在C中获取sizeof一个void（*）（）函数指针

Question

To aid with some precise memory allocation that I'm doing in C , I'm trying to get the sizeof a function pointer for a function with return type void that takes no parameters.

However, when I do sizeof (void (*)()) , I generate a compiler warning:

function declaration isn't a prototype [-Wstrict-prototypes]

How do I get the size that I'm looking for?

Answer 1

这是一个旧样式函数定义，它缺少参数列表，所以添加它们：

sizeof( void(*)(void) )

Answer 2

For readability purposes, I recommend declaring a type for the signature of any function you'll want have some pointer to:

 typedef void my_sigt(void);

then, to declare a pointer to such a function, code:

 my_sigt* funptr;

and to get its size code sizeof(my_sigt*) (or sizeof(funptr) if you have such a variable funptr )

BTW, I am not sure that the C99 standard guarantees that every function pointer has the same size (or has the same size as some data pointer). But POSIX requires that (in particular, to be able to use dlsym for dynamic linking of such functions).