A no-op device for function pointer tables? - c

void table_no_op()
{
// this is for function table elements that do nothing,
// fills space between states, use less of them
return;
}
I am currently using this to define a "zero" in a function pointer table, where the input index is supposed to do nothing. Is it okay or something glaringly wrong?

While there's nothing wrong with the no-op per se (in general, this is called Null Object Pattern), I would be worried about the function declaration - i.e. does every function in the table take 0 arguments and return void?
A counterexample would be OpenGL where you often retrieve a pointer to function and cast it to the desired type yourself - but casting a void->void pointer to something else, e.g. (int, int)->int would be undefined behavior and likely cause crash (or uninitialized return value, or else).
So, if the functions in the table are homogeneous - go for it. If not - better do something else.
EDIT: You can only do 2 things with a function pointer - cast it to a different function pointer; and call, but only with the original type.
See http://blog.frama-c.com/index.php?post/2013/08/24/Function-pointers-in-C for details. Raymond Chen has another example here - http://blogs.msdn.com/b/oldnewthing/archive/2011/05/06/10161590.aspx
EDIT2: However, you may make a number of no_ops (noop_IntInt_Int, noop_IntDouble_Double and so on... then if you match the types every time, that might work)

Related

Alternative "null" value

It seems common to have the tail pointer at the end of a linked list be null (0).
What if I want to have two possible different "tails"?
My use case is a big integer representation that supports two's complement: I want to have a tail corresponding to "the rest of this number is zeros" and "the rest of this number is ones", where I can tell them apart just by performing a pointer equality.
It seems like this should be common enough to have standard practice, but it's hard to think of exactly what to search for. It seems somewhat arbitrary that we only get one "forbidden" pointer value (that will give a useful-ish error when accidentally dereferenced).
Options seem to include:
Use some arbitrary second value (like 1, or 0xdeadbeef). This seems evil. For one thing, I guess it needs to be aligned? Also, I will have obscure bugs if malloc happens to allocate a real linked list cell at the same address. Is there some region of memory malloc is guaranteed not to use?
Call malloc with a dummy non-zero size. This seems more sensible, but ideally I would have the pointer value be const, rather than requiring initialisation.
Take the address of something arbitrary, like a function defined in the file. This seems very evil, but does seem to lack any practical disadvantages (assuming it would work).
Given some ListItem type and a desired to have a ListItem * value that serves as a sentinel (also see sentinel node), we can simply define a ListItem object to serve that purpose:
ListItem SentinelObject;
ListItem * const SentinelValue = &SentinelObject;
This could also be made static if they will be used only in one translation unit.
The named object could be eliminated by using a compound literal:
ListItem * const SentinelValue = & (ListItem) {0};
(The initializer may need adjustment if 0 is not a suitable initailizer for the first member of ListItem.)
Alternately, wasting space could be avoided by overlapping the unused ListItem object with some other object:
union { SomeUsefulType SomeUsefulThing; ListItem SentinelObject; } MyUnion;
ListItem * const SentinelValue = &MyUnion.SentinelObject;
While this gives SomeUsefulThing and SentinelObject the same address, that is unlikely to be a problem given they have different types.

What does void (*var)(void*) in an argument mean? [duplicate]

