Updating a global variable in C

Question

I have a beginners C question. I want in the code below...

include <stdio.h>

void iprint();
int i=0;

int main()
{
  int j;

  for (j=0; j<50; j++)
    {
      iprint(i);
      printf("%d\n",i);
    }
}

void iprint(i)
{
  i +=1;
  //printf("%d\n",i); 
}

... the function "iprint" to update the value of i each time is it called, e.g. update i so that it can be used in main with the value 1 for iteration 2, and 3 for iteration 2 etc.

I accomplished this by changing the code to this:

 include <stdio.h>

int iprint();
int i=0;

int main()
{
  int j;

  for (j=0; j<50; j++)
    {
      i= iprint(i);
      printf("%d\n",i);
    }
}

int iprint(i)
{
  i +=1;
  //printf("%d\n",i); 
  return(i);
}

Do i have to return(i) to make that happen? The reason for asking, is that if i have a lot of functions using i, it's a bit annoying having to pass i between them. If you instead, somehow could update i like you update a global variable in matlab, that'd be great. Is it possible?

This is not even syntactically correct. Does this actually compile? (not to mention using global variables is generally a bad smell ) — ppeterka
– ppeterka, Commented Sep 23, 2013 at 10:11
possible duplicate of Pointers in C and how to pass local variables as pointers — Dariusz
– Dariusz, Commented Sep 23, 2013 at 10:36

Vamsavardhana Vijay · Accepted Answer · 2016-04-20 00:37:47Z

7

Use a pointer to point to the global variable. Change the pointer value. Thats it

answered Apr 20, 2016 at 0:37

Vamsavardhana Vijay

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2 Comments

Yohan Obadia Over a year ago

That's probably the best answer in my opinion since it assess the underlined question of how to modify a variable from outside the scope of a function.

Raul Chiarella Over a year ago

Hey. Can u give us a example code ? :v In theory this is the solution, but i just did that and it did'nt work...

Some programmer dude · Accepted Answer · 2013-09-23 10:11:50Z

6

The problem with the first is that you pass the variable as an argument to the function, so when the function modifies the variable it's only modifying its own local copy and not the global variable. That is, the local variable i shadows the global variable i.

Not to mention that you don't actually declare the argument properly, so your program should not even compile.

answered Sep 23, 2013 at 10:11

Some programmer dude

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4 Comments

PP. Over a year ago

And you could modify this behaviour by passing a pointer to i, e.g. void iprint( int *i ) { *i += 1 } and calling iprint( &i );.

Some programmer dude Over a year ago

@PP. Why pass a global variable to a function at all?

RedX Over a year ago

@JoachimPileborg Maybe to make the function pure? Or to make the dependency visible.

PP. Over a year ago

@JoachimPileborg If the purpose of the function is to modify a variable then that function could be made generic and re-usable by accepting, as an argument, the variable to be modified. It makes software engineers happy because it helps keep coupling low. However you are correct that one need not do this in this context.

Klas Lindbäck · Accepted Answer · 2013-09-23 10:17:52Z

2

You don't need to pass global variables as parameters. If you declare a parameter or local variable with the same name as the global variable you will hide the global variable.

include <stdio.h>

void iprint();
int i=0;

int main()
{
  int j;

  for (j=0; j<50; j++)
    {
      iprint();
      printf("%d\n",i);
    }
}

void iprint()
{
  i +=1;  /* No local variable i is defined, so i refers to the global variable.
  //printf("%d\n",i); 
}

answered Sep 23, 2013 at 10:17

Klas Lindbäck

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Comments

Himanshu Pandey · Accepted Answer · 2013-09-23 10:31:03Z

0

you could have incremented the value of i in main function itself. By the way change the function to

int iprint(int i){
/*you have to mention the type of arguemnt and yes you have to return i, since i  
variable in this function is local vaiable when you increment this i the value of  
global variable i does not change.
*/
return i+1;
}

the statement

i=iprint(i); //this line updates the value of global i in main function

This happens like this because you are passing value in function by 'pass by value' method where a copy of variable is made. When you increment the i iprint method copy of global variable i is incremented. Global variable remains intact.

answered Sep 23, 2013 at 10:31

Himanshu Pandey

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Comments

Manav Akela · Accepted Answer · 2019-02-06 13:34:17Z

0

Must try this code

#include <stdio.h>
int i=0;
void iprint()
{
  i =i+1;
  //printf("%d\n",i); 
}
int main()
{
    int j;
    for (j=0; j<50; j++)
    {
      iprint();
      printf("%d\n",i);
    }
}

answered Feb 6, 2019 at 13:34

Manav Akela

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