I am a newbie in C. My problem is quite simple. Below is my code. I expect it to increase req_id by 1 and then pint out 1. However, the result is 0.
typedef uint32_t req_id_t;
typedef struct view_stamp_t{
req_id_t req_id;
}view_stamp;
struct consensus_component_t{
view_stamp highest_seen_vs;
};
typedef struct consensus_component_t consensus_component;
static void view_stamp_inc(view_stamp vs){
vs.req_id++;
return;
};
int main()
{
consensus_component* comp;
comp = (consensus_component*)malloc(sizeof(consensus_component));
comp->highest_seen_vs.req_id = 0;
view_stamp_inc(comp->highest_seen_vs);
printf("req id is %d.\n", comp->highest_seen_vs.req_id);
free(comp);
return 0;
}
When you call functions in C, parameters are passed by value, not by reference. So vs in view_stamp_inc is a copy of comp->highest_seen_vs. Incrementing req_id in the copy has no effect on the original structure.
You need to pass the address of the structure.
static void view_stamp_inc(view_stamp *vs) {
vs->req_id++;
return;
}
...
view_stamp_inc(&comp->highest_seen_vs);
To change the original object passed as an argument to a function it should be passed to the function by reference.
For example
static void view_stamp_inc(view_stamp *vs){
vs->req_id++;
};
//...
view_stamp_inc( &comp->highest_seen_vs );
Related
I am working with a framework where structures are defined with underscore first and then structure that is used is created as pointer to that internal structure.
Passing it as argument require additional steps of defining local pointer within the function first.
What is use of this comparing with just declaring structure and using it directly?
//.h
typedef struct _tFoo
{
int value;
} _tFoo;
typedef _tFoo* tFoo;
//.c
void tFoo_setFoo(tFoo* const f, int value)
{
_tFoo* foo = *f;
foo->value = value;
}
What is benefit of having code above comparing with
//.h
typedef struct _tFoo
{
int value;
} tFoo;
//.c
void tFoo_setFoo(tFoo* const f, int value)
{
f->value = value;
}
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'm trying to learn c using learncodethehardway c book. In ex19 I have the following code:
int Monster_init(void *self)
{
Monster *monster = self;
monster->hit_points = 10;
return 1;
}
int Monster_attack(void *self, int damage)
{
Monster *monster = self;
printf("You attack %s!\n", monster->proto.description);
monster->hit_points -= damage;
if(monster->hit_points > 0) {
printf("It is still alive.\n");
return 0;
} else {
printf("It is dead.\n");
return 1;
}
}
Object MonsterProto = {
.init = Monster_init,
.attack = Monster_attack
};
This is the Object structure:
typedef struct {
char *description;
int (*init)(void *self);
void (*describe)(void *self);
void (*destroy)(void *self);
void *(*move)(void *self, Direction direction);
int (*attack)(void *self, int damage);
} Object;
And this is the Monster structure:
struct Monster {
Object proto;
int hit_points;
};
I'm having a tough time wrapping my head around the Monster_init and Monster_attack functions. I have a MonsterProto variable of type Object defined and inside there .init is set to the Monster_initfunction and .attack is set to the Monster_attack function.
I think I understand the notion of void in terms of declaring a function that has side effects but doesn't need to return something. What I don't understand is what exactly is the void *self pointer pointing at and why does it allow me to call a function with no arguments? What is the purpose of the self pointer?
I didn't want to include too much code here but if this is not enough context to answer the question, then you can find all the code here.
I appreciate any pointers in the right direction; nu pun intended :)
This code seems to be effectively implementing a kind of object-oriented approach.
self is the address of the struct Monster that you pass to those functions. Each of those functions operates on an individual object, and passing in the pointer to that object is how they know which one to work on.
This:
.init = Monster_init,
is not "calling a function with no arguments" - the init member of your struct is a pointer to a function returning an int and accepting a single void * parameter, and that line assigns the address of Monster_init() to it. This way, if you have a pointer to an object, you can call int n = myobject->proto.init(&myobject); or similar without knowing which actual function gets called. With a different object, you might be calling a different function with the same line of code.
