In the firmware that I am writing, I created a type of variable. Similar to what is below:
struct SitemMenu {
unsigned int id;
char * text;
void * submenu
}
typedef struct SitemMenu TitemMenu;
Be any function:
void functionX () {
...
}
If I create this variable:
TitemMenu itemWhatever;
and do:
itemWhatever.submenu = &function (X);
Can I call functionX doing:
(*itemWhatever.submenu)();
I did something similar to this and the compiler give this answer:
error: (183) function or function pointer required
Yes you can, but not quite the way you've written it.
A function pointer is not declared in quite the same way as a 'normal' pointer.
What you need is:
struct SitemMenu {
unsigned int id;
char * text;
void (* submenu)(void); // this is a function pointer, as opposed to the 'normal' pointer above
};
typedef struct SitemMenu TitemMenu;
TitemMenu itemWhatever;
then, if you have some function declared with the same parameters and return type, like:
void functionX(), then you can do:
itemWhatever.submenu = functionX;
itemWhatever.submenu();
Related
I declared a struct like this one :
typedef struct s_data {
char buff[2048];
int len;
void *func[10];
struct data *next;
} t_data;
In my code, when passing a *data, I assigned some functions (just giving one so it is more understandable)
void init_data(t_data *data)
{
data->len = 0;
data->func[0] = &myfirstfunctions;
//doing the same for 9 others
}
My first function would be something taking as argument *data, and an int.
Then, I try to use this function in another function, doing
data->func[0](data, var);
I tried this and a couple of other syntaxes involving trying to adress (*func[0]) but none of them work. I kind of understood from other much more complex questions over there that I shouldn't store my function like this, or should cast it in another typedef, but I did not really understand everything as I am kind of new in programming.
void* can only be used reliably as a generic object pointer ("pointer to variables"). Not as a generic function pointer.
You can however convert between different function pointer types safely, as long as you only call the actual function with the correct type. So it is possible to do just use any function pointer type as the generic one, like this:
void (*func[10])(void);
...
func[0] = ((void)(*)(void))&myfirstfunction;
...
((whatever)func[0]) (arguments); // call the function
As you might note, the function pointer syntax in C is horrible. So I'd recommend using typedefs:
typedef void genfunc_t (void);
typedef int somefunc_t (whatever*); // assuming this is the type of myfirstfunction
Then the code turns far easier to read and write:
genfunc_t* func [10];
...
func[0] = (genfunc_t*)&myfirstfunction;
...
((somefunc_t*)func[0]) (arguments);
If all of your functions will have the same signature, you can do this like:
#include <stdio.h>
typedef void (*func)(void *, int);
struct s_data {
char buff[2048];
int len;
func f[10];
struct s_data *next;
};
static void
my_first_function(void *d, int x)
{
(void)d;
printf("%d\n", x + 2);
}
static void
init_data(struct s_data *data)
{
data->len = 1;
data->f[0] = my_first_function;
}
int
main(void)
{
struct s_data d;
init_data(&d);
d.f[0](NULL, 5);
return 0;
}
If your functions have different signatures, you will want to either use a union, or perhaps you will need several different members of the struct to store the function pointers.
The problem is that you haven't actually declared an array of function pointers. What you actually did is an array of pointers to void.
The syntax of declaring a pointer to function is as following:
function_return_type (*pointer_name)(arg1_type,arg2_type,...);
Then you can create an array of pointers to functions:
function_return_type (*arr_name[])(arg1_type, arg2_type,...)
Therefore, the declaration of your structure should look like this:
typedef void (*pointer_to_function)(void *, int);
struct s_data {
char buff[2048];
int len;
pointer_to_function array_of_pointeters[10];
struct s_data *next;
};
Good luck:)
I am learning C and trying to pass structure to call back function. Gone through online resources but unable to pass structure to call back function. Here is my code.
// myvariables.h
struct callbackStruct
{
int a;
int b;
int c;
};
extern struct callbackStruct callbackStructObject;
typedef void (*callback)(struct callbackStruct);
extern void callback_reg(callback pointerRefCallback);
// Operations.c
struct callbackStruct callbackStructObject;
void callback_reg(callback pointerRefCallback) {
(*pointerRefCallback)(callbackStructObject);
}
// main.c
struct callbackStruct myCallbackStruct1;
void my_callback(struct callbackStruct myCallbackStruct) {
printf("A value:%d" + myCallbackStruct.a);
}
int main()
{
callback ptr_my_callback = my_callback(myCallbackStruct1);
callback_reg(ptr_my_callback);
return 0;
}
Can anyone resolve this scenario?
