访问传递给函数的结构成员变量

Question

I have a struct stat variable named s inside a struct I defined myself as follows:

struct myStruct {
  struct stat s;
};

and I want to find the difference between the st_mode between two myStruct objects, so my logic is to point to that struct and use a '.' for its member variable.

int func(const void *a, const void *b)
{ 
  return a->s.st_mode - b->s.st_mode;
}

However, there are issues with this implementation:

error: request for member 's' in something not a structure or union
warning: dereferencing 'void *' pointer [enabled by default] 

What do I do to correct this?

Answer 1

The parameters to your function are of type const void* not of type struct myStruct* .

Do you know for sure that they are in fact pointers to struct myStruct s? If so you can do a cast:

return ((struct myStruct*)a)->s.st_mode - ((struct myStruct*)b)->s.st_mode

but it's safer to redefine your function to only accept pointers of the correct type thus:

int func(const struct myStruct* a, const struct myStruct* b)

and then keep the rest unchanged.

Answer 2

Your function should be declared as taking pointers to struct myStruct , otherwise the compiler has no way of inferring the types of a and b .

int func(const struct myStruct *a, const struct myStruct *b)
{ 
  return a->s.st_mode - b->s.st_mode;
}

Alternatively, you may use casting:

int func(const void *a, const void *b)
{ 
  return (const struct myStruct*)a->s.st_mode - (const struct myStruct*)b->s.st_mode;
}