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#include<stdio.h>
#include<conio.h>
# define swap(a,b) temp=a; a=b; b=temp;
void main( )
{
int i, j, temp;
i=5;
j=10;
temp=0;
if( i > j)
swap( i, j );
printf( "%d %d %d", i, j, temp);
}

Answers were Sorted based on User's Feedback



#include<stdio.h> #include<conio.h> # define swap(a,b) temp=a; a=b; b=temp; void mai..

Answer / rohit751

Hi all...the main mistake in the program is using the semi
colons in the macro definition. Try this..
#include<stdio.h>
#include<conio.h>
# define swap(a,b) temp=a, a=b, b=temp;
void main( )
{
int i, j, temp;
i=5;
j=10;
temp=0;
if( i < j)
swap( i, j );
printf( "%d %d %d", i, j, temp);
}

Is This Answer Correct ?    8 Yes 0 No

#include<stdio.h> #include<conio.h> # define swap(a,b) temp=a; a=b; b=temp; void mai..

Answer / ashish rajvanshi

My dear frnds its a simple question
here the code of macro inserted in the program at the time of preprocessing and then the calculation will performed.
and then it will be calculated.

void main()
{
int i,j, temp;
i=5;
j=10;
temp=0;
if(i>j)
temp=a;
a=b;
b=temp;
printf("%d %d %d" , i , j , temp);
}

for further detail contect at(ashish.raj.mrt@gmail.com)

Is This Answer Correct ?    3 Yes 1 No

#include<stdio.h> #include<conio.h> # define swap(a,b) temp=a; a=b; b=temp; void mai..

Answer / ashishrajvanshi01

On compiling i got ans 10, 0, 0. I did not understand the concept. Please expalin this to me.

Is This Answer Correct ?    5 Yes 4 No

#include<stdio.h> #include<conio.h> # define swap(a,b) temp=a; a=b; b=temp; void mai..

Answer / vinodhini

The output will be
5100

Is This Answer Correct ?    1 Yes 0 No

#include<stdio.h> #include<conio.h> # define swap(a,b) temp=a; a=b; b=temp; void mai..

Answer / harish

it enters into swap function only when i>j.

but i contains 5 and j contains 10.

can u plz explain this

Is This Answer Correct ?    0 Yes 0 No

#include<stdio.h> #include<conio.h> # define swap(a,b) temp=a; a=b; b=temp; void mai..

Answer / abhijeet

my frnd, C is dumb compiler. It doesn't know what actually
we mean.
Here, in this program, we have defined 1 macro. Ok
this macro wil be inserted and expanded into code.
this gives the unpredicted result. U try and find out what i
mean to say.

Is This Answer Correct ?    1 Yes 1 No

#include<stdio.h> #include<conio.h> # define swap(a,b) temp=a; a=b; b=temp; void mai..

Answer / shweta

#include<stdio.h>
#define fun(a) #a#a#a
void main(){
printf("%u",fun(5+9));
}

Is This Answer Correct ?    1 Yes 1 No

#include<stdio.h> #include<conio.h> # define swap(a,b) temp=a; a=b; b=temp; void mai..

Answer / kumaravel

chance of providing answer is not possible. empty screen become view.

Is This Answer Correct ?    0 Yes 0 No

#include<stdio.h> #include<conio.h> # define swap(a,b) temp=a; a=b; b=temp; void mai..

Answer / manish soni tagore collage jai

my dear frd....
here some hapen.
#define swap(a,b) temp=a;a=b;b=temp;----------(1)
is was as function of swap two no;(but it is differnt frm
funtion.)
so
at i>j;
enter swap(i,j);
so jump at eq(n) ---(1) and call swap function'
and


temp=5;


i=10;
j=0;

you see three variable.
but it printf i,j to convert swap.
bcoz it have two arguments.
and print temp=0;
becoz temp have 0;

Is This Answer Correct ?    0 Yes 7 No

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