如何创建类型安全的枚举? [英] How to create type safe enums?

问题描述

在 C 中使用枚举实现类型安全是有问题的，因为它们本质上只是整数.并且枚举常量实际上被标准定义为 int 类型.

To achieve type safety with enums in C is problematic, since they are essentially just integers. And enumeration constants are in fact defined to be of type int by the standard.

为了实现一点类型安全，我用这样的指针做技巧:

To achieve a bit of type safety I do tricks with pointers like this:

typedef enum
{
  BLUE,
  RED
} color_t;

void color_assign (color_t* var, color_t val) 
{ 
  *var = val; 
}

因为指针的类型规则比值更严格，所以这可以防止这样的代码:

Because pointers have stricter type rules than values, so this prevents code such as this:

int x; 
color_assign(&x, BLUE); // compiler error

但它不会阻止这样的代码:

But it doesn't prevent code like this:

color_t color;
color_assign(&color, 123); // garbage value

这是因为枚举常量本质上只是一个 int 并且可以隐式分配给枚举变量.

This is because the enumeration constant is essentially just an int and can get implicitly assigned to an enumeration variable.

有没有办法编写这样的函数或宏color_assign，即使对于枚举常量也能实现完整的类型安全?

Is there a way to write such a function or macro color_assign, that can achieve complete type safety even for enumeration constants?

推荐答案

可以通过一些技巧来实现这一点.给定

It is possible to achieve this with a few tricks. Given

typedef enum
{
  BLUE,
  RED
} color_t;

然后定义一个不会被调用者使用的虚拟联合，但包含与枚举常量同名的成员:

Then define a dummy union which won't be used by the caller, but contains members with the same names as the enumeration constants:

typedef union
{
  color_t BLUE;
  color_t RED;
} typesafe_color_t;

这是可能的，因为枚举常量和成员/变量名称驻留在不同的命名空间中.

This is possible because enumeration constants and member/variable names reside in different namespaces.

然后制作一些类似函数的宏:

Then make some function-like macros:

#define c_assign(var, val) (var) = (typesafe_color_t){ .val = val }.val
#define color_assign(var, val) _Generic((var), color_t: c_assign(var, val))

这些宏的调用方式如下:

These macros are then called like this:

color_t color;
color_assign(color, BLUE); 

说明:

• C11 _Generic 关键字确保枚举变量的类型正确.但是，这不能用于枚举常量 BLUE，因为它属于 int 类型.
• 因此辅助宏 c_assign 创建了一个虚拟联合的临时实例，其中指定的初始化语法用于将值 BLUE 分配给名为 蓝色.如果不存在这样的成员，代码将无法编译.
• 然后将相应类型的联合成员复制到枚举变量中.

• The C11 _Generic keyword ensures that the enumeration variable is of the correct type. However, this can't be used on the enumeration constant BLUE because it is of type int.
• Therefore the helper macro c_assign creates a temporary instance of the dummy union, where the designated initializer syntax is used to assign the value BLUE to a union member named BLUE. If no such member exists, the code won't compile.
• The union member of the corresponding type is then copied into the enum variable.

我们实际上不需要辅助宏，我只是为了可读性而拆分了表达式.写起来也一样好

We actually don't need the helper macro, I just split the expression for readability. It works just as fine to write

#define color_assign(var, val) _Generic((var), 
color_t: (var) = (typesafe_color_t){ .val = val }.val )

<小时>

示例:

color_t color; 
color_assign(color, BLUE);// ok
color_assign(color, RED); // ok

color_assign(color, 0);   // compiler error 

int x;
color_assign(x, BLUE);    // compiler error

typedef enum { foo } bar;
color_assign(color, foo); // compiler error
color_assign(bar, BLUE);  // compiler error

<小时>

编辑

显然，上面的内容并不能阻止调用者简单地输入color =garbage;.如果您希望完全阻止使用枚举的这种分配的可能性，您可以将它放在一个结构中，并使用带有 "opaque type" 的私有封装的标准过程:

Obviously the above doesn't prevent the caller from simply typing color = garbage;. If you wish to entirely block the possibility of using such assignment of the enum, you can put it in a struct and use the standard procedure of private encapsulation with "opaque type":

颜色.h

#include <stdlib.h>

typedef enum
{
  BLUE,
  RED
} color_t;

typedef union
{
  color_t BLUE;
  color_t RED;
} typesafe_color_t;

typedef struct col_t col_t; // opaque type

col_t* col_alloc (void);
void   col_free (col_t* col);

void col_assign (col_t* col, color_t color);

#define color_assign(var, val)   
  _Generic( (var),               
    col_t*: col_assign((var), (typesafe_color_t){ .val = val }.val) 
  )

颜色.c

#include "color.h"

struct col_t
{
  color_t color;
};

col_t* col_alloc (void) 
{ 
  return malloc(sizeof(col_t)); // (needs proper error handling)
}

void col_free (col_t* col)
{
  free(col);
}

void col_assign (col_t* col, color_t color)
{
  col->color = color;
}

main.c

col_t* color;
color = col_alloc();

color_assign(color, BLUE); 

col_free(color);

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