I recall I saw somewhere some code which used to have a struct as a base class, and a C++ class as a derived class
struct Base_Struct
{
}
class Derived : Base_Struct
{
...
}
And the point is that a pointer to Base_Struct* was passed from the C++ files to some C files which then managed to use some function pointers in Base_Struct.
My question is: if I pass Base_Struct* to a C file, will the C code be able to use the Base_Struct completely? What about the derived class?
If I pass
Base_Struct*
to a C file, will the C code be able to use theBase_Struct
completely?
If it's a standard-layout class, and the C compiler uses the same ABI for such classes as the C++ compiler, then it can access all the data members. Obviously, it couldn't access any member functions or static members, since such things don't exist in C and would have to be left out of any C-compatible definition of the structure.
What about the derived class?
You couldn't define that class in a C program, so it couldn't do anything interesting with the pointer.
Maybe it can work for the base class. But that is on very special case:
But it will never directly for the derived class.
But that create so many restriction that I would provide a C level API to allocate an manipulate those pointers defined as alias on "void*" (I'm not even sure you can do such aliases this in pure C).
Something like:
typedef void* BaseStructPtr;
BaseStructPtr AllocateBase(/* constructor params */);
BaseStructPtr AllocateDerived(/* constructor params */);
TypeOfField1 GetField1(BaseStructPtr ptr);
TypeOfField2 GetField2(BaseStructPtr ptr);
/* etc... */
I do not see any problem to derive from a structure declared in C:
C/C++
#include <stdio.h>
#ifdef __cplusplus
extern "C" {
#endif
typedef struct {
int a;
int b;
} X;
void print(struct StructX* x) {
printf("%d, %d\n", x->a, x->b);
}
#ifdef __cplusplus
} // extern "C"
#endif
C++
#include <iostream>
struct Y : StructX {
int c;
Y()
: c(3)
{
a = 1;
b = 2;
}
void print() {
::print(this);
std::cout << c << std::endl;
}
};
int main() {
Y y;
y.print();
}
Having the structure declared in C++ will complicate things. Now the struct might (or will) have information C is not aware of and(lets say) a memcpy could introduce undefined behavior. To avoid this you would have to wrap the functionality of the C++ structure in C:
C/C++
#ifdef __cplusplus
extern "C" {
#endif
typedef struct Z; // An opaque type (for X) to ensure minimum type safety.
extern int a(Z*);
extern int b(Z*);
#ifdef __cplusplus
} // extern "C"
#endif
C++
extern "C" {
int a(Z* z) {
return ((X*)z)->a;
}
int b(Z* z) {
return ((X*)z)->b;
}
} // extern "C"
In your sample yes a C compiler will be able to use the Base_Struct* pointer and access members. You can use the same struct definition in C and CPP files without problems. As long as it's the Cpp who cast the struct for the C file the compiler will do the job.
void Give_Typed_Pointer_to_C_Function(Base_Struct* pStruct) {...}
void Give_Pointer_to_C_Function(void* pStruct) {...}
Base_Struct A;
Derived B;
Give_Typed_Pointer_to_C_Function(&A); // OK
Give_Typed_Pointer_to_C_Function(&B); // OK
Give_Pointer_to_C_Function(&A); // OK
Give_Pointer_to_C_Function(&B); // wrong
Give_Pointer_to_C_Function((Base_Struct*)&B); // OK
For this to work you have to use the same alignment and packing of struct but as you pass pointer I suppose you are same project so it should be already be the case.
You can even do more, if you keep struct in same order
struct S1 { int x;
int y;
};
class C1 { int x;
int y;
int add(int,int);
};
void Give_Typed_Pointer_to_C_Function(S1 *pStruct);
C1 object;
Give_Typed_Pointer_to_C_Function((S1*)&object); // OK
// an safer alternative as you can't have to be sure that S1 and C1 match
struct C1 { int x;
int y;
#ifdef __cplusplus
int add(int,int);
#endif
};
If you put Cpp keyword like class/public/private and methods in struct with defines. And if you don't use inheritance it will work as well.
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