C: Passing any function to a function - c

I would like to know how I would go about passing any function to a function, as in a generic function pointer that can take any function whatsoever, The goal of this is to make a destructor system, so basically storing the function and calling it with it's arguments also stored later down the line,
Something like:
Defer(SDL_DestroyWindow, Window);
I already handled the arguments, but I don't know how to manage the function pointer part of this, Thank you!
Edit: I added more info ...
typedef struct {
void** args;
} IM_Defer_Resource;
/* Defer & Whatnot */
IM_Stack* IM_Defer_Stack;
void IM_Defer_Init() {
IM_Defer_Stack = IM_Stack_Init();
}
void IM_Defer(/* What to put here? */) {
}
void IM_Defer_All() {
while(IM_Defer_Stack->size) {
IM_Defer_Resource* resource = IM_Stack_Pop(IM_Defer_Stack);
if(!resource) continue;
/* What to do */
}
}
I don't have the actual functions of defer, but I did copy every argument into the stack and can pop them successfully, I don't know how to implement the variadic function calling though
Edit2:
After receiving some input: I think this would be more feasible:
Defer(SDL_DestroyWindow, "SDL_Window*", window);
I am brainstorming how this would be possible, but I would appreciate some input
Edit3:
/* Defer & Whatnot */
typedef struct {
char** types;
void** args;
int count;
} IM_Defer_Resource;
IM_Stack* IM_Defer_Stack;
void IM_Defer_Init() {
IM_Defer_Stack = IM_Stack_Init(IM_Get_Stack_Type(IM_Defer_Resource));
}
void IM_Defer_Internal(void* var, int n, ...) {
char* type;
void* arg;
va_list args;
va_start(args, n);
IM_Defer_Resource resource;
int count = n / 2;
resource->types = calloc(count, sizeof(char*));
resource->args = calloc(count, sizeof(void*));
resource->count = count;
for(count > 0; n -= 1) {
type = va_arg(args, char*);
resource->types[count-1] = type;
arg = va_arg(args, void*);
resource->args[count-1] = arg;
}
IM_Stack_Push(IM_Defer_Stack, &resource);
}
void IM_Defer_All() {
while(IM_Defer_Stack->size) {
IM_Defer_Resource* resource = IM_Stack_Pop(IM_Defer_Stack);
if(!resource) continue;
/* I have a char* and a void* to the resource, Now what? */
free(resource->types);
free(resource->args);
}
}
This is what I came up with, but I am wondering how I can conver that char* into a type...

As I said in comment a big problem is that when declaring a variadic function the undeclared parameters are subject to the default argument promotions. This means that you can find the passed arguments different from that intended by the function, that will eventually lead to exceptions. What you want to do is feasible, but really very complex.
One solution, but limited because requires a lot of coding, could be:
#include <stdarg.h>
#include <stdio.h>
typedef enum { fn1, fn2, fn3, /*....*/} e_fn;
void multi_fun(e_fn fn, ...)
{
va_list ap;
int j;
va_start(ap, fn); /* Requires the last fixed parameter (to get the address) */
switch(fn)
{
case fn1:
{
//suppose prototype for fn1 to be void fn1_fn(int, float, struct mystruct *);
int this_int = va_arg(ap, int);
float this_float = va_arg(ap, float);
struct mystruct *this_struct = va_arg(ap, struct mystruct *);
fn1_fn(this_int, this_float, this_struct);
break;
}
case fn2:
{
...
}
}
va_end(ap);
}

You should take a look at Fake Function Framework (fff) on GitHub. They've done this using macros for caching mock functions. MIT Licensed. However, just like #Frankie_C said, this requires a LOT of code. The header file that defines all of the macros is around 6K LOC. And functions are still limited to 20 arguments.

Related

In C language, is there any way to use callback with arbitrary / variable arguments?

