GATE CSE 2012 | Question: 49

Question

Consider the following C code segment.

int a, b, c = 0; 
void prtFun(void); 
main()
{ 
    static int a = 1;       /* Line 1 */
    prtFun(); 
    a += 1;
    prtFun();
    printf(“ \n %d %d ”, a, b);
}

void prtFun(void)
{
    static int a = 2;       /* Line 2 */
    int b = 1;
    a += ++b;
    printf(“ \n %d %d ”, a, b);
}

What output will be generated by the given code segment if:

Line 1 is replaced by auto int $a = 1$;

Line 2 is replaced by register int $a = 2$; 

• A. $\begin{array}{ll}  \text{3}  & \text{1} \\ \text{4}  & \text{1} \\ \text{4}  & \text{2} \\ \end{array}$
• B. $\begin{array}{ll}  \text{4}  & \text{2} \\ \text{6}  & \text{1} \\ \text{6}  & \text{1} \\ \end{array}$
• C. $\begin{array}{ll}  \text{4}  & \text{2} \\ \text{6}  & \text{2} \\ \text{2}  & \text{0} \\ \end{array}$
• D. $\begin{array}{ll}  \text{4}  & \text{2} \\ \text{4}  & \text{2} \\ \text{2}  & \text{0} \\ \end{array}$

Comments

Link to program execution: https://ideone.com/t5t6e

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Register is used for faster computation

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Will be useful for static variables : https://gateoverflow.in/2176/gate-cse-2012-question-48?show=386635#c386635

Answer 1

main
$a=1$

$prtFun()$
$a=2$
$b=1$
$a= a  + \text{++}b = 2+2 = 4$
$b = 2$
printf $\rightarrow  4 \ 2$
back to main
$a = a+1 \rightarrow 1+1 \rightarrow 2$

$prtFun()$
$a=2$ //previous a is lost
$b=1$
$a= a  + \text{++}b = 2+2 = 4$
$b = 2 $
printf $\rightarrow  4 \ 2$

back to main
$a = 2$
$b = 0$ (initial value of global b. in $prtFun$ local b is only updated)
printf $\rightarrow 2 \ 0$

Answer is D.

Comments

Small correction I think In second prtfun() record a=2 not 1 @sankarnarayanan.p.n

And thanks anu007 for detail reference.

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@MRINMOY_HALDER, Scope - tells where this identifier can be available to access and Life Time -  says how long this identifier will live in the program irrespective of availability for accessing or not.

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can you please share the code after the changes suggested in question are done. I’m confused about which lines are referenced as line 1 and line 2.

Answer 2

Ans D

Answer 3

I tried the same and I am getting C
Here is my code:

#include<stdio.h>

int a, b, c = 0;
void prtFun(void);
main() {
    static int a = 1;       /* Line 1 */
    prtFun();
    a += 1;
    prtFun();
    printf("\n%d %d", a, b);
    }

     void prtFun(void)
     {
         static int a = 2;       /* Line 2 */
         int b = 1;
         a += ++b;
        printf("\n%d %d", a, b);
        }
 

Comments

This can clear all the doubts:

http://web.archive.org/web/20130927234242/http://itee.uq.edu.au/~comp2303/Leslie_C_ref/C/CONCEPT/storage_class.html

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best link .. thank you :)

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Static keyword which declares locally to a function,doesnot update global variable. Only the variable in main will update the value of global variable . Even if we simply run this code

https://ideone.com/aXzwNc

the output clearly shows static keyword not updated global variable value

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can you please share the code after the changes suggested in question are done. I’m confused about which lines are referenced as line 1 and line 2.

Answer 4

Static is special but Register and Auto has nothing special in terms of Scope. So a simple execution follows.

Comments

There is something special about register variables  too, they speed up program execution but wont affect output generated in anyway.!!

Answer 5

Shortcut to question. 

register keyword causes variabe to be stored in regster no effect extra. 
we removed static keyword from lin2

prtFun has no inputs, no static variables. So each time we call it executes as new. So both time calls it will give same output. And first 2 times prtFun's print stmt executd and at last main()'s print executed. 

So first two ines of output SHOULD be same 
Only option D qualifies this condition. 

Answer 6

 Since we completely removed the static keyword from prtFun(), the function no longer remembers anything from its previous executions. Every single time you call prtFun(),  it starts completely fresh from scratch!