How to know if a variable is a function pointer in C?

Question

I have a function of type void(*func)(int) and I get a pointer to this function, how can I know for sure that a pointer is really of this type?

if (typeof(pointer) == void(*func)(int)) {
    //do something     
}

Is that correct?

edit - I get the pointer from a function as an argument and I want to check the type of the pointer inside this function.

If you have a `void (*pointer)(int)`, then its type *is* `void (*)(int)`. There's nothing to check. Why do you think it would be any different? — Kevin, May 08 '20 at 15:59
C is a statically typed language. By the time it gets into the variable it is a pointer of the specified type. — Raymond Chen, May 08 '20 at 16:05
Does this answer your question? [How do I check if a variable is of a certain type (compare two types) in C?](https://stackoverflow.com/questions/6280055/how-do-i-check-if-a-variable-is-of-a-certain-type-compare-two-types-in-c) — RobertS supports Monica Cellio, May 08 '20 at 16:06
@DeanTaler: you can accept one of the answers by clicking on the grey checkmark below its score — chqrlie, May 11 '20 at 19:56

score 4 · Answer 1 · edited May 08 '20 at 16:21

4

if you are getting your function pointer dynamically , so try to use union to and a flag variable to get it's type without checking something complex

#define type1 1
#define type2 2
union FN
{
void ....
int ....
....
}

typedef struct{
int type;

union FN fn;
} Function;

edited May 08 '20 at 16:21

Ardent Coder

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answered May 08 '20 at 16:04

1

Interesting. Could you add some sample code to your answer to make it more complete and not just a single sentence answer? – pastaleg May 08 '20 at 16:07
I´m not quite sure what you mean but note that the OP wants to have the check at runtime, not compile-time. – RobertS supports Monica Cellio May 08 '20 at 16:10
check it after editing ,it's useful for more dynamically ,but to be honest i don't know where he wants to get the pointer function from – May 08 '20 at 16:15
but I want to chek the pointer inside a fucntion that gets the pointer, so it looks more complicate to do it in this way, is there no easier option? – Dean Taler May 08 '20 at 16:17
`void(*func)(int)` is like checking if `type==TYPE1` – May 08 '20 at 16:20

score 3 · Answer 2 · answered May 08 '20 at 16:04

typeof not a part of the C standard. It is a GNU extension.

If you want to be sure that something is of a particular type, you can handle it at compile-time with _Generic introduced in C11:

_Generic((pointer), 
    void(*)(int): /* insert your code here */,
)

If said pointer is not of type void(*)(int), then it will fail to compile.

chqrlie · Answer 3 · 2020-05-09T18:55:12.350

If your function gets the pointer as an argument and has this prototype:

void my_function(void (*func)(int)) {
    // How can I test if `func` really points to a function taking an `int`?
}

Then there is not much you can do inside the body of my_function to verify what func really points to.

you can test if it is a null pointer:
```
if (!func) { /* NULL was passed */ }
```

you can test if it points to a known function:

if (func == exit) { /* haha! (*func)(1) will abort */ }

beyond the above, it is impossible to tell what was actually passed to my_function: it could be an actual function with the correct prototype, or another function implicitly or explicitly cast to the expected type or even some other scalar object with a cast.

If you want to write a macro whose expansion depends on the type of an expression, you can use the new C11 _Generic construction:

#define IS_VOID_FUNC_OF_INT(func)  _Generic(func, void(*)(int): 1, default: 0)

But using this inside the body of function my_function above will always evaluate to 1: only the defined type of func is tested by _Generic, not that of the original expression used as an argument to the function.

How to know if a variable is a function pointer in C?

3 Answers3