Implications of typedef void FOO vs. #define FOO void in function signatures [duplicate]

Question

While going through some source code that heavily mixes C and C++, I came across the following (slightly modified to protect the work of the company, the meaning remains the same):

/*
 * Typedefs of void are synonymous with the void keyword in C,
 * but not in C++. In order to support the use of MY_VOID
 * in place of the void keyword to specify that a function takes no
 * arguments, it must be a macro rather than a typedef.
 */
#define MY_VOID void

What is the difference between typedef void MY_VOID and #define MY_VOID void in this specific context?

---

I don't believe this is a duplicate of this question because it asks specifically about the implications in regards to function signatures, rather than a much more general "what's the difference".

Answer 1

A simple test program in C++ demonstrates the difference:

typedef void VOID;

void f(VOID) {}

int main()
{
    f();
}

On compilation (as C++), it gives these error:

prog.cpp:5:8: error: '<anonymous>' has incomplete type
 void f(VOID) {}
        ^
prog.cpp:5:12: error: invalid use of 'VOID {aka void}'
 void f(VOID) {}
            ^
prog.cpp: In function 'int main()':
prog.cpp:9:7: error: too few arguments to function 'void f(<type error>)'
     f();
       ^
prog.cpp:5:6: note: declared here
 void f(VOID) {}
      ^

which explains what the comment means in your code. In particular, it seems the typedef VOID attempts to be a type different from void, when it is used as parameter type.

Answer 2

The comment explains the difference. Given an alias for void:

typedef void MY_VOID;

If you try to use this instead of void to indicate that a function takes no parameters:

int f(MY_VOID);

C will allow this, but C++ won't.

So, if you really want to make life difficult for yourself by writing code that (a) is valid in both languages and (b) uses an alias for this particular use of void, that alias will have to be a macro.