This question already has answers here:
How do function pointers in C work?
(12 answers)
Closed 1 year ago.
I am doing a C exercise where we are given prototypes, and we have to turn them into functions.
I stumbled onto the following prototype:
void ft_lstdelone(t_list *lst, void (*del)(void*))
This function is supposed to delete a node from a linked list, but I don't understand what is happening in the second argument "void (*del)(void*)".
The description of the exercise has:
Parameters -
#1 The element to free.
#2 The address of the function used to delete the content.
Description - Takes as a parameter an element and frees the
memory of the element’s content using the function
’del’ given as a parameter and free the element.
The memory of ’next’ must not be freed.
I take from this that I have to use a function as an argument, but it doesn't make any sense.
Can someone help me unwrap what this means?
It actually makes a lot of sense. It is a function pointer, that is, ft_lstdelone will find the location of the element. If it was the first element, then the new first element will be its next. If it was the last element, then its previous will point to null. Otherwise its previous will point to its next.
However, how will you remove the actual item? Will you call free? It seems to be a logical approach first, but if you think it through, you will realize that it's a naive approach. If your list contains elements of composite types, then calling free on it will free the address of your data, but the inner members of your data will still hold some allocated data. This raises the possible issue of memory leak, so, depending on the data type you have at your item, you will have to free it in different manners. This is only known at the time when you call ft_lsdelone, because at that point you know the data type of your data.
So, in order to be able to cope with the different scenarios, the function separates the concerns. This function is only concerned in the manner of how an item is to be removed from a list, but will not worry about its deallocation. Instead, it trusts its caller with the deallocation and expects to get a function that is assumed to handle the deallocation properly.

Changing a pointer as a result of destroying an "object" in C

As part of a course I am attending at the moment, we are working in C with self-developed low level libraries, and we are now working in our final project, which is a game.
At a certain point, it seemed relevant to have a struct (serving as a sort of object) that held some important information about the current game status, namely a pointer to a player "object" (can't really call the simulated objects we are using actual objects, can we?).
It would go something like this:
typedef struct {
//Holds relevant information about game current state
state_st currstate;
//Buffer of events to process ('array of events')
//Needs to be pointers because of deallocating memory
event_st ** event_buffer;
//Indicates the size of the event buffer array above
unsigned int n_events_to_process;
//... Other members ...
//Pointer to a player (Pointer to allow allocation and deallocation)
Player * player;
//Flag that indicates if a player has been created
bool player_created;
} Game_Info;
The problem is the following:
If we are to stick to the design philosophy that is used in most of this course, we are to "abstract" these "objects" using functions like Game_Info * create_game_info() and destroy_game_info(Game_Info * gi_ptr) to act as constructors and destructors for these "objects" (also, "member functions" would be something like update_game_state(Game_Info * gi_ptr), acting like C++ by passing the normally implicit this as the first argument).
Therefore, as a way of detecting if the player object inside a Game_Info "instance" had already been deleted I am comparing the player pointer to NULL, since in all of the "destructors", after deallocating the memory I set the passed pointer to NULL, to show that the object was successfully deallocated.
This obviously causes a problem (which I did not detect at first, and thus the player_created bool flag that fixed it while I still was getting a grasp on what was happening) which is that because the pointer is passed by copy and not by reference, it is not set to NULL after the call to the "object" "destructor", and thus comparing it to NULL is not a reliable way to know if the pointer was deallocated.
I am writing this, then, to ask for input on what would be the best way to overcome this problem:
A flag to indicate if an "object" is "instanced" or not - using the flag instead of ptr == NULL in comparisons to assert if the "object" is "instanced" - the solution I am currently using
Passing a pointer to the pointer (calling the functions with &player instead of only player) - would enable setting to NULL
Setting the pointer to NULL one "level" above, after calling the "destructor"
Any other solution, since I am not very experienced in C and am probably overlooking an easier way to solve this problem.
Thank you for reading and for any advice you might be able to provide!
I am writing this, then, to ask for input on what would be the best way to overcome this problem: …
What would be the best way is primarily opinion-based, but of the ways you listed the worst is the first, where one has to keep two variables (pointer and flag) synchronized.
Any other solution…
Another solution would be using a macro, e. g.:
#define destroy_player(p) do { /* whatever cleanup needed */; free(p), p = NULL; } while (0)
…
destroy_player(gi_ptr->player);

