Compile-time assertion to determine if pointer is an array

问题

Currently, I have the following block of code to make safe string copying (it works):

#define STRCPY(dst, src) do { assert(((void*)(dst)) == ((void*) & (dst))); \
                              strlcpy(dst, src, sizeof(dst)); } while (0)

So it accepts the construction like:

const char *src = "hello";
char dest[5];
STRCPY(dest, src); //hell

And denies the following:

void (char *dst) {
    STRCPY(dst, "heaven"); //unknown size of dst
}

The problem is that the block of code creates an assertion. Is there a way to perform this check on compilation time?

So I want to have the error on compilation (like creating an array with negative size) instead of crashing code if it possible.

回答1:

For compile-time assertion on whether or not dst is an array, I would use @Lundin solution (or _Static_assert) in case C11 is available.

Else, I would use the following:

#define BUILD_BUG_ON_NON_ARRAY_TYPE(e) (sizeof(struct { int:-!!(((void*)(e)) != ((void*) & (e))); }))

Which basically takes your compile-time evaluated expression ((void*)(dst)) == ((void*) & (dst)) but instead of using it in run time assert just use it in a compile-time assertion manner.

So, overall I would change your macro into:

#define STRCPY(dst, src) do { BUILD_BUG_ON_NON_ARRAY_TYPE(dst); \
                              strlcpy(dst, src, sizeof(dst)); } while (0)

回答2:

If standard C is available, then you can do this:

#define STRCPY(dst, src)                                        \
  _Generic(&(dst),                                              \
           char(*)[sizeof(dst)]: strlcpy(dst,src,sizeof(dst)) )

Explanation:

You can't use a _Generic expression on an array type, because it is not one of the special cases that is exempt from the "array decay" rule (C17 6.3.2.1 §3). So by simply using _Generic((dst), ... in my example, dst would end up as a char* when the expression is evaluated, and then we would lose the information of its original type.

But if we take the address of the array using &, we do utilize one of those special cases and the array decay doesn't happen. Instead we end up with an array pointer, meaning that _Generic will have to check for an array pointer of the expected type and size: char(*)[sizeof(dst)].

As a side-note/safety concern, I never use do-while(0) macros and discourage them, but that's another story.

回答3:

I've found that my post was closed as duplicated for a while and followed the link mentioned. The solution is only for GCC, but it's fine for me, because I have night builds on GCC too. So the pre-draft of the code is:

#if defined(__GNUC__)
#define BUILD_BUG_ON_ZERO(e) (sizeof(struct { int:-!!(e); }))
#define __same_type(a, b) __builtin_types_compatible_p(typeof(a), typeof(b))
#define __must_be_array(a) BUILD_BUG_ON_ZERO(__same_type((a), &(a)[0]))
#define STRCPY(dst,src) do{__must_be_array(dst);strlcpy(dst, src, sizeof(dst));}while(0)
#else
#define STRCPY(dst,src) do{strlcpy(dst, src, sizeof(dst));}while(0)
#endif

回答4:

If you want to check it the only way I can think of is:

assert((sizeof(dst)) != (sizeof(void*)));

but it only work if the size of the array is different than the size of the pointer on OPs system

来源：https://stackoverflow.com/questions/53831489/compile-time-assertion-to-determine-if-pointer-is-an-array

标签

assertion