static keyword in c:
Keyword static is used for declaring static variables in c. This modifier is used with all data types like int, float, double, array, pointer, structure, function etc.
Important points about static keyword:
1. It is not default storage class of global variables. For example, analyze the following three programs and its output.
(a)
#include<stdio.h>
int a;
int main(){
printf("%d",a);
return 0;
}
Output: 0
(b)
#include<stdio.h>
static int a;
int main(){
printf("%d",a);
return 0;
}
Output: 0
(c)
#include<stdio.h>
extern int a;
int main(){
printf("%d",a);
return 0;
}
Output: Compilation error
At first glance if you will observe the output of above three codes you can say default storage class of global variable is static. But it is not true. Why? Read extern storage class.
2. Default initial value of static integral type variables are zero otherwise null. For example:
#include <stdio.h>
static char c;
static int i;
static float f;
static char *str;
int main(){
printf("%d %d %f %s",c,i,f,str);
return 0;
}
Output: 0 0 0.000000 (null)
3. A same static variable can be declared many times but we can initialize at only one time. For example:
(a)
#include <stdio.h>
static int i; //Declaring the variable i.
static int i=25; //Initializing the variable.
static int i; //Again declaring the variable i.
int main(){
static int i; //Again declaring the variable i.
printf("%d",i);
return 0;
}
Output: 25
(b)
#include <stdio.h>
static int i; //Declaring the variable
static int i=25; //Initializing the variable
int main(){
printf("%d",i);
return 0;
}
static int i=20; //Again initializing the variable
Output: Compilation error: Multiple initialization variable i.
4. We cannot write any assignment statement globally. For example:
#include <stdio.h>
static int i=10; //Initialization statement
i=25; //Assignment statement
int main(){
printf("%d",i);
return 0;
}
Output: Compilation error
Note: Assigning any value to the variable at the time of declaration is known as initialization while assigning any value to variable not at the time of declaration is known assignment.
(b)
#include <stdio.h>
static int i=10;
int main(){
i=25; //Assignment statement
printf("%d",i);
return 0;
}
Output: 25
(5) A static variable initializes only one time in whole program. For example:
#include <stdio.h>
static int i=10;
int main(){
i=5;
for(i=0;i<5;i++){
static int a=10; //This statement will execute
//only time.
printf("%d",a++);//This statement will execute
//five times.
}
return 0;
}
Output: 10 11 12 13 14
(6)If we declared static variable locally then its visibility will within a block where it has declared. For example:
#include<stdio.h>
int main(){
{
static int a=5;
printf("%d",a);
}
//printf("%d",a); variable a is not visible here.
return 0;
}
Output: 5
7. If declared a static variable or function globally then its visibility will only the file in which it has declared not in the other files. For example:
(a)
#include<stdio.h>
static float a=144.0f; //global to all function
int main(){
{
printf("%d",a); //variable a is visible here.
//printf("%d",b); variable b is not visible here.
}
printf("%d",a); //variable a is visible here.
//printf("%d",b); variable b is not visible here.
return 0;
}
static int b=5; //Global to only calculation function
void calculation(){
printf("%d",a); //variable a is visible here.
printf("%d",b); //variable b is visible here.
}
(b) Consider a c program which has written in two files named as one.c and two.c:
//one.c
#include<conio.h>
static int i=25;
static int j=5;
void main(){
clrscr();
sum();
getch();
}
//two.c
#include<stdio.h>
extern int i; //Declaration of variable i.
extern int j; //Declaration of variable j.
/**
Above two lines will search the initialization statement of variable i and j either in two.c (if initialized variable is static or extern) or one.c (if initialized variable is extern)
*/
extern void sum(){
int s;
s=i+j;
printf("%d",s);
}
Compile and execute above two file one.c and two.c at the same time:
In Turbo c compiler
Step 1: Write above two codes in the file named as one.c and two.c (You can give any name as you like) and save it.
Step 2: In Turbo c++ IDE click on Project -> Open project menu as shown in following screen dump.
Step 3: After Clicking on open project you will get following screen:
In Open project File text field write any project name with .prj extension. In this example I am writing project name as CProject.PRJ. Now press OK button.
Step 4: After pressing OK button you will get following screen:
Now click on Project -> Add item menu.
Step 5: After clicking Add item you will get following screen:
In the name text field write down all c source code file one by one i.e. first write one.c and click on Add button
Then write two.c and click on Add button and so on
Step 6: At the end click on Done button. After clicking on done button you will get following screen:
At the lower part of window you can see project name, list of files you have added etc.