I would like to send callbacks with different signatures for the same function. Somenthing like this:
#include <stdio.h>
#include <stdarg.h>
void a(int pa) {}
void b(int pb1, float pb2) {}
// exec implementation
int main() {
exec(a, 1);
exec(b, 1, 2.3);
}
I thought of using something like:
void exec(void (*func)(...), ...) {
int arg1;
float arg2;
va_list valist;
va_start(valist, size);
arg1 = va_arg(valist, int);
if (size == 1) {
(*func)(arg1);
va_end(valist);
return;
}
arg2 = va_arg(valist, float);
if (size == 2) {
(*func)(arg1, arg2);
va_end(valist);
return;
}
}
But obviously it doesn't work :(
The usual solution to making callback function interfaces flexible with respect to data provided to the function is to give the callback signature a void * parameter (possibly in addition to other parameters). Arbitrary data can be provided via such a parameter. Something like this:
void exec(void (*func)(void *), void *data) {
func(data);
}
struct s2 {
int i;
float f;
};
void func1(void *data) {
int i = *(int *)data;
// ...
}
void func2(void *data) {
struct s2 s = *(struct s2 *)data;
// ...
}
int main(void) {
int i = 42;
struct s2 s = { .i = 17, .f = 3.14 };
exec(func1, &i);
exec(func2, &s);
}
HOWEVER, It is possible to do something more like you describe, where the callback functions genuinely have different signatures, by specifying the callback type without a prototype. In that case, there are still at least these caveats:
If the callback functions themselves are defined with prototypes (as they should be) then the parameter types should not be any that are altered by the default argument promotions. So, pointers, ints, doubles, but not floats or short ints or chars (not an exhaustive list). If you wanted to support other parameter types then you would need to cast the function pointer before calling the function, as described later.
The callback functions cannot be variadic.
If the front-end is variadic, then it needs to be told at runtime, somehow, what the actual number and types of the arguments are.
Furthermore, there will need to be explicit calls to the callback functions, with correct arguments, so there can be only a fixed set of predetermined callback signatures supported.
For example, that might look something like this:
enum sig { INT, INT_DOUB };
void exec(void (*func)(/* no prototype */), enum sig cb_sig, ...);
void a(int pa) {}
void b(int pb1, double pb2) {}
int main(void) {
exec(a, INT, 1);
exec(b, INT_DOUB, 1, 2.3);
}
void exec(void (*func)(/* no prototype */), enum sig cb_sig, ...) {
va_list valist;
va_start(valist, cb_sig);
switch (cb_sig) {
case INT: {
int i = va_arg(valist, int);
func(i);
break;
}
case INT_DOUB: {
int i = va_arg(valist, int);
double d = va_arg(valist, double);
func(i, d);
break;
}
default:
assert(("Can't be reached", 0));
}
va_end(valist);
}
It is possible that that would elicit a few warnings, such as about a function declaration that does not provide a prototype, and about calling a (declared, but) unprototyped function. Since you know the signatures by the time you execute the calls, however, you could get rid of the latter kind of warning via appropriate casting. For example,
// ...
case INT: {
int i = va_arg(valist, int);
((void (*)(int))func)(i);
break;
}
// ...
You could change the callbacks to take a single va_list argument:
void a(va_list args)
{
int pa = va_arg(args,int);
}
void b(va_list args)
{
int pb1 = va_arg(args,int);
double pb2 = va_arg(args,double);
}
And have your other function pass the va_list along.
void exec(void (*func)(va_list), ...)
{
va_list valist;
va_start(valist, func);
func(valist);
va_end(valist);
}
You can use va_args to solve this.
#include <stdio.h>
#include <stdint.h>
#include <stdarg.h>
#define exec_func(func, ...) func(__VA_ARGS__)
long func(char *a, int b, long c, long d)
{
printf("a: %s, b: %d, c: %ld, d: %ld\n", a, b, c, d);
return c + d;
}
int main()
{
printf("c + d: %ld\n", exec_func(func, "test", 10, 1000, 1000));
}

Call a function using a pointer and pass the pointer that can point to the function along in the parameters

