restrict type qualified pointers

Hello all,

I want to give a try to "restrict" type-qualifier. I have looked for examples on google and I see the format is used: "int * restrict p";

However when I write this I get an error: "p is undeclared"

I am using GNU GCC Compiler with Code::Blocks. How can I solve this problem. Is it possible me to adapt the compiler to C99 standarts, if I can, will it solve my problem?

compiler command line option "-std=c99"

Thank you very much for your answer. I have looked for it a bit and I can use C99 standarts now. By the way I couldn't understand the restrict qualifier:

Code:

---

#include <stdio.h>
int main(void){

int a=5;
int *restrict p=&a;
int *q;
q=p;
printf("%d",*q);

return 0;
}

---

I wrote this code. I assumed the pointer q cannot access the memory stored by p but it does. What I understood was that from my research but it seems I was wrong. Could you explain me the restrict qualifier?

you don't use the restrict qualifier that way. it's only for use on function parameters.

Could you explain me a bit how to use it?

I did a try:

Code:

---

#include <stdio.h>

int f(int *restrict x,int *restrict y, int *restrict z);

int main(void){
int a=5;
int b=3;
int *restrict p=&a;
int *restrict q=&b;
int *restrict r=&b;
int result=f(p,q,r);
printf("%d",result);
return 0;
}

int f(int *restrict x,int *restrict y, int *restrict z){
*y+=*z;
*x+=*z;
printf("%d\n",*x);
printf("%d\n",*y);
printf("%d\n",*z);
return (*x)*(*y)*(*z);
}

---

I did a mistake intentionally. Although I declared p and q as restrict both point to the same object.

I assumed the compiler will assume z was unchanged -because I declared it as restrict- and I will get a wrong result. However it works perfectly. So I don't understand how the compiler can be optimised with restrict if it doesn't assume it is changed?

you cannot use restrict in the body of a function. it is only for function parameters.

this is ok:

Code:

---

int f(int *restrict x,int *restrict y, int *restrict z);

---

this is not:

Code:

---

int main(void){
  int a=5;
  int b=3;
  int *restrict p=&a;
  int *restrict q=&b;
  int *restrict r=&b;
  int result=f(p,q,r);
  printf("%d",result);
  return 0;
}

---

Thank you very much for your correction. However the code below still doesn't do what I assume.

Code:

---

#include <stdio.h>

int f(int *restrict x,int *restrict y, int *restrict z);

int main(void){
int a=5;
int b=3;
int *p=&a;
int *q=&b;
int *r=&b;
int result=f(p,q,r);
printf("%d",result);
return 0;
}

int f(int *restrict x,int *restrict y, int *restrict z){
*y+=*z;
*x+=*z;
printf("%d\n",*x);
printf("%d\n",*y);
printf("%d\n",*z);
return (*x)*(*y)*(*z);
}

---

According to the wikipedia article, I think I should get a wrong result because the value of "z" will be loaded once.

restrict - Wikipedia, the free encyclopedia

However, it doesn't seem like this.

this may be a case of undefined behavior. literally anything could happen with the variables involved.

When you engage in undefined behavior the results is NOT defined!!

Tim S.

The last two answers really have explained to me. So I can say I shouldn't trust the code above to be always true.

Thank you very much for your answers.

Quote:

---

Originally Posted by Elkvis

you cannot use restrict in the body of a function. it is only for function parameters.

---

(bold mine)

Since when?

Quote:

---

Originally Posted by SirPrattlepod

(bold mine)

Since when?

---

since GCC complains if you try it in a function body.

Quote:

---

Originally Posted by Elkvis

since GCC complains if you try it in a function body.

---

This works:

Code:

---

// gcc -std=c99 -Wall -Wextra restrict.c

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
  int * restrict p = malloc(sizeof(int));
  *p = 42;
  printf("%d\n", *p);
  free(p);
  return 0;
}

---

it's possible that I forgot to add -std=c99 to my compiler command. I'll have to try it again when I have access to a C compiler again on monday.