by you is:

typedef enum
{
MIN = 0,
NAME  = MIN,
ADDRESS,
BIRTHDATE,
MAX = BIRTHDATE
};
int NumOfElement = MAX-MIN+1;

或直接BIRTHDATE + 1（如果Enum从0开始）。

Or directly BIRTHDATE + 1 (if Enum start with 0).

以你问的方式... 否。

没有可以提供这样的功能（要执行的代码）结果，因为标准C ++有无反射。

标准RTTI专为支持dynamic_cast而定制，并且在枚举上没有任何作用。



如果我们考虑一些语言扩展（比如C ++ / CLI，这是一个不被C ++标准的所有者ANSI接受的ECMA标准......），可能会增加反射，从而增加反射能力在运行时查询源的状态。



当然可以有一些技巧，比如

In the way you asked ... NO.
There is no function (code to be executed) that can give such a result, since standard C++ has no reflection.
The standard RTTI is tailored to support dynamic_cast and have no role on enums.

Things may be different if we consider some language extensions (like C++/CLI, that is an ECMA standard not accepted by ANSI that is the owner of the C++ standard ...) that can add reflection and hence the capability to query at runtime the state of the source.

Of course there can be tricks like

enum E
{
    a,b,c, count;
};

导致a = 0，b = 1，c = 2，count = 3。但这与你提出的问题有所不同（可能是不合适的，出于特定目的）。

That leads to a=0, b=1, c=2, count=3. But that's a different thing respect to what you asked (may be not adequate, for certain purpose).

我同意SA，但是在进行通用编程和考虑未来的可扩展性时，我们总是在枚举中添加eENUMLast，这样如果我们添加/删除任何枚举值，eENUMLast总是保持最后。

I am agree with SA, however while doing generic programming and thinking about future extensibility, we always add eENUMLast in enum, so that if we add/delete any enumerated value, eENUMLast always remain last.

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