Say that I have a pointer to function theFunc. theFunc takes along a pointer that can point to any function with the same parameter list as theFunc, so the function called can set the passed pointer to NULL or a different function.
Using it would look like this:
while (funcPtr != NULL)
{
funcPtr(&funcPtr);
}
Would defining this be impossible?
Yes, it's doable.
The simple way:
void (*fptr_t)(void*);
The function pointer is data, even though it point to non-data. Therefore a pointer to function pointer can be converted to void* without relying on compiler extensions.
This solution lacks type safety. However, it can be improved.
Currently, it is possible to declare a function taking unspecified number of parameters. It allows to form an incomplete function type. For example:
int foo();
declares a function that returns int and takes unspecified parameters. To have a function taking no parameters use int foo(void).
This allows to declare a function taking a pointer to pointer to incomplete function type:
int foo(int (**)());
// call
int (*fptr)(int (**)()) = foo;
fptr(&fptr);
As mentioned in other answers typedef-ing function types makes the code cleaner.
typedef int foo_aux_f();
typedef int foo_f(foo_aux_f**);
foo_f *fptr = &foo;
fptr(&fptr);
It is possible to improve type safety by nesting the declaration of function types deeper and deeper.
typedef int foo_aux0_f();
typedef int foo_aux1_f(foo_aux0_f**);
typedef int foo_aux2_f(foo_aux1_f**);
typedef int foo_aux3_f(foo_aux2_f**);
typedef int foo_f(foo_aux3_f**);
foo_f fptr = &foo;
fptr(&fptr);
The perfect recursive type would be reached with infinite chain of declaration but in practice 2-3 levels are sufficient.
With some abuse of the syntax of typedef keyword it is possible to squeeze the declaration of this type:
typedef int foo_aux0_f(),
foo_aux1_f(foo_aux0_f**),
foo_aux2_f(foo_aux1_f**),
foo_aux3_f(foo_aux2_f**),
foo_f(foo_aux3_f**);
Unfortunately ... or fortunately, this trick will likely not work in upcoming C23 because the old function declarations without prototypes are planned to be removed from the language making () mean no arguments rather then unspecified number of argument.
Yes, you can pass pointer to pointer to function. The syntax is much easier if you use typedefs.
typedef void somefunc(void);
void func1(void)
{
printf("Func1\r");
}
void func2(void)
{
printf("Func2\r");
}
void swapfunction(somefunc **ptr)
{
if(*ptr == func1) *ptr = func2;
else *ptr = func1;
}
int main(void)
{
somefunc *ptr = NULL;
swapfunction(&ptr);
ptr();
swapfunction(&ptr);
ptr();
}
You can also use function return value:
typedef void somefunc(void);
void func1(void)
{
printf("Func1\r");
}
void func2(void)
{
printf("Func2\r");
}
somefunc *swapfunction(somefunc *ptr)
{
if(!ptr) return func1;
else if (ptr == func1) return func2;
else return NULL;
}
int main(void)
{
somefunc *ptr = NULL;
while(ptr = swapfunction(ptr))
{
ptr();
}
}
Ref your github comment, suggest you use a structure instead of type casting pointers to function pointers, etc. It's not exactly what you are requesting, but kind of.
The code will then look like:
#include <stdio.h>
struct funcArgStruct
{
void (*state)(struct funcArgStruct *);
// int extra_data; // optional
};
typedef struct funcArgStruct funcArg;
void start (funcArg *ptr);
void task1 (funcArg *ptr);
void stop (funcArg *ptr);
/* Implementation of an fsm. */
int main()
{
funcArg ptr_, *ptr = &ptr_;
ptr->state = start;
// ptr->extra_data = 0; // optional
while (ptr->state != NULL)
{
ptr->state(ptr);
}
return 0;
}
void start (funcArg *ptr)
{
ptr->state = task1;
}
void stop (funcArg *ptr)
{
ptr->state = NULL;
}
void task1 (funcArg *ptr)
{
ptr->state = stop;
}
This sort of works:
#include <stdio.h>
void *a(void)
{
printf("Calling a()\n");
return NULL;
}
void *b(void)
{
printf("Calling b()\n");
return a;
}
void *c(void)
{
printf("Calling c()\n");
return b;
}
int main(void)
{
void *(*funcPtr)(void) = &c;
while (funcPtr) {
funcPtr = funcPtr();
}
}
I don't really see good uses, especially in passing the pointer to the function itself as an argument (which I why I omitted it) but whatever floats your boat. You can of course replace the arguments to whatever you need.
You could add a typedef to help out a bit with a type:
typedef void *(*myfunc)(void);
Then you could do the following:
myfunc funcPtr = &c;
// instead of: void *(*funcPtr)(void) = &c;
I don't think any of this is particularly elegant, but it should work.
Note that it doesn't matter if you assign c or &c to myfunc, or whether you return a or &a from one of the functions.

