Alternative function pointer syntax accepted by compilers with C11 standard - c

Question

I am not sure about this function pointer type declaration syntax, but it works. The syntax is just like declaring a regular old function.
typedef struct Element GetChildCallback(void *userdata, usize parent, usize child_no);
I couldn't find any information on how standard-compliant it is or what drawbacks it could possibly have.
I thought this only worked for typedefs, so I took a step further and found out this also works for regular function parameters:
extern inline void dmpstrn(const char *t, usize n, int printer(const char *fmt, ...));
inline void dmpstrn(const char *t, usize n, int printer(const char *fmt, ...)) {
usize len = strlen(t) > n ? n : strlen(t);
printer("\"");
for (usize i = 0; i < len; i += 1) {
if (t[i] == '\n')
printer("\\n");
else
printer("%c", t[i]);
}
printer("\"\n");
}
// ...
int main() {
dmpstrn("Hello\nworld", UINT64_MAX, printf);
}
This however doesn't work for variables
int printer(const char *fmt, ...) = printf; // Invalid
It's as if the it isn't the function pointer but the actual function, but what does it mean?

Answer 1

Function declarations used as function parameters are implicitly adjusted by the compiler to pointers to the function types.
From the C Standard (6.7.6.3 Function declarators (including prototypes))
8 A declaration of a parameter as ‘‘function returning type’’ shall be
adjusted to ‘‘pointer to function returning type’’, as in 6.3.2.1
So for example these two function declarations
void f( void( int ) );
and
void f( void ( * )( int ) );
declare the same one function.
On the other hand (6.3.2.1 Lvalues, arrays, and function designators)
4 A function designator is an expression that has function type.
Except when it is the operand of the sizeof operator65) or the unary &
operator, a function designator with type ‘‘function returning type’’
is converted to an expression that has type ‘‘pointer to function
returning type’’
Here is a demonstrative program.
#include <stdio.h>
typedef void F( int );
F display;
void g( F );
void g( F *f )
{
int x = 10;
f( x );
}
void display( int x )
{
printf( "x = %d\n", x );
}
int main(void)
{
F *fp = display;
g( fp );
return 0;
}
The program output is
x = 10
Investigate the program.
Pay attention to that for example this typedef
typedef void F( int );
may be equivalently rewritten like
void typedef F( int );

Answer 2

typedef struct Element GetChildCallback(void *userdata, usize parent, usize child_no);
This defines GetChildCallback as the actual "function type", which is different from the pointer to function type. This can be used to declare a function or form the pointer to function type, but not to define a function:
GetChildCallback MyButtonCB; /* OK, declares function */
GetChildCallback MyResizeCB {} /* ERROR, bad syntax */
struct Element MyButtonCB(void *, usize, usize) {} /* OK */
GetChildCallback cb1 = MyButtonCB; /* ERROR, bad function definition */
GetChildCallback* cb2 = MyButtonCB; /* OK */
void add_callback(Button* button, GetChildCallback* cb); /* OK */
When a function type is used as a function parameter, it silently becomes the pointer to function type, just like an array parameter type becomes the element pointer type. So your example
inline void dmpstrn(const char *t, usize n, int printer(const char *fmt, ...)) {
means just the same as
inline void dmpstrn(const char *t, usize n, int (*printer)(const char *fmt, ...)) {
including the fact that printer in the function body is a pointer to function, so for example (&printer) does not act the same as (printer).