如果函数参数超出范围,则强制编译错误

在C中,枚举类型与整数无法区分.很烦人.

我能想到的唯一前进方法是使用结构而不是枚举的kludgy解决方法.结构是生成性的,因此数以千计是截然不同的.如果调用约定合理(AMD64),则不会有运行时开销.

这是一个使用结构的示例,它可以获得您想要的编译时错误.Kludgy,但它有效:

#include 
enum hundred_e {
    VALUE_HUNDRED_A = 100,
    VALUE_HUNDRED_B
};

enum thousand_e {
    VALUE_THOUSAND_A = 1000,
    VALUE_THOUSAND_B
};

struct hundred { enum hundred_e n; };
struct thousand { enum thousand_e n; };

const struct hundred struct_hundred_a = { VALUE_HUNDRED_A }; 
const struct hundred struct_hundred_b = { VALUE_HUNDRED_B }; 
const struct thousand struct_thousand_a = { VALUE_THOUSAND_A }; 
const struct thousand struct_thousand_b = { VALUE_THOUSAND_B }; 

void print_hundred(struct hundred foo)
{
    switch (foo.n) {
        case VALUE_HUNDRED_A:     printf("hundred:a\n");     break;
        case VALUE_HUNDRED_B:     printf("hundred:b\n");     break;
        default: printf("hundred:error(%d)\n", foo.n); break;
    }
}

void print_thousand(struct thousand bar)
{
    switch (bar.n) {
        case VALUE_THOUSAND_A:     printf("thousand:a\n");     break;
        case VALUE_THOUSAND_B:     printf("thousand:b\n");     break;
        default: printf("thousand:error(%d)\n", bar.n); break;
    }
}

int main(void)
{

    print_hundred(struct_hundred_a);
    print_hundred(struct_thousand_a);  /* Want a compile error here */

    print_thousand(struct_thousand_a);
    print_thousand(struct_hundred_a);  /* Want a compile error here */

    return 0;
}