What is the "right" way to work with struct pointers and a C interface?

I'm trying to setup a system where I pass around a pointer to a structure while hiding the definition of the structure from the end user. I have two options that seem to work, but I don't know if I'm making this harder than it needs to be, missing a trade off, or just doing something really stupid. I am stuck with C for any approach and can't use C++. Additionally, this will eventually need to talk to a Fortran program through, and I'm trying to make that as straightforward as possible.
I have a little utility to demonstrate the concept. Option one uses a void pointer to a pointer so that I can return a status integer from the function, if necessary. However, I don't like having to malloc before the call as I'm concerned about the Fortran side of things. This may be unfounded as I haven't done that demo, yet. Option two just returns a void pointer from the function, but I lose the ability to do a status return that way. With both versions, I do have a custom free function, even if not necessarily with the exact current implementation. The struct has it's own void pointer that will be defined based off the option input, and it will need to free that as part of the teardown process.
#include <stdio.h>
#include <stdlib.h>
struct State
{
int type;
void *data;
};
int Init1(int option, void **state);
void* Init2(int option);
void printState(void *state);
void free1(void **state);
void free2(void *state);
void* allocateData(int option);
int main(int argc, char *argv[])
{
void **ps1;
void *s2;
int ret;
ps1 = malloc(sizeof(void*));
ret = Init1(1, ps1);
printState(*ps1);
free1(ps1);
s2 = Init2(2);
printState(s2);
free2(s2);
return 0;
}
int Init1(int option, void **state)
{
(*state) = malloc(sizeof(struct State));
struct State* ret = *state;
ret->type = option;
return 0;
}
void free1(void **state)
{
free(*state);
free(state);
}
void* Init2(int option)
{
struct State* ret = malloc(sizeof(struct State));
ret->type = option;
return ret;
}
void free2(void *state)
{
free(state);
}
void printState(void *state)
{
struct State* data = state;
printf("Type : %d\n", data->type);
}
Look to the FILE type in stdio.h as an example. You can expose the type name without exposing its definition:
/**
* State.h
*/
#ifndef STATE_H
#define STATE_H
/**
* Create a typedef name for the *incomplete* type "struct State"
*/
typedef struct State STATE;
/**
* Define your interface
*/
void Init1( int, STATE ** );
STATE *Init2( int );
void printState( STATE * );
void sFree( STATE * );
void sFree2( STATE ** );
#endif
Then you complete the definition of the type in the implementation file:
/**
* State.c
*/
#include "State.h"
#include <stdlib.h>
...
/**
* Complete the type definition
*/
struct State {
int type;
void *data;
};
/**
* Implement the interface
*/
int Init1( int option, STATE **s )
{
*s = malloc ( sizeof **s ); // type definition is complete at this
if ( *s ) // point so we can use sizeof
{
(*s)->type = option;
}
return *s != NULL; // I'm *assuming* you want to return true (1)
} // if the allocation is successful
...
Now, when it comes to interoperating with Fortran ... I can't be that much help. I did that, once, on a VAX, 30-some-odd years ago, and it didn't involve opaque types like this.

In pure C how can I pass a variable number of parameters into a function?

