Typedeffing a function (NOT a function pointer)

Question

typedef void int_void(int);

int_void is a function taking an integer and returning nothing.

My question is: can it be

Answer 1

It can be used in the following cases (out of the top of my head):

• generic code:

  boost::function<int_void> func;

• other typedefs:

  typedef int_void* int_void_ptr;

• declarations:

  void add_callback(int_void* callback);

There may be others.

Answer 2

I think it's legal - the following demonstrates its use:

typedef void f(int);

void t( int a ) {
}

int main() {
    f * p = t;
    p(1); // call t(1)
}

and actually, this C++ code compiles (with g++) & runs - I'm really not sure how kosher it is though.

#include <stdio.h>

typedef void f(int);

void t( int a ) {
    printf( "val is %d\n", a );
}

int main() {
    f & p = t;   // note reference not pointer
    p(1);
}

Answer 3

What happens is that you get a shorter declaration for functions.

You can call test, but you will need an actual test() function.

You cannot assign anything to test because it is a label, essentially a constant value.

You can also use int_void to define a function pointer as Neil shows.

---

Example

typedef void int_void(int);

int main()
{
    int_void test; /* Forward declaration of test, equivalent to:
                    * void test(int); */
    test(5);
}

void test(int abc)
{
}

Answer 4

You are not declaring a variable; you are making a forward declaration of a function.

typedef void int_void(int);
int_void test;

is equivalent to

void test(int);

Answer 5

This should work, no casting required:

void f(int x) { printf("%d\n", x); }

int main(int argc, const char ** argv)
{
    typedef void (*int_void)(int);
    int_void test = f;
    ...
 }

A function's name "devolves" into a function pointer anytime you use the function's name in something other than a function call. If is is being assigned to a func ptr of the same type, you don't need a cast.

The original

typedef int_void(int);

is not useful by itself, without using a pointer to the type. So the answer to your question is "no, you can't use that typedef without a pointer".

Answer 6

Pointers to functions are values in C/C++. Functions are not.