Writing a function pointer in c

Question

I was recently reading a code, and found that a function pointer is written as :

int (*fn_pointer ( this_args ))( this_args )

I usually encounter a function pointer like this :

return_type (*fn_pointer ) (arguments);

Similar thing is discussed here:

// this is a function called functionFactory which receives parameter n
// and returns a pointer to another function which receives two ints
// and it returns another int
int (*functionFactory(int n))(int, int) {
printf("Got parameter %d", n);
int (*functionPtr)(int,int) = &addInt;
return functionPtr;
}

Can somebody tell me what is the difference and how does this work ?

can you please post the exact code? a compilable sample?

Iharob Al Asimi
– Iharob Al Asimi

2015-02-18 12:38:33 +00:00
Commented Feb 18, 2015 at 12:38 — Iharob Al Asimi
– Iharob Al Asimi, Commented Feb 18, 2015 at 12:38
This might help stackoverflow.com/questions/840501/…

Arun A S
– Arun A S

2015-02-18 12:42:52 +00:00
Commented Feb 18, 2015 at 12:42 — Arun A S
– Arun A S, Commented Feb 18, 2015 at 12:42
@ iharob: My code is longer. I just updated.

Shivendra Mishra
– Shivendra Mishra

2015-02-18 12:51:26 +00:00
Commented Feb 18, 2015 at 12:51 — Shivendra Mishra
– Shivendra Mishra, Commented Feb 18, 2015 at 12:51

Community · Accepted Answer · 2017-05-23 12:29:42Z

int (*fn_pointer ( this_args ))( this_args );

declares fn_pointer as a function that takes this_args and returns a pointer to a function that takes this_args as argument and returns an int type. It is equivalent to

typedef int (*func_ptr)(this_args);
func_ptr fn_pointer(this_args);

Let's understand it a bit more:

int f1(arg1, arg2);  // f1 is a function that takes two arguments of type   
                     // arg1 and arg2 and returns an int.

int *f2(arg1, arg2);  // f2 is a function that takes two arguments of type  
                      // arg1 and arg2 and returns a pointer to int.  

int (*fp)(arg1, arg2); // fp is a pointer to a function that takes two arguments of type  
                       // arg1 and arg2 and returns a pointer to int.  

int f3(arg3, int (*fp)(arg1, arg2)); // f3 is a function that takes two arguments of  
                                        // type arg3 and a pointer to a function that 
                                        // takes two arguments of type arg1 and arg2 and 
                                        // returns an int.  

int (*f4(arg3))(arg1, arg2); // f4 is a function that takes an arguments of type   
                             // arg3 and returns a pointer to a function that takes two 
                             // arguments of type arg1 and arg2 and returns an int

How to read int (*f4(arg3))(arg1, arg2);

          f4                           -- f4
        f3(   )                        -- is a function
        f3(arg3)                       --  taking an arg3 argument
       *f3(arg3)                       --   returning a pointer
     (*f3(arg3))(    )                 --   to a function
    (*f3(arg3))(arg1, arg2)            --     taking arg1 and arg2 parameter
  int (*f3(arg3))(arg1, arg2)           --     and returning an int

So, finally a home work :). Try to figure out the declaration

void (*signal(int sig, void (*func)(int)))(int);

and use typedef to redefine it.

Grzegorz Szpetkowski · Accepted Answer · 2015-02-18 13:10:59Z

3

From cdecl (which is a handy helper tool to decipher C declarations):

int (*fn_pointer ( this_args1 ))( this_args2 )

declare fn_pointer as function (this_args1) returning pointer to function (this_args2) returning int

Hence the former is a function, that returns pointer to function, while the latter:

return_type (*fn_pointer ) (arguments);

is an ordinary "pointer to function".

Read more about undestanding complex declarations from the Clockwise/Spiral Rule article.

edited Feb 18, 2015 at 13:10

answered Feb 18, 2015 at 12:44

Grzegorz Szpetkowski

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1 Comment

Wintermute Over a year ago

fn_pointer is not a pointer in this declaration (just a function).

Iharob Al Asimi · Accepted Answer · 2015-02-18 13:04:56Z

2

This

int (*fn_pointer ( this_args1 ))( this_args2 )

declares a function that takes as arguments this_args1 and returns a function pointer of type

int (*fn_pointer)(this_args2)

so it's just a function that returns a function pointer.

answered Feb 18, 2015 at 13:04

Iharob Al Asimi

53.1k6 gold badges66 silver badges100 bronze badges

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