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’’.