C/C++ struct question

Question

Suppose you have a .cpp file (that is, compiled using a C++ compiler like MSVC). In that file, you define a struct in the following way:

struct Pixel
{
   float x, y;
};

In the same file, you have a line of code that will call a C function, that requires a C struct equal to Pixel. If you write:

Pixel my_pixel
// set my_pixel to something
c_func(&my_pixel);

will it work? I mean, the C++ compiler will create the object my_pixel, but it will pass it to a function that is compiled as C code (i have only a .lib of that library).

Answer 1

The reason David Schzwartz says you need an extern "C" block is that without the extern "C" block, the compiler will "mangle" the name of the C function you are calling at the point you call it. If you are calling a C function and not a C++ function, the function's definition in your library will not have a mangled name, so your executable will fail to link.

That's what you want if the function you are calling is written in C++, as name mangling allows for function name overloading. The types of each of a function's parameters are compactly encoded in the mangled function name.

Name mangling was originally provided in C++ to allow C++ object files to be linked with legacy linkers, rather than having to provide a C++-specialized linker that has explicit support for overloaded functions.

C doesn't permit function name overloading, so C function names are never mangled. To provide a prototype in C++ for a single C function you do this:

extern "C" Foo( int theInt );

If you have a whole header file full of C function prototypes and you want to #include that header from a C++ source, enclose the #include in an extern C block

extern "C" {
    #include "Foo.h"
}

Answer 2

If the header file is correct, it will work, assuming the C compiler and the C++ compiler use compatible calling conventions. Make sure the header file has an appropriate extern "C" block that contains the function definition.