what does void(*) void and int(*) int mean in C?

Question

Can anyone explain these two lines of code in C:

void (*pfs)(void) = &fs;        
long int (*pfact)(int) = &fact; 

Answer 1

To make these declarations more clear

void (*pfs)(void)=&fs;
long int (*pfact)(int)=&fact; 

you can introduce typedef names for function declarations as for example

typedef void FUNC1( void );
typedef long int FUNC2( int );

and then write

FUNC1 *pfs = &fs;
FUNC2 *pfact = &fact; 

So the original declarations declare pointers to functions of the specified types and initialize them with addresses of the given functions.

Here is a demonstrative program

#include <stdio.h>

typedef void FUNC1( void );
typedef long int FUNC2( int );

void fs( void )
{
    puts( "Hello Islacine" );
}

long int fact( int x )
{
    return x;
}

int main(void) 
{
    FUNC1 *pfs = &fs;
    FUNC2 *pfact = &fact;

    pfs();

    printf( "sizeof( long int ) = %zu\n", sizeof( pfact( 0 ) ) );

    return 0;
}

Its output might look like

Hello Islacine
sizeof( long int ) = 8

Take into account that instead of

    FUNC1 *pfs = &fs;
    FUNC2 *pfact = &fact;

or instead of

    void (*pfs)(void)=&fs;        
    long int (*pfact)(int)=&fact; 

you could even write

    FUNC1 *pfs = fs;
    FUNC2 *pfact = fact;

or

    void (*pfs)(void) = fs;        
    long int (*pfact)(int) = fact; 

because in expressions with rare exceptions a function designator is converted to pointer to function.

You could even write :)

    FUNC1 *pfs = *****fs;
    FUNC2 *pfact = *****fact;

or

    void (*pfs)(void) = *****fs;        
    long int (*pfact)(int) = *****fact; 

From the C Standard (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’’.

Answer 2

This syntax defines a function pointer. You can use them like this:

void a() {
    //do something
}

int main() {
    void (*functionPointer)();
    functionPointer = a;
    //or
    functionPointer = &a;

    //call the funcion pointer as you would a regular function
    functionPointer();
}

The main reasons to use function pointers are to make callback functions and "jump-tables " (This is in quotes because function pointer arrays are not really jump tables. Real jump tables can only be made in assembly language).

Answer 3

what does void(* ) (void) and int(* ) (int) mean in C?

They just mean pointer to function taking void argument and returning void.

void (*pfs)(void)=&fs; 

pfs is an pointer to function taking void as an argument and returning void. This has initialized with function of same signature type i.e fs here.

long int (*pfact)(int)=&fact; 

pfact is an pointer to function taking int as an argument and reutring long int. After the assignment pfact is pointing to function fact.

Additional Note:

There are tools to read complicated declaration. One of them is https://cdecl.org/. Also as others have pointed out better way to work with pointer to function is typdef them .

Answer 4

They are function pointers. In your example:

void (*pfs)(void)=&fs;

You are creating a variable named "pfs". This variable has the type of a function that takes in no arguments and returns nothing.

The following makes your examples much more clear.

return_type (*variable_name) (argument_list)

Now the right side of the = sign is pretty easy if you understand the first half.

All we do is assign the location of the function "fs" to our function pointer (variable) "pfs".

Here is a full example to demo the explanations above.

#include <stdio.h>
void my_print_method(int number) {
    printf("The number is: %d", number);
}
int main() {
    // Declare our function pointer variable
    void (*printer)(int);

    // Assign our function pointer 'printer' to the method 'my_print_method'
    printer = &my_print_method

    // We can also assign the method to our function pointer as such:
    printer = my_print_method

    // Use our function pointer as if it were a function
    (*printer)(55);

    // Keep in mind that that in C we can simply write the function call as such:
    printer(55);

}