C const pointer to const struct array as function argument

Question

How can i make Alt.1 to work as intended by passing a pointer of an array and get the requested reference of an array in Alt.1 ?

struct mystruct
{
    int id1;
    int id2;
};

const struct mystruct local_struct[] = {
    {0, 55},
    {1, 66},
};

// Alt.1 i like to make this work (not working)
int get_reference_1(const struct mystruct *s){

   s = local_struct;
   return 0;
}

// Alt.2 works perfect but i like to use the return as status as in Alt.1.
const struct mystruct *get_reference_2(){
   return local_struct;
}

int main()
{
  struct mystruct *s = NULL;

  // Alt.1
  if(get_reference_1(s))
     /* Expected Fail*/
  else
     /* Expected Success*/

  // Alt.2
  s = get_reference_2()
  if(!s)
     /* Expected Fail*/
  else
     /* Expected Success*/

  return 0;
}

Maybe i'm thinking wrong and i need to pass a double pointer?

Edit: Corrected with 'const'. Edit2: Updated header.

There are many duplicates. You apparently still don't understand quite well how pointers work, please read more about it. The reason it doesn't work is because s is local to get_referenec_1. You have to pass a pointer to the poitner you want to update. — Iharob Al Asimi
– Iharob Al Asimi, Commented Feb 16, 2018 at 14:50
why would tou call a global struct "local_struct"? Are you trying to confuse the enemy? — Eugene Sh.
– Eugene Sh., Commented Feb 16, 2018 at 14:50
@EugeneSh. This is just a short example of my problem. I'm working in a bigger system which i cannot include everything. Sorry for the confusion. — Zäta
– Zäta, Commented Feb 16, 2018 at 15:07

Sergey Kalinichenko · Accepted Answer · 2018-02-16 15:01:44Z

3

s = local_struct; is changing a local variable - it won't change the one in main. Pass the address of the variable and make changes to the original variable dereferencing it.

int get_reference_1(struct mystruct **s){

   *s = local_struct;
   return 0;
}

Calling it would be

  if(get_reference_1(&s))
     /* Expected Fail*/
  else
     /* Expected Success*/

Also you are making the compiler complain by assigning a const variable to non-const one. Here the local_struct is a constant struct declared in your code. Solution check whether you are doing the right thing - is this assignment necessary? You could also add const qualifiers as needed:

int get_reference_1(const struct mystruct **s){
   *s = local_struct;
   return 0;
}
...
const struct mystruct *s = NULL;

In the worst case drop the const qualifier.

edited Feb 16, 2018 at 15:01

Sergey Kalinichenko

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answered Feb 16, 2018 at 14:48

user2736738

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6 Comments

user2736738 Over a year ago

Right now I have terrible net. My edits are slow.. I apologize.

user2736738 Over a year ago

@dasblinkenlight.: const struct mystruct *a is basically saying that a is a pointer to a const. Modifying the value of a struct will be illegal. But the pointer is free. We can make it point anywhere. So we can only read through this pointer. If we want to create a linked node by modifying some member of a struct instance then it will complain. (via p we will access it). Thanks for the edit.

Sergey Kalinichenko Over a year ago

Since OP's local_struct is marked const, we have no alternative but to produce a const pointer to it. Otherwise we might accidentally modify local_struct through a pointer to it, causing undefined behavior. The only way to keep s modifiable is to remove const from local_struct.

user2736738 Over a year ago

@dasblinkenlight. Is the right thing to say is pointer to const struct? Just working my way through writing more clearly.

Zäta Over a year ago

@dasblinkenlight, the purpose is to avoid modification, therefor the const.

|

amar_1995 · Accepted Answer · 2018-02-16 15:03:16Z

2

Here you go what you want

struct mystruct
{
    int id1;
    int id2;
};

 struct mystruct local_struct[] = {
    {0, 55},
    {1, 66},
};

// Alt.1 i like to make this work (not working)
int get_reference_1(struct mystruct **s){

   *s = local_struct;
   return 0;
}

// Alt.2 works perfect but i like to use the return as status as in Alt.1.
struct mystruct *get_reference_2(){
   return local_struct;
}

int main()
{
  struct mystruct *s = NULL;

  // Alt.1
  if(get_reference_1(&s))
  {  
      /* Expected Fail*/
  }
  else
  {
      /* Expected Success*/
  }   

  // Alt.2
  s = get_reference_2();
  if(!s)
  {   
      /* Expected Fail*/

  }
  else
  {
      /* Expected Success*/
  }   

  return 0;
}

It will execute successfully.

answered Feb 16, 2018 at 15:03

amar_1995

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1 Comment

Zäta Over a year ago

Thanks for your quick reply, this worked as well as the proposed one before.

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