Function Pointer Initialization in C/C++

Question

As I was going through a ESP IDF's documentation; I saw that a function pointer was initialized in a certain way that does not make sense to me.

typedef void *app_driver_handle_t;

app_driver_handle_t app_driver_light_init();
app_driver_handle_t app_driver_button_init();

Etc.

I thought that in order to initialize a function pointer, you must do it the following way:

app_driver_handle_t = app_driver_button_init();

Sorry for my beginner level questions.

It would do wonders if someone could explain this.

Thanks

Where is the function pointer? I see only a pointer to `void` mentioned, and there are no objects of that type in your code. — JaMiT, Oct 01 '22 at 18:05
I think the reason this does not make sense to you is that what you see is not what is actually written. Since we can only see what is actually written, answers require guesses of what you see. It might be a good idea to go into your interpretation of this code in more detail. For example, what do you think `app_driver_handle_t` is? What about `app_driver_light_init`? I suspect one or both of these are off, and yet you asked your question with the implicit assumption that your interpretation of these is correct. — JaMiT, Oct 01 '22 at 18:12
`app_driver_handle_t` is an alias for `void*` and not for a function pointer. Refer to a [good c++ book](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) which are also available as PDFs for free. — Jason, Oct 01 '22 at 18:21

score 0 · Accepted Answer · answered Oct 01 '22 at 18:09

Let's break down the code you're looking at.

typedef void *app_driver_handle_t;

This is not a function pointer. This is a void pointer, which means it can point to basically any values. And this is a type, not a value, so app_driver_handle_t does not actually contain any pointers at all; it's merely a name that's synonymous with void*.

app_driver_handle_t = app_driver_button_init();

Given the typedef above, this syntax is never valid. You're setting a type equal to what is presumably a function call. You can't assign to types. Full stop.

What you can do is declare variables and assign them the result of function calls.

app_driver_handle_t my_variable;
my_variable = app_driver_button_init();

or you can do it in one line.

app_driver_handle_t my_variable = app_driver_button_init();

Finally, these last two lines.

app_driver_handle_t app_driver_light_init();
app_driver_handle_t app_driver_button_init();

These are also not function pointers. These are function prototypes. They're a promise to the compiler, saying, "I will eventually define two functions called app_driver_light_init and app_driver_button_init. These two functions will take no arguments and will return a void*". There's still no function pointer happening here. The functions, when they're eventually defined, will return a void*, which, again, is not a function pointer but a pointer to void.

score 0 · Answer 2 · answered Oct 01 '22 at 18:09

This is effectively same as

void * app_driver_light_init();
void * app_driver_button_init();

These are not function pointers, they are functions that return void*.

A typedef'd fucntion pointer would look like this:

typedef void* (*app_driver_light_init)();

Function Pointer Initialization in C/C++

2 Answers2