+ 1

Function pointer

can anybody provide me some sniptes of function pointer?

16th Jun 2018, 7:23 AM
Azam
Azam - avatar
12 Answers
+ 3
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);
16th Jun 2018, 8:24 AM
Vlad Serbu
Vlad Serbu - avatar
+ 8
~ 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. ;-]
17th Jun 2018, 12:44 PM
Babak
Babak - avatar
+ 6
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
16th Jun 2018, 10:00 PM
Babak
Babak - avatar
+ 5
~ 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
17th Jun 2018, 12:51 PM
Baptiste E. Prunier
Baptiste E. Prunier - avatar
+ 4
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.
16th Jun 2018, 11:35 AM
Babak
Babak - avatar
+ 4
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.
16th Jun 2018, 12:59 PM
Babak
Babak - avatar
+ 3
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; }
16th Jun 2018, 8:14 AM
Baptiste E. Prunier
Baptiste E. Prunier - avatar
+ 1
You can't remove it in the prototype of f1 C++ Soldier (Babak)
16th Jun 2018, 6:01 PM
Baptiste E. Prunier
Baptiste E. Prunier - avatar
0
actually I want pass a function as parameter to a 2nd function. 1st function will be called in 2nd function.
16th Jun 2018, 8:17 AM
Azam
Azam - avatar
0
ok I'll try to solve my problem with this solution. thanks a lot😊
16th Jun 2018, 8:27 AM
Azam
Azam - avatar
0
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
16th Jun 2018, 12:28 PM
Baptiste E. Prunier
Baptiste E. Prunier - avatar
0
Type in a code to declare a function that takes and returns void pointers. void * foo(void *);
7th Oct 2022, 6:55 AM
Deepak Solanke
Deepak Solanke - avatar