Function pointer

Question

can anybody provide me some sniptes of function pointer?

Answer 1

int foo(int arg);
int bar(int arg);

int callfunc(int (*func) (int), arg){
return func(arg);
}
int (*x) (int);
x = foo;
callfunc(x, 3); //same as foo(3);
x = bar;
callfunc(x, 3); //same as bar(3);

Answer 2

~ swim ~
Thanks for taking your time and made it for us as clear as possible. Personally, I don't know anything about C++ when it comes to fine details!. But most of the times I do my best to provide something useful. Thanks again. ;-]

Answer 3

As I already mentioned, the context is totally different. Just to make sure that I inadvertently didn't something wrong, I put together the following code as a proof that the * sign is optional even in class interface where you place all method prototypes. I even test it on three different compiler (G++, MinGW, MSVC++) with separate .h and .cpp sources. Hopefully your doubt is vanished Baptiste E. Prunier .

https://code.sololearn.com/c9fSx9vhpr3H/?ref=app

Answer 4

~ swim ~ Thanks for clearing the subject ! I did not knew it was possible to remove the '*' when declaring the type
I come from C and I never saw it that way. It might work in C too, I'll do some experimenting on it

Answer 5

Azam

class A {
public:

void f1(int (*fx)(int n), int val) {

int dx = (*fx)(val);
// ...
}
};

class B {
public:

static int f(int n) {

return n * n + 2 * n + 1;
}
};

int main() {

A a;
int number = 5;
a.f1(B::f, number);
}

Btw, * sign in (*fx) is optional.

Answer 6

Baptiste E. Prunier

The context is different.
In my case, since I merely pass around function's name as a pointer, there's no need to declare it in main or anywhere else.

Answer 7

a function pointer is declared like that :
return_type (*variable_name)(list of parameters);
you can then assign it a function. Let's do an example :
#include <stdio.h>
int max(int a, int b){
return a > b ? a : b;
}
int main(){
int (*m)(int, int);

m = max;
// or
m = &max;

printf("%d\n",m(1,2));

return 0;
}

Answer 8

You can't remove it in the prototype of f1 C++ Soldier (Babak)

Answer 9

actually I want pass a function as parameter to a 2nd function.
1st function will be called in 2nd function.

Answer 10

ok I'll try to solve my problem with this solution.
thanks a lot😊

Answer 11

just to add to what C++ Soldier (Babak) said, * is optional when you USE fx, and only when you use it, it is mandatory when you declare it

Answer 12

Type in a code to declare a function that takes and returns void pointers.

void * foo(void *);