How can i pass (and access) using C, not c++, variable parameters into a function?
void foo(char* mandatory_param, char* optional_param, char* optional_param2...)
thanks
/fmsf
Use stdarg.h
You need to use va_list and then use the macros va_start, va_arg, and va_end.
For more information, see http://www.acm.uiuc.edu/webmonkeys/book/c_guide/2.10.html
It sounds like you are looking for varargs.
#include <stdarg.h>
void foo(const char *fmt, ...)
{
va_list argp;
va_start(argp, fmt);
int i = va_arg(argp, int);
// Do stuff...
va_end(argp);
}
Read about Variable Arguments in C
#include <stdarg.h>
void do_sth (int foo, ...)
{
int baz = 7; /* "baz" argument */
const char *xyz = "xyz"; /* "xyz" argument */
/* Parse named parameters */
va_list ap;
va_start (ap, foo);
for (;;) {
const char *key = va_arg (ap, char *);
if (key == NULL) {
/* Terminator */
break;
} else if (strcmp (key, "baz") == 0) {
baz = va_arg (ap, int);
} else if (strcmp (key, "xyz") == 0) {
xyz = va_arg (ap, char *);
} else {
/* Handle error */
}
}
va_end (ap);
/* do something useful */
}
do_sth (1, NULL); // no named parameters
do_sth (2, "baz", 12, NULL); // baz = 12
do_sth (3, "xyz", "foobaz", NULL); // xyz = "foobaz"
do_sth (4, "baz", 12, "xyz", "foobaz", NULL); // baz = 12, xyz = "foobaz"
Variadic functions and arguments assignment in C/C++
In a language that does not support optional parameters directly, there are a few ways to achieve a similar effect. I will list them in order from the least versatile to the most:
Create multiple overloads of the same function. As I recall, you cannot do this in C.
Use variadic functions. Just Google this: http://www.google.com/search?q=variadic+function+c
I recommend this: Create a "params" or "args" class (or struct in C), like this:
)
// untested C code
struct FooArgs {
char * mandatory_param;
char * optional_param;
char * optional_param2;
// add other params here;
};
and then make your method call take in a single argument:
// untested
void foo(struct fooArgs * args)
This way, as needs change, you can add parameters to fooArgs without breaking anything.
I have a solution that does not use VA_LIST in pure C. However, it works at 32bits only. Here, what happens is that each parameter of the call stack occupies as many bytes according to its type. It is possible to create a structure with a size larger than 4 or 8 bytes, so just align all the parameters in this structure.
int printf(void*,...);
typedef struct{
char p[1024];
}P_CALL;
int soma(int a,int b){
return a+b;
}
void main(){
P_CALL
call;
char
*pcall=(void*)&call;
int
(*f)()=soma,
res;
*(int*)pcall=1;
pcall+=sizeof(void*);
*(int*)pcall=2;
pcall+=sizeof(void*);
res=f(call);
printf("%d\n",res);//3
}

