What does a typedef with parenthesis like "typedef int (f)(void)" mean? Is it a function prototype?

It's a typedef to a function type. The intent is to use it for function pointers, but in this case the syntax to use it would be:

int bar(void);

fc_name* foo = bar; /* Note the * */

Update: As mentioned in the comments to Jonathan Leffler's answer, the typedef can be used to declare functions. One use could be for declaring a set of callback functions:

typedef int (callback)(int, void*);

callback onFoo;
callback onBar;
callback onBaz;
callback onQux;

The first parentheses are superfluous - it is equivalent to:

typedef int fc_name(void);

I don't think this does anything useful, though I can't get GCC to complain about it on its own.

This means that fc_name is an alias for a function type that takes no arguments and returns an int. It isn't directly all that useful, though you can declare, for example, the rand() function using:

fc_name rand;

You cannot use the typedef in a function definition.

A pointer to function typedef would read:

typedef int (*fc_name)(void);

---

This code shows that the typedefs without the asterisk are not function pointers (addressing a now-deleted alternative answer):

static int function(void)
{
    return 0;
}

typedef int   fc_name1 (void);
typedef int  (fc_name2)(void);
typedef int (*fc_name3)(void);

fc_name1 x = function;
fc_name2 y = function;
fc_name3 z = function;

When compiled, 'gcc' says:

gcc -Wextra -Wall -pedantic -c -O x.c
x.c:10:1: error: function ‘x’ is initialized like a variable
x.c:11:1: error: function ‘y’ is initialized like a variable

---

And this code demonstrates that you can indeed use fc_name *var = funcname; as suggested by jamesdlin:

static int function(void)
{
    return 0;
}

typedef int   fc_name1 (void);
typedef int  (fc_name2)(void);
typedef int (*fc_name3)(void);

fc_name1  x_0 = function;
fc_name1 *x_1 = function;
fc_name2  y_0 = function;    // Damn Bessel functions - and no <math.h>
fc_name2 *y_1 = function;    // Damn Bessel functions - and no <math.h>
fc_name3  z   = function;

Using y0, y1 generates GCC warnings:

x.c:12:11: warning: conflicting types for built-in function ‘y0’
x.c:13:11: warning: built-in function ‘y1’ declared as non-function

---

And, building on the comment from schot:

static int function(void)
{
    return 0;
}

typedef int   fc_name1 (void);
typedef int  (fc_name2)(void);
typedef int (*fc_name3)(void);

fc_name1  x_0 = function;   // Error
fc_name1 *x_1 = function;   // x_1 is a pointer to function
fc_name1  x_2;              // Declare int x_2(void);
fc_name1 *x_3 = x_2;        // Declare x_3 initialized with x_2

fc_name2  y_0 = function;   // Damn Bessel functions - and no <math.h>
fc_name2 *y_1 = function;   // Damn Bessel functions - and no <math.h>
fc_name1  y_2;              // Declare int y_2(void);
fc_name1 *y_3 = x_2;        // Declare y_3 initialized with y_2

fc_name3  z   = function;

Interesting - the dark corners of C are murky indeed.