I have doubts about the use of Volatile Keyword in function pointers.
I don't know if in this case, function pointers should be volatile
I use function pointers so that the same function, for example sendCommand() can use different uart's functions.
My code would be like this
//Function pointer
void (*volatile ptr_uart_putc) (unsigned char);
//Set function pointer
void uartConfig( void(*aPtr_uart_putc)(unsigned char) ){
ptr_uart_putc = aPtr_uart_putc ;
}
void sendCommand(unsigned char aCommand){
ptr_uart_putc(aCommand);
}
So in main I would do:
main(){
uartConfig(uart0_putc);
sendCommand('a');
uartConfig(uart1_putc);
sendCommand('b');
}
Related
#include <stdio.h>
void test1(){ printf("test1\n"); }
void test2(){ printf("test2\n"); }
void test3(){ printf("test3\n"); }
void test4(){ printf("test4\n"); }
void (*hv_ptr[2])() = {test1,test2};
void (*dev_ptr[2])() = {test3,test4};
void (**ptr[2])() = {hv_ptr, dev_ptr};
int main() {
ptr[0][0];
ptr[0][1];
ptr[1][0];
ptr[1][1];
return 0;
}
I have declared the array of pointers to function pointers. But with this code, functions are not getting called. Kindly let me know what is going wrong here.
ptr is an array of pointers to pointers to functions, so ptr[0] is a pointer to a pointer to a function, so ptr[0][0] is a pointer to a function.
Thus, in the expression statement ptr[0][0];, the expression is just a pointer to a function. It does not contain a function call.
ptr[0][0](); is a call to the function pointed to by ptr[0][0].
What I would like to do is :
a simple function "register_function" allowing a user to indicate a callback function to call (that this user has implemented)
when calling that function with "call_callback", to be able to pass arguments like a buffer for example
Here is my code:
static void (*callback)(int *buffer, int size) = NULL;
void register_callback(void (*ptr)())
{
(*callback) = ptr;
}
void call_callback(int *buffer, int size)
{
(*callback)(buffer, size);
}
The problem is that I have a compile-time error in the register_callback declaration.
In register_callback() the assignment is just:
void register_callback(void (*ptr)( int *buffer, int size ))
{
callback = ptr;
}
Please note that I also added the parameter declarations to the definition of ptr so the compiler can check for the correct function pointer type passed (assuming you have an identical prototype)
I was reading this code on a website. I am fairly new to programming so please explain in a bit more detail.
#include <stdio.h>
// A normal function with an int parameter
// and void return type
void fun(int a)
{
printf("Value of a is %d\n", a);
}
int main()
{
// fun_ptr is a pointer to function fun()
void (*fun_ptr)(int) = &fun;
/* The above line is equivalent of following two
void (*fun_ptr)(int);
fun_ptr = &fun;
*/
// Invoking fun() using fun_ptr
(*fun_ptr)(10);
return 0;
}
Doubts-
I am not able to understand this type of declaration and assignment void (*fun_ptr)(int) = &fun;
I mean that if we declare a data type, then we do it like int a; and assign it as a=10; but here we are assigning it by writing (*fun_ptr)(10);. Kindly help.
Instead of this record
(*fun_ptr)(10);
you could just write
fun_ptr(10);
That is it is a function call of the function fun pointed to by the function pointer fun_ptr due to the initialization of that pointer in its declaration by the function address
void (*fun_ptr)(int) = &fun;
In turn this declaration could be written simpler like
void (*fun_ptr)(int) = fun;
because a function designator (in this case fun) used in expressions as for example an initializer is implicitly converted to pointer to the function.
You could use a typedef alias for the function type the following way
typedef void Func( int );
In this case the above declaration of the function pointer could look simpler like
Func *fun_ptr = fun;
Here is your program rewritten using a typedef for the function type of the function fun.