Accessing members of the struct via void *

The solution consists of two parts, one is a static library that receives instances of struct from the user of the library. Library doesn't know what will be the type of structs, all it knows there will be two function pointers to it with a specific name.
Library Code
pre-compiled library has no way of knowing types of user structs, hence receiving via void*
void save(void *data) {
// library will save/cache user's object
data->registered(); // if register successful
}
void remove(void *data) {
// library will remove the object from memory
data->remove(); // if removed successful
}
User of the Library Code
struct Temp { // random order of fields
void (*custom1)();
void (*registered)();
void (*custom2)();
void (*remove)();
void (*custom3)();
}
void reg() {
printf("registered");
}
void rem() {
printf("removed");
}
void custom1() {}
void custom2() {}
void custom3() {}
var temp = malloc(struct Temp, sizeof(struct Temp));
temp->registered = reg;
temp->remove = rem;
temp->custom1 = custom1; // some custom functions
temp->custom2 = custom2;
temp->custom3 = custom3;
// calling library code
save(temp);
remove(temp);
Q. Is there a way for the Library to know how to iterate and go through member fields and see if there's a pointer to such function and call it available.
Is there a way for the Library to know how to iterate and go through member fields and see if there's a pointer to such function and call it available.
No there is not.
Your best bet is to create a structure in the library that has these members, and pass that structure instead of void*.
As #immibis said, there is no way for this to work (i.e. no way for the compiler to justify compiling such code) if the compiler does not know what the types of the data being passed to the function are.
Since you wanted to pass the objects along to the library without storing information about the type of each object in the library, you can fake polymorphism in C, by doing the following:
callback.h
#ifndef _CALLBACK_H_
#define _CALLBACK_H_
typedef struct {
void (*registered)();
void (*removed)();
} ICallback;
#endif _CALLBACK_H_
pre_comp.h
#ifndef _PRE_COMP_H_
#define _PRE_COMP_H_
#include "callback.h"
void save(ICallback* data);
void remove(ICallback* data);
#endif /* _PRE_COMP_H_ */
precomp.c
#include <stdlib.h> /* NULL */
#include "callback.h"
#include "pre_comp.h"
void save(ICallback *data) {
if (NULL != data && NULL != data->registered) {
data->registered(); // if register successful
}
}
void remove(ICallback *data) {
if (NULL != data && NULL != data->removed) {
data->removed(); // if removed successful
}
}
main.c
#include <stdio.h>
#include "pre_comp.h"
#include "callback.h"
struct Temp {
ICallback base; // has to be defined first for this to work
void (*custom1)();
void (*custom2)();
void (*custom3)();
};
// calling library code
void reg() {
puts("registered");
}
void rem() {
puts("removed");
}
int main() {
struct Temp data = {{reg, rem}};
save((ICallback*)&data);
remove((ICallback*)&data);
}
compiling
gcc pre_comp.c main.c
output
registered
removed
If the library has 0 information about the possible struct types, then you
cannot do it. The library has to get somehow the information or the offsets.
The only way I can think of is:
All register member have the same prototype
Pass the offset to the function.
I created an example of this
#include <stdio.h>
#include <stddef.h>
#include <stdint.h>
// function that does not know anything about any struct
void reg(void *data, size_t offset)
{
uintptr_t *p = (uintptr_t*) (((char*) data) + offset);
void (*reg)() = (void(*)()) *p;
reg();
}
struct A {
int c;
void (*reg)();
};
struct B {
int b;
int c;
void (*reg)();
};
void reg_a()
{
printf("reg of A\n");
}
void reg_b()
{
printf("reg of B\n");
}
int main(void)
{
struct A a;
struct B b;
a.reg = reg_a;
b.reg = reg_b;
reg(&a, offsetof(struct A, reg));
reg(&b, offsetof(struct B, reg));
return 0;
}
This prints:
$ ./c
reg of A
reg of B
I run it with valgrind and I did not get any errors nor warnings. I'm not sure if
this violates somehow strict aliasing rules or yields undefined behaviour
because of the uintptr_t* conversions, but at least it seems to work.
I think however, the more cleaner solution is to rewrite the register (btw. register
is a keyword in C, you cannot use that for a function name) function to
accept a function pointer and possible parameters, something like this:
#include <stdio.h>
#include <stdarg.h>
void reg(void (*func)(va_list), int dummy, ...)
{
if(func == NULL)
return;
va_list ap;
va_start(ap, dummy);
func(ap);
va_end(ap);
}
void reg1(int a, int b)
{
printf("reg1, a=%d, b=%d\n", a, b);
}
void vreg1(va_list ap)
{
int a = va_arg(ap, int);
int b = va_arg(ap, int);
reg1(a, b);
}
void reg2(const char *text)
{
printf("reg2, %s\n", text);
}
void vreg2(va_list ap)
{
const char *text = va_arg(ap, const char*);
reg2(text);
}
int main(void)
{
reg(vreg1, 0, 3, 4);
reg(vreg2, 0, "Hello world");
return 0;
}
This has the output:
reg1, a=3, b=4
reg2, Hello world
Note that reg has a dummy parameter. I do that because the man page of
stdarg says:
man stdarg
va_start():
[...]
Because the address of this argument may be used in the va_start() macro,
it should not be declared as a register variable, or as a
function or an array type.
You can take an approach similar to qsort and pass function pointers in addition to a void pointer to the structure.
Here is the function prototype for qsort, which is a function that can be used to sort arrays of any type:
void qsort(void *base, size_t nmemb, size_t size, int (*compar)(const void *, const void *));
It takes a function pointer that performs the comparison because without it qsort wouldn't know how to compare two objects.
This can be applied to your task with a function prototype like this:
int DoFoo(void *thing, void (*register)(void *), void (*remove)(void *))
This function takes a void pointer to your struct and then two functions that it can call when it needs to register or remove that struct. Having the functions be members of the struct is not required and I generally do not recommend it. I recommend reading up on qsort because it is does something similar to what you are trying to do.

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