Step7: To compile the two files press Alt+F9 and to run the above program press Ctrl+F9
Note: To close the project click on Project -> Close project.
Output: Compilation error: Unknown symbol i and j.
Hence we can say variable i and j which has initialized into two.c is not visible in file one.c. This example proves visibility of globally declared static variable is file.
Note: In the above example function sum which was declared and defined in two.c has also storage class extern. So we can call from other file (one.c).If it will static then we cannot call function sum since static storage class is only visible to the file where it has declared.
(8)If we static variable has declared locally or globally its scope will always whole the program. For example:
(a) //locally declaration of static variable
#include<stdio.h>
void visit();
int main(){
int i=0;
{ //Opening inner block
static int a=5; //locally declaration
XYZ:; //Label of goto statement
printf("%d ",a);
a++;
i++;
} //closing inner block.
visit();
/* printf("%d",a); Variable a is not visible here but
it is alive. */
if(i<5)
goto XYZ;
return 0;
}
void visit(){
}
Output: 5 6 7 8 9
Explanation: When program control will come out of inner block where variable a has declared then outside of inner block variable a is not visible but its scope is outside the program i.e. variable a hasn’t dead .If with help of goto statement control again comes inside the inner block it prints previous incremented values which was not possible in case of auto or register variables.
(b)
//Locally declarations of variable
There are two c source code files:
//one.c
#include<stdio.h>
#include<conio.h>
void main(){
int i;
for(i=0;i<3;i++){
{
static int a=5;
printf("%d\n",a);
a++;
}
visit();
}
getch();
}
//two.c
#include<stdio.h>
void visit(){
printf("Don’t disturb, I am learning storage class");
/* printf("%d",a); Variable a is not visible here but
It is alive. */
}
Now compile and execute both files together:
Output:
5
disturb, I am learning storage class
6
disturb, I am learning storage class
7
disturb, I am learning storage class
Explanation: When control goes to another file and comes even that variable didn’t dead and it prints previous incremented value.
Note: In both examples if you will declare static variable globally you will get same output.
9. A static variables or functions have internal linkage. An internal linkage variables or functions are visible to the file where it has declared.
23 comments:
explanation is just fab.
In 8th point (b)part,you have called visit in one.c file but in two.c file you have not declared the extern keyword before its defination...can u plz tel me why?
in the 8th point we r learning the scope of static at out side of the files..
in the 8th point the visit() function must be declared extern in two.c .....correct me if i am wrong..??
Thanks.
reaaly nice 1....commendable
#include
static int i; //Declaring the variable i.
static int i=25; //Initializing the variable.
static int i; //Again declaring the variable i.
int main(){
static int i; //Again declaring the variable i.
printf("%d",i);
return 0;
}
Output: 25
her if we compile output we should get 0 only ..becz in main block locally i of course static is not initialised...so it intialissed by default as zero....comparing local and global local variable got more preference than global static variable(i=25)..so compulsory we will get zero as output....
plz comment to this comment ..to conform it correct analysis or not
i checked in vc++,it gives the error while compiling /**error 'i' : redefinition; different storage class**/
Ya thats correct actually it gives output as o.
that was a complete tutorial!! thanks brother!!
Yes you are correct in gcc I do get the same.
vijay
brilliant explanation :D
All functions have extern as the default storage class, so no you don't need to write it.
yes correct
Why its giving output as : 43
Can anyone please explain it to me?
Thanx in advance ?
#include
static int i; //Declaring the variable i.
static int i=25; //Initializing the variable.
static int i; //Again declaring the variable i.
int main(){
int i; //Again declaring the variable i.
printf("%d",i);
return 0;
}
int i; //Again declaring the variable i.
There is no static keyword in front of the declaration. It is "another" i, an automatic, uninitialized i. The content of the memory area where it is placed is accidentally 43.
3. A same static variable can be declared many times but we can initialize at only one time. For example:
(a)
#include
static int i; //Declaring the variable i.
static int i=25; //Initializing the variable.
static int i; //Again declaring the variable i.
int main(){
static int i; //Again declaring the variable i.
printf("%d",i);
return 0;
}
Output: 25
Here Output should be zero, not 25.
Here, it gives an output 25 but not zero because it has been initialized as 25 although it has been declared again as a local variable.
But concepts are wrong... pls try to execute the prg your self u will find lots of error
I was wondering if you could tell me what this means as it is the name of a withdrawal from my bank account.
OUTPUTON.C 39.95_V
The explanation is awesome.....thanks a lot
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