How to reference a variable outside it's block scope in C

daYz · 10-02-2007, 07:04 AM

Hi

I want to make use of a variable outside of the for loop in C from where it is initialized. I wonder what a common solution is for that situation. I am trying to achieve this with the static specifier but that does not work the way I implement it:

Code:

#include <stdio.h>

main()
{

       int i;
       
       for(i=0;i<5;i++){
              
       static int k = 1;                                    
                        
}

printf("%s", k);       

return 0;
                         
}

I get an "'k' undeclared" error.

So what is a common solution in this situation, and what would be the proper way of using the static specifier in this case?

Thanks

virtualCoder · 10-02-2007, 07:47 AM

Hi,

Well first of all this line seems to be incorrect:
printf("%s", k);
since %s is used for strings, and k is an integer.

Second, well you can refer to a variable outside the scope of a for loop by declaring it
outside the scope of a for loop.
So for example:

Code:

// Declare i and k.
int i, k;


for (i=0;i<5;i++) {
// Scope of for loop in here.
  k = 1;
}

// Use i and k outside scope of for loop since they existed outside in the first place.
printf("i = %d, k = %d\n", i, k);  // Will print "i = 5, k = 1".

Hope that clears your problem.

bigearsbilly · 10-02-2007, 07:59 AM

declarations should come first in a C block i.e. function.
you can't declare in a loop

daYz · 10-02-2007, 08:04 AM

Thanks guys

That solves my problem.

rjlee · 10-02-2007, 08:57 AM

Actually, in ANSI C you can declare a variable inside a loop or block:

Code:

#include <stdio.h>

int main() {
  int i;
  for (i=0; i < 10; ++i) {
    int j=i;
    printf("%d\n", j);
  }
  return 0;
}

This compiles and runs without warnings:

Code:

$gcc test.c -o ./test -Wall -ansi && ./test
0
1
2
3
4
5
6
7
8
9

The important thing to remember is only to declare variables at the start a block, and they are not visible outside of that block.

Of course, there is an argument that if you can put a variable inside a block inside a method then that block should probably be a method in itself.

—Robert J Lee

daYz · 10-02-2007, 12:46 PM

Thanks rjlee.

I have run into another problem. When I declare an array, I am unable to initialize it later on. Normally I just did something like

Code:

int i[3] = { 1, 2, 3 };

, but when I declare i first now, like

Code:

int i[3];

, I am unable to intialize it afterwards. So can someone explain to me how this can be done? I know it can be done like this when you intitialize a variable with just one integer:

Code:

int i;

Code:

i = 1;

, but I don't understand why this can't be done when using an array.

rstewart · 10-02-2007, 01:10 PM

Hi,

Code:

int main()
{
    int i[3];

    i[0] = 1;
    i[1] = 2;
    i[2] = 3;
}

--  OR  --
int main()
{
 
   int i[3];
   int j;

   for ( j = 0; j < 3; j++ )
      i[j] = j;
}

You get the idea.

devn · 10-03-2007, 02:54 AM

Hi,
Try is, if it helps your case anyhow...

Code:

#include <stdio.h>

int main()
{

int i[3];

i[1]=1200; // assign 1200 to 2nd index of array
printf("%d",i[1]);

return 0;
}

I guess this is what you already know, but anyway, since all the entities within array are treated as single intergers, just attached together (actually addresses of all the integers with array are allocated at runtime to link them in an integer), you can manipulate any index of the array just by pointing your index at it.

daYz · 10-03-2007, 03:48 AM

Thanks guys

It is clear to me now.

Regards,

Ben