pointers to function question

say I have two compare functions one is a compareInt(char* s1, char* s2) and compareString(char* rp1, char* rp2)

my instructor used this typedef:

typedef int (*COMPAREFUNC)(void*, void*);
COMPAREFUNC compare;

can someone explain what this typedef means??

here's what he does when we wan't to compare int:

Code:

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compare = (COMPAREFUNC) &compareInt

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and if we want to compareString

Code:

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compare = (COMPAREFUNC)& compareString

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I don't get all this, can someone help me

Code:

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

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This creates a typedef that says "type COMPAREFUNC is a function pointer to a function that returns an int and takes 2 arguments, the first being of type void* and the second of type void*".

Quote:

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Originally Posted by -EquinoX-

COMPAREFUNC compare;

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What this does is define a variable that holds the address to such a function that COMPAREFUNC is a typedef for.

Code:

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compare = (COMPAREFUNC) &compareInt

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Code:

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compare = (COMPAREFUNC)& compareString

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What these do is just assign the specific functions addresses to the pointers so you can call them later via the pointer.
However, I would suggest you drop the (COMPAREFUNC) part because it shouldn't be needed. It can also mask a compile error if you try to assign the address of an incompatible function.

It's so you can do this:

Code:

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compare = compareInt;
compare(a_thing, another_thing); //calls compareInt
compare = compareString;
compare(something, something_else); //calls compareString

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http://www.cprogramming.com/tutorial...-pointers.html

Quote:

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Originally Posted by robwhit

What are comparetInt and compareString?

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some functions with the following prototypes

Code:

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int compareInt(char* s1, char* s2); /* I do not really know why int comparator wants char* as parameters */
int compareString(char* rp1, char* rp2);

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Yeah, I missed that. :) Equinox: See my edit. :)

so basically when I do :

COMPAREFUNC compare;

it just basically takes the address of the typedef that I already declared before??

Not even close, especially since a typedef is not something that can have an address. You are declaring a variable of type COMPAREFUNC.

It's just a variable with a type. You aren't assigning anything.

Code:

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compare = (COMPAREFUNC) &compareInt
compare = (COMPAREFUNC)& compareString

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Why use the address-of operator here? I would think that would generate the address of the address of the functions.

Todd

Nope.
someFunc and &someFunc is the same thing.

Quote:

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Why use the address-of operator here? I would think that would generate the address of the address of the functions.

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No. its just pointer to function, but the address operator is optional, because the function is automatically casted to the pointer to function type in this context

This isn't a good example of abstraction because it relies on char * and void * having the same alignment and representation (which is guaranteed). Change it to any other pointer type, and you broke your code.

Hence also why casting to a function pointer is bad because it masks such errors.
You should always use a correct function type. Adapt the actual function declaration.

so COMPAREFUNC here is just like a type? something like an int

Correct.

yes. a pointer to a function returning an int and taking two void pointers as parameters.

so can have a type that returns something and takes something as the parameter? this seems weird to me..

no, the type is a pointer. the type that the pointer points to is a function. that function is a type that returns an int and takes two void *s as parameters.

Code:

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#include <stdio.h>

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

/* ^
 * |Notice the similarity in these two signatures.
 * |Same return type, same type and number of arguments.
 * v
 */

int func(void*param1,void*param2);

int main()
{
        COMPAREFUNC my_func_pointer;
        char a=1,b=0,c=0,d=0;

        func(&a,&b); /* prints "in func" */
        my_func_pointer = func; /* my_func_pointer now points to func */
        my_func_pointer(&c,&d); /* prints "in func" */

        return 0;
}

int func(void *param1, void *param2)
{
        puts("in func");
        return 0;
}

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okay so we can do a typedef for such thing? not only for int, float, etc.. but also something that takes a parameter and returns something?

well, you can do it for a /pointer/ to a function (or a pointer to a pointer to a function or...), but I don't think you can do it for a function.

see my edit in my last message for an example.

yeah I understand your example... but yet still I am not still sure somehow

Wow, talk about complicated stuff. I'll save this thread to my hard drive. I never know when I could use answers to an extremely intelligent question like this one :)

Quote:

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Originally Posted by -EquinoX-

yeah I understand your example... but yet still I am not still sure somehow

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What is it specifically you don't understand?

The typedef simply means you can write this

COMPAREFUNC my_func_pointer;

As opposed to this
int (*my_func_pointer)(void*, void*);

One you can live with, but the syntax is heavy on the parentheses and verbose with all the parameter types, which can be hard on the fingers if you need lots of them. The typedef saves you from all that.


Since no one has mentioned it yet, http://www.newty.de/fpt/index.html

maybe I am just not used to the typedef that can take parameters..that's all my confusion is

Quote:

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Originally Posted by -EquinoX-

maybe I am just not used to the typedef that can take parameters..that's all my confusion is

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The typedef doesn't take parameters. However, the type itself is a function pointer, and part of a function pointer is the arguments to the function. Yes, it takes a bit to get used to. If I haven't used function pointers for a bit, I have to look up and/or think about where all the parenthesis and stars belong.

Here's one that takes a bit of thinking.

Code:

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typedef int *(*funcptr)(int (*func1)(void));

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It does get a bit simpler if you do two typedefs:

Code:

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typedef int (*subfunc)(void);
typedef int *(*funcptr)(subfunc func1);

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That's a function pointer (called funcptr) that returns a pointer to int, and takes a function pointer that returns int, which takes no arguments.

--
Mats

Function pointers are obscure, but when you're used to them then you'll see the power :-)

Code:

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static unsigned char (*const GpsHandler[NUM_FUNS][NUM_TPS])(void (*GpsMsgs)(void)) = {...};

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Is and array of function pointers where functions return unsigned char and have an argument which is function pointer which returns void and has void as an argument.

To add another thing into the mix. I usually do:

typedef int (COMPAREFUNC)(void*, void*);
(As opposed to
typedef int (*COMPAREFUNC)(void*, void*);
Note the missing *)

So that when I define a function pointer, I can do:

COMPAREFUNC* pFnc;

It's now easier to see it's actually a pointer.
Now that it's bad to put the * in the type, but I prefer the second way. Either one is possible and both just as good.

thanks guys for helping me.. can you give me some other typedef examples of a function pointer

Typical Windows API:

Code:

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typedef int bool;
#define true 1
#define false 0

bool callback(int id)
{
    printf("%i\n");
    return true;
}

typedef bool (callbackptr)(int);

void example(callbackptr* pcallback)
{
    int i = 0;
    for (; i < 100; i++)
    {
        if (! pcallback(i) ) /* If function returns false, then break the loop */
            break;
    }
}

int main(void)
{
    example(&callback);
    return 0;
}

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Will print 1 - 100 (or should). The callback function can cancel by returning false instead of true.

Quote:

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Originally Posted by -EquinoX-

thanks guys for helping me.. can you give me some other typedef examples of a function pointer

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Did you try the site mentioned earlier?

Quote:

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Originally Posted by Salem

Since no one has mentioned it yet, http://www.newty.de/fpt/index.html

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Elysia

in your example is the function actually taking another function as the paramter?