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'm a new C programmer and I wanted to know how I can pass a struct through to a function. I'm getting an error and can't figure out the correct syntax to do it. Here is the code for it....
Struct:
struct student{
char firstname[30];
char surname[30];
};
struct student person;
Call:
addStudent(person);
Prototype:
void addStudent(struct student);
and the actual function:
void addStudent(person)
{
return;
}
Compiler errors:
line 21: warning: dubious tag declaration: struct student
line 223: argument #1 is incompatible with prototype:
This is how to pass the struct by reference. This means that your function can access the struct outside of the function and modify its values. You do this by passing a pointer to the structure to the function.
#include <stdio.h>
/* card structure definition */
struct card
{
int face; // define pointer face
}; // end structure card
typedef struct card Card ;
/* prototype */
void passByReference(Card *c) ;
int main(void)
{
Card c ;
c.face = 1 ;
Card *cptr = &c ; // pointer to Card c
printf("The value of c before function passing = %d\n", c.face);
printf("The value of cptr before function = %d\n",cptr->face);
passByReference(cptr);
printf("The value of c after function passing = %d\n", c.face);
return 0 ; // successfully ran program
}
void passByReference(Card *c)
{
c->face = 4;
}
This is how you pass the struct by value so that your function receives a copy of the struct and cannot access the exterior structure to modify it. By exterior I mean outside the function.
#include <stdio.h>
/* global card structure definition */
struct card
{
int face ; // define pointer face
};// end structure card
typedef struct card Card ;
/* function prototypes */
void passByValue(Card c);
int main(void)
{
Card c ;
c.face = 1;
printf("c.face before passByValue() = %d\n", c.face);
passByValue(c);
printf("c.face after passByValue() = %d\n",c.face);
printf("As you can see the value of c did not change\n");
printf("\nand the Card c inside the function has been destroyed"
"\n(no longer in memory)");
}
void passByValue(Card c)
{
c.face = 5;
}
The line function implementation should be:
void addStudent(struct student person) {
}
person is not a type but a variable, you cannot use it as the type of a function parameter.
Also, make sure your struct is defined before the prototype of the function addStudent as the prototype uses it.
When passing a struct to another function, it would usually be better to do as Donnell suggested above and pass it by reference instead.
A very good reason for this is that it makes things easier if you want to make changes that will be reflected when you return to the function that created the instance of it.
Here is an example of the simplest way to do this:
#include <stdio.h>
typedef struct student {
int age;
} student;
void addStudent(student *s) {
/* Here we can use the arrow operator (->) to dereference
the pointer and access any of it's members: */
s->age = 10;
}
int main(void) {
student aStudent = {0}; /* create an instance of the student struct */
addStudent(&aStudent); /* pass a pointer to the instance */
printf("%d", aStudent.age);
return 0;
}
In this example, the argument for the addStudent() function is a pointer to an instance of a student struct - student *s. In main(), we create an instance of the student struct and then pass a reference to it to our addStudent() function using the reference operator (&).
In the addStudent() function we can make use of the arrow operator (->) to dereference the pointer, and access any of it's members (functionally equivalent to: (*s).age).
Any changes that we make in the addStudent() function will be reflected when we return to main(), because the pointer gave us a reference to where in the memory the instance of the student struct is being stored. This is illustrated by the printf(), which will output "10" in this example.
Had you not passed a reference, you would actually be working with a copy of the struct you passed in to the function, meaning that any changes would not be reflected when you return to main - unless you implemented a way of passing the new version of the struct back to main or something along those lines!
Although pointers may seem off-putting at first, once you get your head around how they work and why they are so handy they become second nature, and you wonder how you ever coped without them!
You need to specify a type on person:
void addStudent(struct student person) {
...
}
Also, you can typedef your struct to avoid having to type struct every time you use it:
typedef struct student{
...
} student_t;
void addStudent(student_t person) {
...
}
Instead of:
void addStudent(person)
{
return;
}
try this:
void addStudent(student person)
{
return;
}
Since you have already declared a structure called 'student' you don't necessarily have to specify so in the function implementation as in:
void addStudent(struct student person)
{
return;
}