C++ C整型和_泛型.哪些类型是兼容的

发布于12月09日

我想做一个宏来判断给定的值是否是整数，我发现有些整数类型在使用_Generic时是相互兼容的，所以它们不能都作为单独的_Generic情况使用.

比如int和int32_t

或

signed char和int8_t

what is the rule on that? Is there a solution that w或ks with all platf或ms/compilers?

我当前的代码是:

#define is_integer(exp) \
_Generic(exp, \
    int8_t: 1, \
    int16_t: 1, \
    int32_t: 1, \
    int64_t: 1, \
    uint8_t: 1, \
    uint16_t: 1, \
    uint32_t: 1, \
    uint64_t: 1, \
    default: 0)

However, it doesn't w或k with char (goes to default), but w或ks with signed char (goes to int8_t).

Should I add char as a separate _Generic case? I'm afraid that some implementation typedefs int8_t to char so it won't w或k, 或 are they not allowed to?

推荐答案

有没有适用于所有平台/编译器的解决方案？

(u)intN_t是optional种类型.

在使用通常的(u)intN_t种类型的实现上，请考虑两阶段_Generic.

#define is_integer2(exp) \
_Generic(exp, \
    int8_t: 1, \
    int16_t: 1, \
    int32_t: 1, \
    int64_t: 1, \
    uint8_t: 1, \
    uint16_t: 1, \
    uint32_t: 1, \
    uint64_t: 1, \
    default: 0)

#define is_integer1(exp) \
_Generic(exp, \
    char: 1, \
    signed char: 1, \
    unsigned char: 1, \
    ...  // 2 shorts, int, unsigned, 2 longs
    long long: 1, \
    unsigned long long: 1, \
    default: is_integer2(exp))

代码可以在另一个级别中包括(u)intptr_t.

代码可以在另一个级别中包括size_t, ptrdif_t.

代码可以在另一个级别中包括(u)intmax_t.

等.