#include <stdio.h>
typedef void Func( int );
// Function declaration without its definition using the typedef
// This declaration is redundant and used only to demonstrate
// how a function can be declared using a typedef name
Func fun;
// Function definition. In this case you may not use the typedef name
void fun( int a )
{
printf("Value of a is %d\n", a);
}
int main(void)
{
// Declaration of a pointer to function
Func *fun_ptr = fun;
// Call of a function using a pointer to it
fun_ptr( 10 );
return 0;
}
Lets rewrite it a little bit, by using type-aliases and some comments:
// Define a type-alias names fun_pointer_type
// This type-alias is defined as a pointer (with the asterisk *) to a function,
// the function takes one int argument and returns no value (void)
typedef void (*fun_pointer_type)(int);
// Use the type-alias to define a variable, and initialize the variable
// This defines the variable fun_ptr being the type fun_pointer_type
// I.e. fun_ptr is a pointer to a function
// Initialize it to make it point to the function fun
fun_pointer_type fun_ptr = &fun;
// Now *call* the function using the function pointer
// First dereference the pointer, to get the function it points to
// Then call the function, passing the single argument 10
(*fun_ptr)(10);
Hopefully it makes things a little clearer what's going on.
Following is the meaning of two statments.
void (*fun_ptr)(int) = &fun; this is called declaring and initializing the fun_ptr in the same line , this is same as doing int a = 10;
(*fun_ptr)(10); is not assignment statement, it is invoking the function fun through function pointer fun_ptr.
you can also use typedef to create a new user defined out of function pointer and use as shown in above answer.
This is an advanced topic if you are new to programming, fun_ptr is a pointer to a function.
The declaration:
void (*fun_ptr)(int) = fun;
Means fun_ptr is a pointer to a function taking an int argument and returning void, initialize if with a pointer to the function fun. (you don't need the &)
The line:
(*fun_ptr)(10);
does not assign anything, it calls the function pointed to by fun_ptr, but is way to complex
fun_ptr(10);
As fun_ptr points to fun this is equivalant to `fun(10).
Using a function pointer has its use eg in a sort function where the comparison function is passed in as a function pointer so the sorting can be different between calls.
achieves the same thing and is much easier on the eyes.
I'm new to C and im currently learning about pointers.
I'm not sure why I am getting an error with the following sections of code in regards to pointers :
char ch;
char** pointer;
pointer = &ch;
and
int function1(void)
{
return 42.0;
}
void function2(void)
{
void (*pointer)(int);
pointer = &function1;
...
}
Any help will be appreciated :)
The very first problem is that you are using a double pointer in char** pointer ,as you are not storing the address of some other pointer so you should use char *pointer instead.
Then your function1 has return type as int but you are returning a float value ,although it won't give you any error but it can create some logical issues in your program,so better to properly write the return type in function definition and its prototype.
Then the next problem is in the function2,your function1 returns int but does not take any arguments but your function pointer return void and take int ,so you should better modify this to
int (*pointer)(void);
and then store the address of function1 in pointer ,it will work fine.
* is a single pointer and ** is a pointer to pointer.
So instead of
char** pointer;
It should be:
char* pointer;
In the second case, the function pointer prototype is not matching the prototype of the function it is pointing to.
So instead of
void (*pointer)(int);
it should be:
int (*pointer)(void);
you second section have some mistakes
you function1() return int and not take args
but your fucntion ptr return void and take int
so change it to:
int (*pointer)(void);
pointer = &function1;
Is it possible to declare some function type func_t which returns that type, func_t?
In other words, is it possible for a function to return itself?
// func_t is declared as some sort of function pointer
func_t foo(void *arg)
{
return &foo;
}
Or would I have to use void * and typecasting?
No, you cannot declare recursive function types in C. Except inside a structure (or an union), it's not possible to declare a recursive type in C.
Now for the void * solution, void * is only guaranteed to hold pointers to objects and not pointers to functions. Being able to convert function pointers and void * is available only as an extension.
