Adding Two numbers with using only one variable in C - c

Question

I was trying to create a program that inputs two number and outputs their sum. For this I must have to use two variables. I was just curious whether this can be done by using only one variable.
Note : user has to input two numbers.
#include<stdio.h>
int main()
{
int a, b;
scanf("%d%d",&a,&b);
printf("%d",(a+b));
return 0;
}

Answer 1

#include <stdio.h>
int main ( void )
{
int a[3];
scanf("%d", &a[0]); /* first number */
sscanf("%d", &a[1] ); /* second number */
a[2] = a[0] + a[1];
printf("sum is %d\n", a[0] + a[1] );
printf("sum stored in a[%d] is %d\n", 2, a[2] );
return 0;
}

Answer 2

Technically one variable, a pointer:
#include<stdio.h>
int main() {
int *nums = malloc(2 * sizeof(int));
scanf("%d%d",nums, (nums + sizeof(int)));
printf("%d",(*nums + *(nums + sizeof(int))));
return 0;
}
But no there isn't really an elegant way to use one variable for two inputs.

Answer 3

Note that I've considered the question like a challenge or a puzzle. Do not consider this answer good C practice. Obviously the cleanest way to make a sum of 2 values from input is to use 2 variables. I still find the challenge interesting though.
#include <stdio.h>
#include <math.h>
int main()
{
int a;
printf("%g", fmin((scanf("%d", &a), a), 1.0/0.0 + rand()) + fmin((scanf("%d", &a), a), 1.0/0.0 + rand()));
return 0;
}
Works with negative values.
I'm using the comma operator which executes both expressions but only return the second one. So (scanf("%d", &a), a) is like calling scanf("%d", &a) and returns a. I pass this result through a function (any function) as I want to prevent updating the value (to sum it with the new a). I have no idea if your compiler will call the left or right part of the big expression first but it doesn't matter as both are doing the same thing. Whichever executes first will be the first value from input.
fmin(x, 1.0/0.0 + rand()) makes sure nothing is inlined by the compiler. 1.0/0.0 is Infinity and would never be returned in fmin() in our case. Compiler would inline this to x normally but adding + rand() to Infinity (which is still Infinity) seems to prevent it.
You can even do it by declaring "0" variable by using argc:
#include <stdio.h>
#include <math.h>
int main(int a)
{
printf("%g", fmin((scanf("%d", &a), a), 1.0/0.0 + rand()) + fmin((scanf("%d", &a), a), 1.0/0.0 + rand()));
return 0;
}
I've used this to test: https://www.onlinegdb.com/online_c_compiler

Answer 4

Adding Two numbers with using only one variable in C
Create a helper function with the 1 variable.
#include <stdio.h>
int scan_int(void) {
int a;
if (scanf("%d", &a) == 1) {
return a;
}
return 0;
}
int main(void) {
printf("Sum %d\n", scan_int() + scan_int());
return 0;
}
Note that scan_int() + scan_int(), code could call either the left or the right scan_int() first (or in parallel). Fortunately + is commutative, so it makes no difference here.
The "trick" here is that there exist in sequence or in parallel, a 1st_call_scan_int::a and 2nd_call_scan_int::a. Still only one variable in code.

Answer 5

OK, there's been quite a few interesting answers, but weirdly nobody has thought of the obvious way to store 2 ints in a single variable - structs:
#include<stdio.h>
typedef _in struct {
a int
b int
} inp;
int main(void)
{
inp input;
scanf("%d%d",&input.a,&input.b);
printf("%d",input.a+input.b);
return 0;
}

Answer 6

int main(void) {
int *num = malloc(sizeof(int)*2);
scanf("%d %d", num, num+1);
printf("%d\n", num[0] + (num[1]));
}
int *num = malloc(sizeof(int)*2); //two int space
scanf("%d %d", num, num+1); // num (pos 0), num+1 (pos1)
printf("%d\n", num[0] + (num[1])); //the sum of the positions

Answer 7

Only one variable - no tricks. As many numbers can be added as you want :)
#include <stdio.h>
int ScanAndAdd(void)
{
int a;
if(scanf("%d", &a) != 1) return 0;
return a + ScanAndAdd();
}
int main()
{
printf("%d\n", ScanAndAdd());
return 0; /**/
}

Answer 8

One way to do it is.
#include <stdio.h>
int x;
int enter(){
scanf("%d",&x);
return x;
}
int main()
{
x=enter()+enter();
print("sum of two number is %d",x);
return 0;
}
Another way to do it..
#include <stdio.h>
int main()
{
int x;
scanf("%d",&x);
printf("next no. ");
x= x*(scanf("%d",&x))+x;
printf("%d",x);
return 0;
}
Although the second one is not consistent, in some compiler it works perfectly and in some, it doesn't

Answer 9

#include <stdio.h>
int main()
{
long long int buffer=0;
scanf("%d",(int*)&buffer);
scanf("%d",(int *)&buffer+1);
printf("\nsum is %d\n",*((int*)&buffer)+*((int*)&buffer+1));
return 0;
}

Answer 10

Try this on for size. It uses fixed width C99 types to guarantee that memory alignment works as intended. It even does the 32-bit arithmetic in a uint64_t type to prevent overflow issues. This will work with any of the other fixed-width integer types, signed or unsigned, with trivial modification.
#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
int main (void)
{
uint64_t a;
scanf ("%" SCNu32 "%" SCNu32, (uint32_t*)&a, (uint32_t*)&a + 1);
printf ("%" PRIu64 "\n", (uint64_t)((uint32_t*)&a)[0] +
((uint32_t*)&a)[1]);
}
The above code was modified from my original answer, seen below. I removed the void* cast in (uint32_t*)(void*)&a because it was unnecessary. I also cleaned up the scanf arguments to increase readability, and added a new line to the end of the printf format argument.
#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
int main (void)
{
uint64_t a;
scanf ("%" SCNu32 "%" SCNu32,
&((uint32_t*)(void*)&a)[0], &((uint32_t*)(void*)&a)[1]);
printf ("%" PRIu64, (uint64_t)((uint32_t*)(void*)&a)[0] +
((uint32_t*)(void*)&a)[1]);
}