The type callback is a function pointer type, e.g. a variable of that type is a pointer that points to a function that accepts a struct callbackStruct as single parameter:
typedef void (*callback)(struct callbackStruct);
So when you write (in main):
// ...
callback ptr_my_callback = // HERE
// ...
The expression at HERE must be the address of a function with the correct signature. You write:
my_callback(myCallbackStruct1);
Since my_callback is a function that returns nothing (void) and the above expression calls this function, the expression
callback ptr_my_callback = my_callback(yCallbackStruct1);
is not well formed (syntactically, as well as from the perspective of the type system).
Assuming that my_callback is the function that you want to work as a callback, you need to store its address in the pointer ptr_my_callback:
callback ptr_my_callback = &my_callback;
It's kind of unclear what your code is supposed to achieve, though, so I cannot really help you any further.
I have two functions, each taking a pointer to a different type:
void processA(A *);
void processB(B *);
Is there a function pointer type that would be able to hold a pointer to either function without casting?
I tried to use
typedef void(*processor_t)(void*);
processor_t Ps[] = {processA, processB};
but it didn't work (compiler complains about incompatible pointer initialization).
Edit: Another part of code would iterate through the entries of Ps, without knowing the types. This code would be passing a char* as a parameter. Like this:
Ps[i](data_pointers[j]);
Edit: Thanks everyone. In the end, I will probably use something like this:
void processA(void*);
void processB(void*);
typedef void(*processor_t)(void*);
processor_t Ps[] = {processA, processB};
...
void processA(void *arg)
{
A *data = arg;
...
}
If you typedef void (*processor_t)(); then this will compile in C. This is because an empty argument list leaves the number and types of arguments to a function unspecified, so this typedef just defines a type which is "pointer to function returning void, taking an unspecified number of arguments of unspecified type."
Edit: Incidentally, you don't need the ampersands in front of the function names in the initializer list. In C, a function name in that context decays to a pointer to the function.
It works if you cast them
processor_t Ps[] = {(processor_t)processA, (processor_t)processB};
By the way, if your code is ridden with this type of things and switch's all over the place to figure out which function you need to call, you might want to take a look at object oriented programming. I personally don't like it much (especially C++...), but it does make a good job removing this kind of code with virtual inheritance.
This can be done without casts by using a union:
typedef struct A A;
typedef struct B B;
void processA(A *);
void processB(B *);
typedef union { void (*A)(A *); void (*B)(B *); } U;
U Ps[] = { {.A = processA}, {.B = processB} };
int main(void)
{
Ps[0].A(0); // 0 used for example; normally you would supply a pointer to an A.
Ps[1].B(0); // 0 used for example; normally you would supply a pointer to a B.
return 0;
}
You must call the function using the correct member name; this method only allows you to store one pointer or the other in each array element, not to perform weird function aliasing.
Another alternative is to use proxy functions that do have the type needed when calling with a parameter that is a pointer to char and that call the actual function with its proper type:
typedef struct A A;
typedef struct B B;
void processA(A *);
void processB(B *);
typedef void (*processor_t)();
void processAproxy(char *A) { processA(A); }
void processBproxy(char *B) { processB(B); }
processor_t Ps[] = { processAproxy, processBproxy };
int main(void)
{
char *a = (char *) address of some A object;
char *b = (char *) address of some B object;
Ps[0](a);
Ps[1](b);
return 0;
}
I used char * above since you stated you are using it, but I would generally prefer void *.
I listed some example code below and the question is if there is a way for the function_name to access the value of number from struct_name?
typedef struct struct_name {
int number
void (*func)();
} * struct_name_ptr;
void function_name() {
//access number from struct
}
main() {
struct_name_ptr newobject;
newobject->func=&function_name;
newobject->func(); //can it print the value of the number in the structure above?
}
Uh - no.
