calling C++ method from C code

Question

I've read the following article here that explain how to call c++ based functions from pure C code.

The method is best explained with the following example :

// C++ code:
class C {
    // ...
    virtual double f(int);
};
extern "C" double call_C_f(C* p, int i) // wrapper function
{
    return p->f(i);
}

/* C code: */
double call_C_f(struct C* p, int i);
void ccc(struct C* p, int i)
{
    double d = call_C_f(p,i);
    /* ... */
}

However, I fail to understand if where do i define an instance that is pointed by 'p'. notice that in the C code, it is delivered as pointer to opaque struct, and in the C++ code it's assumed to be pointer to valid instance of class C, but i didn't see where this pointer is being initialized.

p.s. Of course that each code block that relate to different language is compiled with an appropriate compiler, and different object file is created.

Bryan Chen · Accepted Answer · 2016-04-18 21:42:00Z

6

You need similar wrapper methods to create and destroy the C++ object.

C++ code:

extern "C" C * create_c()
{
    return new C;
}

extern "C" void destroy_c(C *p)
{
    delete p;
}

C code:

// forward declarations (better to put those in a header file)
struct C;
struct C * create_c();
double call_C_f(struct C* p, int i);
void destroy_c(struct C *p);

struct C * p = create_c();
double result = call_C_f(p, i);
destroy_c(p);
p = NULL;

edited Apr 18, 2016 at 21:42

answered Apr 18, 2016 at 11:04

Bryan Chen

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4 Comments

Margaret Bloom Over a year ago

A nice add-on to this answer would be a paragraph about POD types since the fact that the C class has a virtual member function (preventing it from being a POD) is the real reason behind the need for an opaque struct.

Zohar81 Over a year ago

thanks for the help. but in the Usage part, shouldn't i need to have function definition prior to calling. the definition is from type 'C * create_c()' ... but the C part doesn't recognize class C.

Bryan Chen Over a year ago

@Zohar81 just add struct C; to declare it

Peter Over a year ago

@Bryan - might want to emphasise the fact that the C code can see a declaration, but should not see a definition of struct C - only the C++ code should.

Leandros · Accepted Answer · 2016-04-18 11:57:57Z

3

In the case of your example, p is an object of your class C. You would have to create it in C++ code, and than pass it to C.

You should also declare the pointers to your C++ class as void * in C. The type is unknown, and struct C implies it's a known struct, which is wrong.

Example:

extern "C" {
struct C;
struct C *create_c(void)
{
     return new C;
}

void destroy_c(struct C *ptr)
{
    delete ((C*)ptr);
}
}

edited Apr 18, 2016 at 11:57

answered Apr 18, 2016 at 11:03

Leandros

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11 Comments

Leandros Over a year ago

@BryanChen It does work, that's not an issue. It just gives the wrong impression, that you actually know the type in the C code. Which you don't. It's an opaque type you don't know, hence the semantic is wrong, and void * has the correct semantic.

Margaret Bloom Over a year ago

I find the use of a struct useful actually. As all void pointers are equal, while different struct pointers are not.

Bryan Chen Over a year ago

Opaque pointer means it is a pointer to some unknown type, which is a good use here. void * will just make the code unmaintainable/unreadable.

jdarthenay Over a year ago

If only using void * and if you have several types wrapped, you won't get any warning if using the wrong type because all you have will be void *.

Leandros Over a year ago

@jdarthenay You don't get a warning anyway, void * and struct C* are equal. C is not C++.

|

Antti Haapala · Accepted Answer · 2016-04-18 11:03:01Z

0

You'd create it in C++, most probably with new C(). There is no way of portably creating it in C, so in case the C code must be able to construct new instances, you'd have to write an extern "C" C* construct_C(); factory function for it; likewise for a destructor.

answered Apr 18, 2016 at 11:03

Antti Haapala

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