Get struct's size passed as void to function

I'm changing some codes in a database library. The way it works I send a void pointer, to get the size of it I call a query and using the query I calculate the size of the structure. Now the problem is I receive the struct as params but the function fails before/in the middle of the first fetch. After that I need to clear the structure, but I dont even have the size.
I know the best way is send the size of the structure as a param, but I have thousands and thousands programs already compiled, the library is from 1996, so I need to find a way to calculate the structure size even if the type is void.
One idea I had was to calculate the position of the next element that is not in the structure
0x000010 0x000042
[int|char[30]|int|int][int]
So the size is 32, because the 0x00042-0x000010 is 32.
Is there a way to know when I got out of the structure.
the prototype of the function is
int getData(char* fields, void* myStruct)
I need to find out the structure size.
Sorry if I missed some information, the code is HUGE and unfortunately I cannot post it here.
No, in general there's no way, given a void *, to figure out what you're after. The only thing you can do is compare it against NULL, which of course doesn't help here.
Note that there's nothing in the void * that even says it points at a struct, it could just as well be pointing into the middle of an array.
If you have some global means of recording the pointers before they're passed to getData(), you might be able to implement a look-up function that simply compares the pointer value against those previously recorded, but that's just using the pointer value as a key.

Why do I have to write *myPointerVar only SOMETIMES in Objective-C?

That's an issue I still don't understand.
Sometimes I have to write:
NSString* myVariable;
myVariable = #"Hey!";
Then, for example I define a Structure "DemoStruct" and get an Variable that uses it. Lets say I have a Structure that has x and y vars from type double.
I want to pass this var to a method which then manipulates my var, and I want that this manipulation has effect on the context from which I passed the var to that method. So I need a pointer, right.
I pass it to the method like that:
[someObject someMethod:&myVarThatUsesTheStruct]
that method now looks like that:
- (void)someMethod:(DemoStruct*)myVar {
(*myVar).x += 10;
}
Before the call, the component x of the struct was lets say 1000. Now, 10 is added and it is 1010 after the method call.
But I really really hardly dont get it why I have to use the Asterisk * for myVar in the Method, since I say already in the Method Header that myVar is a POINTER to a DemoStruct. I just pass with &myVarThatUsesTheStruct the memory address.
Can someone explain why this is like it is?
As you say, myVar is a pointer. As such, myVar.x is not correct: it would by a field of a pointer, which has no sense in C/Objective-C.
If you want to access to the variable pointed to by a pointer, you have to add the asterisk: myVar is a pointer, *myVar is the variable pointed to by myVar.
Moreover, in your case, you can use a special construct of C by writing myVar->x, which is strictly equivalent to (*myVar).x.
All of this is standard C, not specific to Objective-C.
About your first example, you don't have to put an asterisk because you change the value of the pointer, not the value of the variable: myVariable is a pointer to an object which at declaration time is assigned the nil value. The next instruction (myVariable = #"Hey!") is an assignment of pointer values: #"Hey!" is a pointer to a NSString object. The value of this pointer (not the value of the pointed constant) is assigned to myVariable, which then points to the object #"Hey!".
Yes, this is diffucult to follow at first time...
* is the dereference operator. All that *myVar means is "Get me the thing that the pointer myVar points to". You need this distinction because you need to tell the computer that you want to change the thing that myVar points to, not myVar itself.
taken from the learning objective c link via the developers portal for iphone devs:
Objective-C supports both strong and weak typing for variables
containing objects. Strongly typed variables include the class name in
the variable type declaration. Weakly typed variables use the type id
for the object instead. Weakly typed variables are used frequently for
things such as collection classes, where the exact type of the objects
in a collection may be unknown. If you are used to using strongly
typed languages, you might think that the use of weakly typed
variables would cause problems, but they actually provide tremendous
flexibility and allow for much greater dynamism in Objective-C
programs.
The following example shows strongly and weakly typed variable
declarations:
MyClass *myObject1; // Strong typing
id myObject2; // Weak typing
Notice the * in the first declaration. In Objective-C, object
references are pointers. If this doesn’t make complete sense to you,
don’t worry—you don’t have to be an expert with pointers to be able to
start programming with Objective-C. You just have to remember to put
the * in front of the variable names for strongly-typed object
declarations. The id type implies a pointer.

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