如何创建类型安全枚举? [英] How to create type safe enums?
问题描述
int
的类型。 要实现一些类型安全我用这样的指针来做点:
typedef枚举
{
BLUE,
RED
} color_t;
void color_assign(color_t * var,color_t val)
{
* var = val;
}
因为指针比值更严格的类型规则,所以这样可以防止这样的代码:
int x;
color_assign(& x,BLUE); //编译器错误
但是它不会阻止这样的代码:
color_t color;
color_assign(& color,123); //垃圾值
这是因为枚举常量本质上只是一个 int
,并可以隐式分配给枚举变量。
有没有办法编写这样一个函数或宏 color_assign
,即使枚举也能实现完整的类型安全常数?
可以通过一些技巧来实现这一点。给定
typedef枚举
{
BLUE,
RED
} color_t;
然后定义一个不会被调用者使用的虚拟联合,但包含相同的成员名称作为枚举常数:
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。
然后制作一些类似功能的宏:
#define c_assign(var,val) (var var,val)= $($)
$ b #define color_assign(var,val)_Generic((var),color_t:c_assign(var,val))
然后这些宏被调用如下:
color_t color;
color_assign(color,BLUE);
说明:
- C11
_Generic
关键字确保枚举变量是正确的类型。但是,它不能用于枚举常量BLUE
,因为它的类型为int
。 - 因此,辅助宏
c_assign
创建一个虚拟联合的临时实例,其中使用指定的初始化器语法分配值BLUE
到名为BLUE
的工会成员。如果没有这样的成员,代码将不会编译。 - 然后将相应类型的union成员复制到枚举变量中。
我们实际上不需要帮助宏,我只是拆分表达式以便阅读。
#define color_assign(var,val)_Generic((var),\
color_t:(var)=(typesafe_color_t){.val = val} .val)
示例:
color_t color;
color_assign(color,BLUE); // ok
color_assign(color,RED); // ok
color_assign(color,0); //编译器错误
int x;
color_assign(x,BLUE); //编译器错误
typedef枚举{foo} bar;
color_assign(color,foo); //编译器错误
color_assign(bar,BLUE); //编译器错误
编辑
显然,上述不能阻止调用者简单地键入 color = garbage;
。如果您希望完全阻止使用枚举这种分配的可能性,则可以将其放在一个结构体中,并使用opaque type的私有封装的标准过程:
color.h
#include< stdlib.h>
typedef枚举
{
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)\
)
color.c
#includecolor.h
struct col_t
{
color_t color;
};
col_t * col_alloc(void)
{
return malloc(sizeof(col_t)); //(需要正确的错误处理)
}
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);
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.
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;
}
Because pointers have stricter type rules than values, so this prevents code such as this:
int x;
color_assign(&x, BLUE); // compiler error
But it doesn't prevent code like this:
color_t color;
color_assign(&color, 123); // garbage value
This is because the enumeration constant is essentially just an int
and can get implicitly assigned to an enumeration variable.
Is there a way to write such a function or macro color_assign
, that can achieve complete type safety even for enumeration constants?
It is possible to achieve this with a few tricks. Given
typedef enum
{
BLUE,
RED
} color_t;
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;
This is possible because enumeration constants and member/variable names reside in different namespaces.
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))
These macros are then called like this:
color_t color;
color_assign(color, BLUE);
Explanation:
- The C11
_Generic
keyword ensures that the enumeration variable is of the correct type. However, this can't be used on the enumeration constantBLUE
because it is of typeint
. - Therefore the helper macro
c_assign
creates a temporary instance of the dummy union, where the designated initializer syntax is used to assign the valueBLUE
to a union member namedBLUE
. If no such member exists, the code won't compile. - The union member of the corresponding type is then copied into the enum variable.
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 )
Examples:
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
EDIT
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":
color.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) \
)
color.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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