A possible solution with structs:
struct func_wrap
{
struct func_wrap (*func)(void);
};
struct func_wrap func_test(void)
{
struct func_wrap self;
self.func = func_test;
return self;
}
Compiling with gcc -Wall gave no warnings, but I'm not sure if this is 100% portable.
You can't cast function pointers to void* (they can be different sizes), but that's not a problem since we can cast to another function pointer type and cast it back to get the original value.
typedef void (*fun2)();
typedef fun2 (*fun1)();
fun2 rec_fun()
{
puts("Called a function");
return (fun2)rec_fun;
}
// later in code...
fun1 fp = (fun1)((fun1)rec_fun())();
fp();
Output:
Called a function
Called a function
Called a function
In other words, is it possible for a function to return itself?
It depends on what you mean by "itself"; if you mean a pointer to itself then the answer is yes! While it is not possible for a function to return its type a function can return a pointer to itself and this pointer can then be converted to the appropriate type before calling.
The details are explained in the question comp.lang.c faq: Function that can return a pointer to a function of the same type.
Check my answer for details.
Assume the function definition
T f(void)
{
return &f;
}
f() returns a value of type T, but the type of the expression &f is "pointer to function returning T". It doesn't matter what T is, the expression &f will always be of a different, incompatible type T (*)(void). Even if T is a pointer-to-function type such as Q (*)(void), the expression &f will wind up being "pointer-to-function-returning-pointer-to-function", or Q (*(*)(void))(void).
If T is an integral type that's large enough to hold a function pointer value and conversion from T (*)(void) to T and back to T (*)(void) is meaningful on your platform, you might be able to get away with something like
T f(void)
{
return (T) &f;
}
but I can think of at least a couple of situations where that won't work at all. And honestly, its utility would be extremely limited compared to using something like a lookup table.
C just wasn't designed to treat functions like any other data item, and pointers to functions aren't interchangeable with pointers to object types.
what about something like this:
typedef void* (*takesDoubleReturnsVoidPtr)(double);
void* functionB(double d)
{
printf("here is a function %f",d);
return NULL;
}
takesDoubleReturnsVoidPtr functionA()
{
return functionB;
}
int main(int argc, const char * argv[])
{
takesDoubleReturnsVoidPtr func = functionA();
func(56.7);
return 0;
}
#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
typedef void *(*fptr)(int *);
void *start (int *);
void *stop (int *);
void *start (int *a) {
printf("%s\n", __func__);
return stop(a);
}
void *stop (int *a) {
printf("%s\n", __func__);
return start(a);
}
int main (void) {
int a = 10;
fptr f = start;
f(&a);
return 0;
}
It is not possible for a function to return itself by value. However it is possible itself to return by a pointer.
C allows to define function types that take undefined number of parameters and those those types are compatible with function types that take defined parameters.
For example:
typedef void fun_t();
void foo(int);
fun_t *fun = foo; // types are fine
Therefore the following function would work.
void fun(void (**ptr)()) {
*ptr = &fun;
}
And below you can find the exemplary usage:
#include <stdio.h>
void fun(void (**ptr)()) {
puts("fun() called");
*ptr = &fun;
}
int main() {
void (*fp)();
fun(&fp); /* call fun directly */
fp(&fp); /* call fun indirectly */
return 0;
}
The code compiles in pedantic mode with no warnings for C89 standard.
It produces the expected output:
fun() called
fun() called
There's a way, you just try this:
typedef void *(*FuncPtr)();
void *f() { return f; }
int main() {
FuncPtr f1 = f();
FuncPtr f2 = f1();
FuncPtr f3 = f2();
return 0;
}
If you were using C++, you could create a State object type (presuming the state machine example usage) wherein you declare an operator() that returns a State object type by reference or pointer. You can then define each state as a derived class of State that returns each appropriate other derived types from its implementation of operator().