ISRO2020-63

Question

What is the output in a $32$ bit machine with $32$ bit compiler?

#include<stdio.h>
rer(int **ptr2, int **ptr1)
{
    int *ii;
    ii=*ptr2;
    *ptr2=*ptr1;
    *ptr1=ii;
    **ptr1*=**ptr2;
    **ptr2+=**ptr1;
}
void main(){
    int var1=5, var2=10;
    int *ptr1=&var1,*ptr2=&var2;
    rer(&ptr1,&ptr2);
    printf("%d %d",var2,var1);
}

• A. $60,70$
• B. $50,50$
• C. $50,60$
• D. $60,50$

Comments

Answer: Option D but I marked option C Because of reading this line wrong way.

printf"%d%d",var2,var1);

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anybody please help it,

ans wil be 50,55. i tested this code in turboc32bit compiler and also in codeblock ide.

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In the calling function from main &ptr1, &ptr2 is used while calling..

But while receiving in called function : **ptr2 ,**ptr1 is used.

You can see the function receiving is reversing the parameters.

So by calling reversibly the pointers I think this exchanges the pointer values ( just by the way they are received)

You might have written rer(**ptr1,**ptr2) that's why you might be getting the answer

Answer 1

$\underline{\textbf{Answer:}\Rightarrow}\;\;{\textbf{(d)}}$

$\underline{\textbf{Explanation:}\Rightarrow}$

Let's assume the addresses in terms of $\mathrm{x100, x200,\dots}$

Now,

$\mathrm{var1} = \underset{\mathrm x100}{\bbox[lightblue,5px,border: 2px solid red]{5}}\;\;$, $\mathrm{var2} = \underset{\mathrm x200}{\bbox[lightblue,5px,border: 2px solid red]{10}}$

$\mathrm{ptr1} = \underset{\mathrm x300}{\bbox[lightblue,5px,border: 2px solid red]{\mathrm x100}}\;\;$, $\mathrm{ptr2} = \underset{\mathrm x400}{\bbox[lightblue,5px,border: 2px solid red]{\mathrm x200}}\;\;$

Now, on calling the function $\underline{\mathbf{rer(\&ptr1, \;\&ptr2)}}$

$\require{cancel}{\mathrm{ptr2} = \underset{\mathrm x500}{\bbox[lightblue,5px,border: 2px solid red]{\cancel{\mathrm x300}_{\mathrm x400}}}\;\;,\mathrm{ptr1} = \underset{\mathrm x600}{\bbox[lightblue,5px,border: 2px solid red]{\cancel{\mathrm x400}_{\mathrm x300}}}\;\;}$

$\mathrm{ptr1} = \underset{\mathrm x700}{\bbox[lightblue,5px,border: 2px solid red]{\mathrm x300}}$

Now,

$\mathrm{**ptr1 \color{red}{*}= ** ptr2}$ means it will assign the product of the values of double pointer $\mathbf{ptr2}$ and $\mathbf{ptr1}$ to the double pointer $\mathbf{ptr1}$

$\therefore \;\;\require{cancel} \mathrm{var1} = \underset{\mathrm x100}{\bbox[lightblue,5px,border: 2px solid red]{\cancel{5}_{50}}}\;\;$, $\mathrm{var2} = \underset{\mathrm x200}{\bbox[lightblue,5px,border: 2px solid red]{10}}$

$\therefore $ After executing statement $\mathbf{**ptr2 \;+= \; **ptr1}$ which can be again simplified as

$\mathrm{**ptr2 = **ptr2 + ** ptr1\; = \; 50 + 10 = \mathbf{60}}$

So, after executing the below statement, output will be:

$\color{magenta}{\textbf{printf("%d%d",var2,var1);}}$

$\mathbf{Output = 60, 50}$

$\therefore \mathbf{(d)}$ is the correct option.

Comments

srestha  JEET the output of the above program differs for the 64-bit machine?

it's useful if some old questions are provided as ref links for the above topic.

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@srestha

You have to multiple pointer 1 with pointer 2.

You mis typed that.

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Originally given:

$\mathbf{**ptr1 \color {red}*= **ptr2}$

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@Hira Thakur Yes, it does.

But I don't think it will vary here.

Let me see if I can find some.