Memory-layout compatibility between C and C++

Question

I'm building a C++ library which uses many functions and struct's defined in a C library. To avoid porting any code to C++, I add the typical conditional preprocessing to the C header files. For example,

//my_struct.h of the C library
#include <complex.h>

#ifdef __cplusplus
extern "C" {
#endif

typedef struct {
  double d1,d2,d3;
#ifdef __cplusplus
  std::complex<double> z1,z2,z3;
  std::complex<double> *pz;
#else
  double complex z1,z2,z3;
  double complex *pz;
#endif
  int i,j,k;
} my_struct;

//Memory allocating + initialization function
my_struct *
alloc_my_struct(double);

#ifdef __cplusplus
}
#endif

The implementation of alloc_my_struct() is compiled in C. It simply allocates memory via malloc() and initializes members of my_struct.

Now when I do the following in my C++ code,

#include "my_struct.h"
...
  my_struct *const ms = alloc_my_struct(2.);

I notice that *ms always have the expected memory layout, i.e., any access such as ms->z1 evaluates to the expected value. I find this really cool considering that (correct me if I'm wrong) the memory layout of my_struct during allocation is decided by the C compiler (in my case gcc -std=c11), while during access by the C++ compiler (in my case g++ -std=c++11).

My question is : Is this compatibility standardized? If not, is there any way around it?

NOTE : I don't have enough knowledge to argue against alignment, padding, and other implementation-defined specifics. But it is noteworthy that the GNU scientific library, which is C-compiled, is implementing the same approach (although their structs do not involve C99 complex numbers) for use in C++. On the other hand, I've done sufficient research to conclude that C++11 guarantees layout compatibility between C99 double complex and std::complex<double>.

Answer 1

C and C++ do share memory layout rules. In both languages structs are placed in memory in the same way. And even if C++ did want to do things a little differently, placing the struct inside extern "C" {} guarantees C layout.

But what your code is doing relies on C++ std::complex and C99 complex to be the same.

So see:

• https://gcc.gnu.org/ml/libstdc++/2007-02/msg00161.html
• C Complex Numbers in C++?