Function pointers

[vkq123]

Hi can somebody explain me what this function pointer means...

void *(*(*fp1)(int))[10];
As I understand this means its is a pointer to a function which takes int as an argument and returns an array of 10 void pointers.

Is it correct?
And can somebody write a simple function of this type? I mean which takes int as an argument and returns array of 10 pointers? (I am not able to do so :( )

[Duthomhas]

It doesn't look valid to me...

If I were to guess, it is an array of ten pointers to function pointers of type void* (*fp1)(int).

[edit] Do you have any examples of how fp1 is used?

[writetonsharma]

pointer to function returning pointer to *array[10] type.


void *(*(*fp1)(int))[10];

these make it of type void** or *void[10] type.

second pointer from left make it pointer to *void[10].

and third one is the function pointer.

hopefully this is what it meant.

[vkq123]

Thanks for the reply guys..... but can anyone write a simple function which this fp1 can point to? I am not able to do so

Please check the code
#include <iostream>

using namespace std;
void ** MyFunc(int k)
{
void ** p = new void *[10] ;

cout << "Inside MyFunc" << endl;

return p;
}
int main(int argc, char* argv[])
{
void *(*(*fp1)(int))[10] = NULL;

fp1 = &MyFunc;
(*fp1)(1);

return 1;
}

ERROR:
error C2440: '=' : cannot convert from 'void **(__cdecl *)(int)' to 'void *(*(__cdecl *)(int))[10]'

this for the line fp1 = &MyFunc;

Can you help me run the code?

[writetonsharma]

your function looks fine..

but the problem is what you have written.. i also tried on the same lines and getting the same error.. how to declare a function of type void* (*(*)(int)[10]

if we cast it.. then the problem can be solved and the function pointer will run fine too.. but i dont know if thats the correct way..

i mean this:

typedef void *(*(*FP1)(int))[10];


fp1 = (FP1)MyFunc;
fp1(10);

by the way from where you have taken this thing??? :D

[Disch]

This is a little absurd but I figured it out:

// easy way .. typedef

typedef void* uglyrettype[10];
uglyrettype* func_a(int v) { /* ... */ }

// harder way, no typedef

void* (*func_b(int k))[10] { /* ... */ }

// functions return a pointer to a pointer to an array of 10 void pointers
//   ie:

uglyrettype* func_a(int v)
{
  uglyrettype* ret = (uglyrettype*)malloc(sizeof(uglyrettype));

  // *ret is 10 void pointers:
  int butt, booty;
  (*ret)[0] = &butt;
  (*ret)[1] = &booty;
  // etc

  return ret;

  // alternatively -- can do something like this
  static void* r2[10];
  return &r2;
}

///-----------------------------
//
void *(*(*fp1)(int))[10];
fp1 = &func_a;             // works 

I tried looking for ways to allocate the return value with new, but couldn't figure it out, so I resorted to malloc. void* x[10] is not the same as void** x for purposes of new, apparently, so it gets tricky.


EDIT:

Hrm... apparently I just tried this:

uglyrettype* r1 = new uglyrettype;    // does not compile
uglyrettype* r2 = new uglyrettype[10]; // does compile 

I don't know if that r2 is what you'd want though?

In any event -- whoever came up with this function pointer needs to be shot.

[writetonsharma]

:o....

damn... thats what i was looking for.. how to declare a function returning *(fun)[10].. never did it coz i can return a **...
hehehe....

but i may not agree on how you have allocated the memory..!!

anyway.. i learned how to make a function return *(fun)[].. :D

good one disch.

[Disch]

but i may not agree on how you have allocated the memory..!!

I don't like it either. But like I say, I couldn't find a way to make new work. uglyrettype* r2 = new uglyrettype[10]; compiled, but I think it allocates 10x more memory than you'd need.

[writetonsharma]

haahahaha.. .i know you also didnt liked it and thats why i didnt highlighted what i wrote...
but we can leave it.. what we wanted is solved though..


but i want to point one thing..
as you have typedef it, its already a pointer so you dont have to do this:

ugly *u;
i think this will do..

ugly    u;//allocated 10 pointers, do you agree?

*(u+0) = (char*)malloc(10);
*(u+1) = (char*)malloc(10);
strcpy(*(u+0),"Hello");
strcpy(*(u+1),"Hello");

printf("%s\n%s\n",*(u+0),*(u+1));

what you say??

[Disch]

as you have typedef it, its already a pointer so you dont have to do this:

The function returns a pointer to a pointer to an array of 10 pointers. So yeah -- you need the extra pointer there.

ugly u;//allocated 10 pointers, do you agree?

Yes -- but that's not what the function returns.

what you say??

You have no chance to survive make your time.

[writetonsharma]

you have put the function part into your pointer..!!!
the function will return and not your data type.. then this will make it **void[10].

You have no chance to survive make your time.

:o


anyway .. forget it..

[Disch]

you have put the function part into your pointer..!!!
the function will return and not your data type.. then this will make it **void[10].

I'm not sure I get what you're asking.

*(void*[10]) <-- what the function returns
void*[10] <-- what 'uglyrettype' is typedef'd as
*(void*[10]) <--- which makes this 'uglyrettype*'

which is why you return 'uglyrettype*' and not 'uglyrettype'

:o

You never heard of Zero Wing / All Your Base?

http://en.wikipedia.org/wiki/All_your_base

Captain: What you say !!
CATS: You have no chance to survive make your time.

[writetonsharma]

ha ha ha ha..!!!


never heard about the game.. though "all your base are belong to us" is a cheat in warcraft..!! hehehehe..

[vkq123]

Thanks guys. :)... i understand this was a very difficult one... and the culprit is none other than Bruce Ackels... in his Thinking in C++... :) ..... Hope u will help me the same way in future too :)