GATE IT 2008 | Question: 50

Question

Consider the C program below. What does it print?

# include <stdio.h>
# define swap1 (a, b) tmp = a; a = b; b = tmp
void swap2 ( int a, int b)
{
        int tmp;
        tmp = a; a = b; b = tmp;
 }
void swap3 (int*a, int*b)
{
        int tmp;
        tmp = *a; *a = *b; *b = tmp;
}
int main ()
{
        int num1 = 5, num2 = 4, tmp;
        if (num1 < num2) {swap1 (num1, num2);}
        if (num1 < num2) {swap2 (num1 + 1, num2);}
        if (num1 > = num2) {swap3 (&num1, &num2);}
        printf ("%d, %d", num1, num2);
}

• A. $5, 5$
• B. $5, 4$
• C. $4, 5$
• D. $4, 4$

Comments

if (num1 > num2) {swap1 (num1, num2);}
if (num1 < num2) {swap2 (num1 + 1, num2);}
if (num1 > = num2) {swap3 (&num1, &num2);}

 Now ans will be option (b) , First if condition is True so macros will be expanded and it will swap the value of num1 and num2 so num1 becomes 4 and num2 becomes 5 , 2nd if condition is False now come to 3 rd if  condition which is true and it's a real swap of num values using the call by reference, num1 becomes 5 and num 2 become 4. 

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K&R Pg.no. 89

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I think the question was edited multiple times as different asnwers over time are using different code (especially the if conditions).

This code is the one I found in gfg too as of the time I'm writing this.

Right answer is this accepted answer, where only statement 3 is getting executed.

        if (num1 < num2) {swap1 (num1, num2);}
        if (num1 < num2) {swap2 (num1 + 1, num2);}
        if (num1 > = num2) {swap3 (&num1, &num2);}

Answer 1

Answer is C.

Only:

if (num1 > = num2) {swap3 (&num1, &num2);}

Statement works, which in turn swaps num1 and num2.

Comments

Hi, could you kindly give a detailed explanation ? I could not understand the solution. Not even from the comments am I clear.

5>4. So 1^st condition is true !. Macro swaps so the 2^nd cond is also true ! swap2 only changes adress so no change. 3^rd cond is true and finally this gives us (5,5)…..

I understand my answer is wrong.
Could you kindly correct my mistakes?

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Bhai,I am not getting this swapping by macro

How it is been performed???

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Right answer but wrong approach. swap1 expands and variable gets swapped. swap2 & swap3 won't execute.

Answer 2

In main program num1=5, num2=4

first condition true. num1=4, num2=5

Now, second condition become true but no affect on values still num1=4, num2=5

third condition fail.

Ans: 4,5

Comments

Is 5<4???? @KunuSwavik

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Wrong! GCC expands swap1 and only this gets executed.

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I believe that using macros it cannot be swap.

the third condition will run.

Answer 3

the swap1( ) function does not create new variables for a and b and temp. so it changes the same a and b. so they become 4 and 5.

swap2( ) function cant change the value of a and b because it is called by value and create different a and b.

the third condition does not get satisfied. so swap3( ) is not called and values remain as 4 and 5 for a and 5 respectively.

Comments

Swap 1 is executing before the birth of a,b ... Then how it changes the *same a and b* ?

Answer 4

Here is my approach:

 

 

Answer 5

The macro function gets subsituted by its defination. So, basically wherever in code swap1(a,b) is used it is replaced with 
tmp = a; a = b; b = tmp;

After substitution the code becomes:
 

int main(){

int num1 = 5, num2 = 4 , tmp;

if(num1 > num2){
    tmp = num1; num1 = num2; num2 = tmp;
}
...

So, first if condition (5 > 4) is satisfied . After swapping the values of num1 and num2 become 4, 5.

The second if condition(4 < 5 .) is satisfied. But as function call is by value, so num1 and num2 don't change.

The third if condition(4 >= 5) fails .

So, final answer is num1 = 4 , num2 = 5. (C)

Answer 6

Option C is the answer.

• num1 = 5
• num2 = 4

1. First if statement: if(num1>num2)
   • 5 > 4 is true, so we execute swap1(num1, num2)
   • swap1 is a macro that expands to:
   • {
int tmp = num1;
num1 = num2;
num2 = tmp;
}
   • This successfully swaps the values
   • After swap1: num1 = 4, num2 = 5
      
2. Second if statement: if(num1<num2)
   • 4 < 5 is true, so we execute swap2(num1 + 1, num2)
   • swap2 receives copies of the values (5 and 5)
   • This function swaps the copies but doesn't affect the original variables
   • After swap2: num1 = 4, num2 = 5 (unchanged)
      
3. Third if statement: if(num1>=num2)
   • 4 >= 5 is false, so swap3 is not executed
      
4. Thus, prints: 4, 5