GATE CSE 2015 | Set 2 | Question: 38

Question

Consider the C program below

#include <stdio.h>
int *A, stkTop;
int stkFunc (int opcode, int val)
{
    static int size=0, stkTop=0;
    switch (opcode) {
        case -1: size = val; break;
        case 0: if (stkTop < size ) A[stkTop++]=val; break;
        default: if (stkTop) return A[--stkTop];
    }
    return -1;
}
int main()
{
    int B[20]; A=B; stkTop = -1;
    stkFunc (-1, 10);
    stkFunc (0, 5);
    stkFunc (0, 10);
    printf ("%d\n", stkFunc(1, 0)+ stkFunc(1, 0));
}

The value printed by the above program is ________.

Comments

in this question the scope of stkTop is local to the function then while calling another time why it is not 0.

someone pls explain this

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Bcoz it is static!

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but cant understand how they change the static value of static int size = 0,stkTop =0;

for swithc statement the static will not be changed

 

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@ashutosh_verma, static variable persists between function calls. read this https://stackoverflow.com/questions/572547/what-does-static-mean-in-c

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the feeling of getting 15😁

Answer 1

Initially stack is empty = -1

stkFunc (-1, 10); this function

case -1: size = val; break; and static size= 10 // size memory declare one time only// and control comes out of switch b/c of break

stkFunc (0, 5); this function run

case 0: if (stkTop < size ) A[stkTop++]=val; break; here stktop is static value so memory declare at compile time only now check if condition 0< 10 true then A[stktop++== A[0+1]=val= 5 i.e. push 5 into stack break comes so control comes outside

stkFunc (0, 10); this comes

case 0: if (stkTop < size ) A[stkTop++]=val; break; same as above make A[stkTop++]= 10 i,e. push 10 into stack and break comes so control comes outside

printf ("%d\n", stkFunc(1, 0)+ stkFunc(1, 0));

this function  

stkFunc(1, 0) this will run

default: if (stkTop) return A[--stkTop] return top of stack which is 10

stkFunc(1, 0) this will run

default: if (stkTop) return A[--stkTop] return top of stack which is 5

printf ("%d\n", stkFunc(1, 0)+ stkFunc(1, 0));= 5+10=15 15 will be printed

Comments

how are A [stacktop++] and A [--stacktop] different? is there any case of post or pore increment? if so then A [--stacktop] should pop 5 first and then stop at index before 0.

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does static variable overrides global variable?

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No, a static variable does not override a global variable. However, it can hide or shadow the global variable within its scope.

but 

Inside the function, the static variable with the same name will always be used instead of the global variable unless you explicitly refer to the global variable (possible in C++ or with workarounds in C)

Answer 2

Answer: $15$

The code is pushing $5$ and $10$ on stack and then popping the top two elements and printing their sum.

http://ideone.com/kIUdQT

Answer 3

By Analyzing the above code it shows that.. this is simply stack implementation with 3 choices:-

Case (-1):-->>[Set Size of the Stack]

Code :-->>size = val

Case (0):-->>[ Push Operation]

                 Step1:- verify stack is full or not?

                 Step2:-  If not full then push the  element  in stkTop And increment stkTop.

                  Code:-->>if (stkTop < size )    //   step1

                  A[stkTop++]=val;  //step 2+3

Case (default):--(here when 1 is as choice)[Pop Operation]

                  step1:-> if stack is not empty thenPop the element from stkTop

                 Step2:->decrement stkTop

                 Code:->>if (stkTop) return A[--stkTop];// step 1+2

 

So simply in this program

1.set stack size 10

2.push 2 elements in top of the stack as

5 and 10

3.pop both ;then add them

So result is 15.

PS:->>But in exam we should directly trace the code and write the answer.

Answer 4

Thus, answer is 15.