Reputation: 44762

Typedeffing a function (NOT a function pointer)

typedef void int_void(int);

int_void is a function taking an integer and returning nothing.

My question is: can it be used "alone", without a pointer? That is, is it possible to use it as simply int_void and not int_void*?

typedef void int_void(int);
int_void test;

This code compiles. But can test be somehow used or assigned to something (without a cast)?

/* Even this does not work (error: assignment of function) */
typedef void int_void(int);
int_void test, test2;
test = test2;

Upvotes: 4

Answers (6)

user14554

Reputation:

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)
{
}

Upvotes: 7

anon

Reputation:

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);
}

Upvotes: 1

Jim Buck

Reputation: 20726

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);

Upvotes: 3

Tim Schaeffer

Reputation: 2636

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