Is 'private' a C keyword?

Question

Are \'private\' or \'public\' keywords in ANSI C (or any other C for that matter), or were they only added in C++ (and Java, C#, ...)?

Answer 1

Neither are C keywords, but some people do the following:

#define public 
#define private static

Update:

For those who think it is a bad idea to do the above, I would agree. But it does explain why someone might think public or private are C keywords.

For those who think it won't compile in C, try this:

#include 
#include 

#define public
#define private static

private void sayHello(void);

public int main(void) {
    sayHello();

    return (EXIT_SUCCESS);
}

private void sayHello(void) {
   printf("Hello, world\n");
}

For those who think it won't compile in C++, yes the above program will.

Update:

Well actually it is undefined behaviour due to this part of the C++ standard:

A translation unit that includes a header shall not contain any macros that define names declared or defined in that header. Nor shall such a translation unit define macros for names lexically identical to keywords.

So the example above and below are not required to do anything sane in C++, which is a good thing. My answer still is completely valid for C (until it is proven to be wrong! :-) ).

In the case of a C++ class with private members, you can do something similar (considered an abuse) like this:

main.c:

#include 
#define private public
#include "message.hpp"

int main() {
    Message msg;

    msg.available_method();
    msg.hidden_method();

    return (EXIT_SUCCESS);
}

message.hpp:

#ifndef MESSAGE_H
#define MESSAGE_H

#include 

class Message {
  private: 
      void hidden_method();

  public: 
      void available_method();
};

inline void Message::hidden_method() {
    std::cout << "this is a private method" << std::endl;
}

inline void Message::available_method() {
    std::cout << "this is a public method" << std::endl;
}

#endif