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Pointer to a const structure, can members still be modified?

I have a structure which I want to pass to some external c code via some callback function that they register with my program. However, I want to pass that structure as read-only. My concern is that they can still modify the structures which I have pointer to inside the original structure that I pass. Explained with small example below:

struct s1
int a;
int b;
;

struct s2
int x;
struct s1 *y;
;

void f(const struct s2 *o)

//o->x=10; //error
o->y->a=20; //no error
o->y->b=30; //no error


int main()

struct s1 o1 = 10, 20;
struct s2 o2 = 30, &o1;
f(&o2);



So, How do I improve my code design so that they cannot modify anything w.r.t the structure I pass?

5 Answers
5

To properly handle this situation, you can only use forward declaration to hide members together with getter and setter functions.

Focus on code below and check:

struct s1

struct s2

struct s1

mylib.c

mylib.h:

#ifndef __MYLIB_H
#define __MYLIB_H

//Create forward declaration only
//Implementation is in .c file
struct s1;

//Create user structure
struct s2
int x;
struct s1* y;
;

int get_a_from_s1(struct s2* s);
void set_a_to_s1(struct s2* s, int a);

#endif /* __MYLIB_H */


mylib.c:

#include "mylib.h"

//Now implement structure
struct s1
int a, b;
;

//Make getter
int
get_a_from_s1(struct s2* s)
return s->y->a;


//Make setter
void
set_a_to_s1(struct s2* s, int a)
s->y->a = a;



main.c:

#include <stdio.h>
#include "mylib.h"

int main(void)
struct s2 s;
int a;

....

s.y->a = 5; //error

//Set s1.a value from s2 structure
set_a_to_s1(&s, 10); //OK

//To view members of s1 inside s2, create member functions
a = get_a_from_s1(&s); //OK

printf("a: %drn", a);

return 0;



Of course, please make sure that ->y is not NULL or you have undefined behavior.

->y

NULL

struct s1

You cannot. Even if you pass the struct s2 by value, you will get in the function a pointer to a non const struct s1, simply because it is what s2 contains per its definition.

struct s2

struct s1

s2

And once you have a pointer to a non const object you can change that object. What I mean here and what other answers mean, is that it is not a language problem - more exactly the language can nothing for you here - but a design problem. If for any reason it is not acceptable that the struct s1 can be changed from f then you have to find a different design where you do not pass a non const pointer to it, be it member of a const struct or not. Here a simple way would be to pass the individual members:

struct s1

f

void f(int x, const struct s1 *y)
y->a = 20; // error



It may not be what you expect, but it is the best I can say for C language.

You can change the second struct declaration like so:

struct s2
int x;
struct s1 const *y;
;


The added const ensures that y is read-only.

const

I would write another function that takes in a const struct s1 * to modify the values of s1.

In f(), const struct s2 *o means in the struct s2 being pointed to by o, the values of its members ie, x and y cannot be changed.

f()

const struct s2 *o

struct s2

o

x

y

With o->y->a=20, you are not modifying the value of y. You are just using that address to modify a member of the structure being pointed to by y and y's value, which is an address, remains the same.

o->y->a=20

y

y

y

So those 2 lines will give no error.

I could see no way to avoid this unless you could make the struct s2 definition to

struct s2

struct s2
int x;
const struct s1 *y;
;


in which case y is a pointer to a constant struct s1.

y

struct s1

See the spiral rule and visit the cdecl website.

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