Find sum of two numbers

[uday.bidkar]

#include<stdio.h>
int main()
{
int i;
printf("Enter 2 numbers\n");

/*
Write here few lines of code so that user gives 2 numbers as
input(on console) and display the addition of those 2 numbers.. and
the condition is no more variable(s) except for 'i' declared in first line
should be used throughout the program.
*/

return 0;
}

Regards,
Uday

[sahana]

hi,
i have used the consecutive memory location of i to store the 2 data.this works.
#include<stdio.h>
#include<conio.h>
void main()
{
int i;
scanf("%u",&i);
scanf("%u",&i+1);
printf("%d",(*(&(i)))+(*(&(i)+1)));
}

[uday.bidkar]

Originally Posted by sahana

hi,
i have used the consecutive memory location of i to store the 2 data.this works.
#include<stdio.h>
#include<conio.h>
void main()
{
int i;
scanf("%u",&i);
scanf("%u",&i+1);
printf("%d",(*(&(i)))+(*(&(i)+1)));
}

The memory pointed to &i+1 is not valid memory that you can use for operations as you have not allocated the same neither asked compiler to allocate it for you on stack. Hence the program may crash because of invalid memory operations. So solution not acceptable!

-Uday

[DEP]

Originally Posted by uday.bidkar

The memory pointed to &i+1 is not valid memory that you can use for operations as you have not allocated the same neither asked compiler to allocate it for you on stack. Hence the program may crash because of invalid memory operations. So solution not acceptable!

-Uday

hi uday,
i have tried a lot to solve this problem and believe me whether you accept it or not pointers is the only way to solve it.

i guess i quite agree with the solution provided by sahana and am unable to figure out any other solution.

so please post any other solution if you have it !!!

[uday.bidkar]

Originally Posted by DEP

hi uday,
i have tried a lot to solve this problem and believe me whether you accept it or not pointers is the only way to solve it.

i guess i quite agree with the solution provided by sahana and am unable to figure out any other solution.

so please post any other solution if you have it !!!

There is a way to solve this. But before i post the answer for this, I would like to relax the condition and wait for few more days.
And the relaxed question is
#include<stdio.h>
int main()
{
/*
Write here few lines of code so that user gives 2 numbers as
input(on console) and display the addition of those 2 numbers.. and the
condition is only one integer variable should be declared and used
throughout the program. You are now free to declare the variable as you want.
*/

}

[DEP]

Originally Posted by uday.bidkar

There is a way to solve this. But before i post the answer for this, I would like to relax the condition and wait for few more days.
And the relaxed question is
#include<stdio.h>
int main()
{
int *a;
cin>>*a;
cin>>*(a+1);
cout<<*a+*(a+1);
}

well uday i am sorry but this is the only solution that i have worked out.and i guess this is the c++ equivalent code of what sahanna had posted.
please please please tell the answer. am dying to know it.

[uday.bidkar]

Originally Posted by DEP

please please please tell the answer. am dying to know it.

Here is answer to the problem when variable is not
already declared.

int main()
{
static int i = 0;
if( !i )
{
i = -1 ;
printf( "The Sum is %d\n", main() + main() ) ;
return 0 ;
}
printf( "Enter a number\n" );
scanf( "%d",&i );
if( !i )
{
i = -1 ;
return 0 ;
}
else
{
return i ;
}
}

Regards,
Uday

[sahana]

wow.that was really smart.
i accept my mistake.well actually i actually got the output ,as i was lucky that the unallocated memory was not a part of anyother program, and didnt overwrite any data of another program.

[Satyajit]

hi i have the programme of sum of 2 numbers
#include<stdio.h>
void main()
{
int a,b,c;
printf("enter any 2 numbers");
scanf("%d%d",&a,&b);
c=a+b;
printf("%d",&c);
}

[The_Big_K]

Originally Posted by Satyajit

hi i have the programme of sum of 2 numbers
#include<stdio.h>
void main()
{
int a,b,c;
printf("enter any 2 numbers");
scanf("%d%d",&a,&b);
c=a+b;
printf("%d",&c);
}

Satyajit, mind reading the first post in this thread? There is a special condition to be followed.

-The Big K-