如何将一个枚举值转换为另一个枚举 [英] How to cast / assign one enum value to another enum

问题描述

我有两个不同的模块有两个枚举，它们具有完全相同的值集。如何投递给另一个人？

  typedef EnumA {
 a_dog = 0，
 a_cat = 1 
}枚举; 
 
 typedef EnumB {
 b_dog = 0，
 b_cat = 1 
}枚举B; 
 
枚举a = a_dog; 
 EnumB b; 
 
 b = a; 
  

这样的作业会导致一个警告：枚举类型与另一个类型混合
我可以通过类型转换避免switch-case，例如

  b =（int）a; 
  

或

  b =（EnumB）a; 
  

解决方案

我从您的问题中创建了一个工作代码。您已经错过了您的类型定义中的枚举。

  typedef枚举EnumA 
 {
 a_dog = 0，
 a_cat = 1 
}枚举; 
 
 typedef枚举EnumB 
 {
 b_dog = 0，
 b_cat = 1 
}枚举B; 
 
 int main（）
 {
 EnumA a = a_dog; 
 EnumB b; 
 
 b =（EnumB）a; 
 printf（％d\\\
，b）; 
 return 0; 
} 
  

代码 b = a 也可以正常工作，没有演员。另外 b =（int）a; 正在工作 - 至少在C11中，因为枚举只是整数。无论如何，嗯，这是一个很好的做法做一个明确的演员。

I have 2 enums in 2 different modules that have exactly same value-set. How can I cast one to another?

typedef EnumA{
a_dog = 0,
a_cat = 1
} EnumA;

typedef EnumB{
b_dog = 0,
b_cat = 1
} EnumB;

EnumA a = a_dog;
EnumB b;

b = a;

Such an assignment is resulting in a warning: enumerated type mixed with another type Can I avoid switch-case by typecasting, like say

b = (int)a;

or

b = (EnumB)a;

解决方案

I made a working code from your question. You have missed the enum from your type definitions.

typedef enum EnumA
{
    a_dog = 0,
    a_cat = 1
} EnumA;

typedef enum EnumB
{
    b_dog = 0,
    b_cat = 1
} EnumB;

int main()
{
    EnumA a = a_dog;
    EnumB b;

    b = (EnumB) a;
    printf("%d\n", b);
    return 0;
}

The code b = a also works properly without the cast. Also b = (int) a; is working - at least in C11, becuse enums are really just integers. Anyway, IMHO it is good practice to make an explicite cast.

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