Give non-declared function to function pointer

Question

I am trying to understand all this HUGE concept of function pointers...

Is it possible to initalize function pointer with non-declared function?

For example, a simple use of function pointer would be:

void f()
{
  printf("Hello world!");
}
//Somewhere in the program
void (*foo)();
foo = &f;

Is there any way to skip the declaration of f? something like that:

void (*foo)();
foo = { printf("Hello world!"); }

In `C`, no. `C++` has the concept of lambdas that fit what you're trying to do. — Jonesinator, Sep 05 '17 at 19:38
In C, you need to make it a named function (probably a `static` one, unless you intend to reuse it from other compilation units) . — Petr Skocik, Sep 05 '17 at 19:51
In C (as well as C++) you haved to declare a name before using it. — too honest for this site, Sep 05 '17 at 19:55

score 4 · Accepted Answer · edited Sep 05 '17 at 20:06

4

What you describe are called lambda functions, a mechanism which is not available in C. You can think of lambda functions as anonymous functions.

edited Sep 05 '17 at 20:06

Jonathan Leffler

answered Sep 05 '17 at 19:39

gsamaras

1 Answers1