如何将带有固定参数的函数传递给C中的另一个方法？

Question

Suppose I have a print Function which takes a void* as an argument:

Print(void *instance)

How can I Set the member of a struct to this method?

struct Foo
{
    void(*Print)(); //this does not match to Print above!
    //void(*Print)(void* Instance); //This would but I cannot change this
}

So how would I have to change this method:

void Init()
{
    struct Foo* k = malloc(sizeof(Foo));

    Foo->Print = &Print(k);   //does not work?!
    //Foo->Pring = &Print;     //would work if types were the same
}

I want to set a function pointer inside a struct with a Fixed argument.

Basically this is a way to implement object oriented code in ANSI-C. The source project is this: Object Orientation in C?!

This would be about changing this piece of C Code:

String* r = New_String("Hello");
r->Free(r); //this works
r->Free();  //this is how it should be!

So that the instance does not have to be passed to the methods themselfes.

Summarized : Goal is to get the Foo instance inside Print called like Foo->Print();

Answer 1

With:

Foo->Print = &Print(k);

you are assigning a function pointer and you are "calling" the function with a parameter. That is wrong. What you can do is:

Foo->Print = Print;
Foo->Print = (void(*)())Print;    // if compiler complains, use a cast

Because Print is a function, the compiler will use its address when assigning it to the function pointer field of the struct. But because your Print function is incompatible with the Print field, you must use a cast to silence the compiler. Your Print function is casted to a "function that can take any number of parameters". The casting is safe here as long as you call it with at least one parameter. Only do this casting when you know what you are doing, or undefined behavior will follow.

Once the function pointer is assigned, you can now call it with your parameter:

Foo->Print(k);