A struct can certainly contain a function pointer. But the function you call wouldn't have any knowledge of the struct. Unless you passed a pointer as a function argument, or made the struct global.
With my limited knowledge of programming, I don't think this is possible. Though the struct contains a function pointer, the address of the function assigned to it is different and I don't think there will be anyway for it to access it unless you pass it as an argument.
Well, two things, struct_name->number should have a value, and it either needs to be in the same scope as &function_name or it needs to be explicitly passed. Two ways to do it:
/* Here is with a global calling struct */
#include<stdio.h>
typedef struct struct_name {
int number;
void (*func)();
} * struct_name_ptr;
struct struct_name newobject = { 0 };
void function_name() {
printf("%d",struct_name);
}
void main() {
struct struct_name_ptr newobject;
newobject->func=&function_name;
newobject->func();
}
/* And one with a modified function_name */
#include<stdio.h>
typedef struct struct_name {
int number;
void (*func)();
} * struct_name_ptr;
void function_name(struct_name) {
printf("%d",struct_name);
}
void main() {
struct struct_name_ptr newobject;
newobject.number = 0;
newobject->func=&function_name;
newobject->func(newobject);
}
No, a pizza won't ever know what the pizza delivery guy, who delivered it, looks like.
A regular function is just an address in memory. It can be called using a function pointer like in this case. In any case: The function won't know how it was called. In particular it won't know that it was called using a function pointer that's part of (a piece of memory corresponding to) some struct.
When using a language with classes like C++, member functions will have a hidden argument which is a pointer to the class instance. That's how member functions know about their data.
You can 'simulate' a simple OOP in plain C, for your example like:
typedef struct {
int number;
void (*func)();
} class;
void function_name(class *this) {
printf("%d",this->number);
}
#define CALL(c,f) c.f(&c)
int main() {
class object={12345,function_name};
CALL(object,func); // voilá
}
I declare a new struct with the name of "Struct"
I have a generic function that takes in an argument "void *data".
void Foo(void *data)
I pass an instance of "Struct" into the generic function.
Struct s;
Foo(&s);
I want to access one of the properties of the struct in the function.
void Foo(void *data) {
char *word = (char*) data.word;
}
It's not allowed because it doesn't recognize data as a valid struct.
I even try to declare the data as the struct type first, and I get an error.
void Foo(void *data) {
Struct s = (Struct) data;
char *word = s.word;
}
I get "conversion to non-scalar type requested".
First of all, you should turn on your compiler's warning flags (all of them). Then you should pass a pointer to your Struct and use something other than struct as a variable name:
Struct s;
Foo(&s);
Then, in Foo:
void Foo(void *data) {
Struct *s = data;
char *word = s->word;
}
You can't convert non-pointer types to and from void* like you're trying to, converting pointer types to and from void* is, on the other hand, valid.
You need to pass a pointer to you struct and get a pointer to the struct inside the function:
Struct struct;
Foo(&struct);
void Foo(void *data) {
Struct* struct = (Struct*) data;
char *word = struct->word;
}
You have to use -> operator when requesting structure member via pointer.
This should work: char *word = (char*) data->word;
You also have to pass the address of the structure to the function. Like this: Foo(&struct);.
Firstly you need to call the function correctly:
Struct s;
Foo(&s);
Notice you're now passing a pointer to the structure.
Now, the function has to know that you're using a Struct (as opposed to something else) - perhaps because of another parameter, or a global variable, or some other reason. Then inside the function you can do:
void Foo(void *data) {
Struct *structpointer = p; /* Note - no need for a cast here */
/* (determine whether data does refer to a pointer then...) */
char *word = structpointer->word;
/* ... then use 'word'... */
}
Data is pointer, so whatever you cast it to must also be a pointer. If you said Struct* myStruct = (Struct*) data, all would be well with the world.
You are mixing pointers and data.
Struct struct defines a data object
void *data expects data to be a pointer.
Call Foo with a pointer to a Struct, and make other necessary changes
Struct struct;
Foo((void*)&struct);
void Foo(void *data) {
Struct *struct = (Struct*)data;
char *word = struct->word;
}
or the more compact form:
Struct struct;
Foo((void*)&struct);
void Foo(void *data) {
char *word = ((Struct*)data)->word;
}