sum (adding 2 numbers ) without plus operator

Question

Can anyone explain the logic how to add a and b?

#include 

int main()
{
     int a=30000, b=20, sum;
     char *p;
             


        

Answer 1

The + is hidden here:

&p[b]

this expression is equivalent to

(p + b)

So we actually have:

(int) &p[b] == (int) ((char *) a)[b]) == (int) ((char *) a + b) == a + b

Note that this technically invokes undefined behavior as (char *) a has to point to an object and pointer arithmetic outside an object or one past the object invokes undefined behavior.

Answer 2

An alternative to the pointer arithmetic is to use bitops:

#include <stdio.h>
#include <string.h>

unsigned addtwo(unsigned one, unsigned two);

unsigned addtwo(unsigned one, unsigned two)
{
unsigned carry;

for( ;two; two = carry << 1)    { 
        carry = one & two; 
        one ^= two;
        } 
return one;
}

int main(int argc, char **argv)
{
unsigned one, two, result;

if ( sscanf(argv[1], "%u", &one ) < 1) return 0;
if ( sscanf(argv[2], "%u", &two ) < 1) return 0;

result = addtwo(one, two);

fprintf(stdout, "One:=%u Two=%u Result=%u\n", one, two, result );

return 0;
}

Answer 3

p[b] is the b-th element of the array p. It's like writing *(p + b).

Now, when adding & it'll be like writing: p + b * sizeof(char) which is p + b.
Now, you'll have (int)((char *) a + b) which is.. a + b.

But.. when you still have + in your keyboard, use it.

---

As @gerijeshchauhan clarified in the comments, * and & are inverse operations, they cancel each other. So &*(p + b) is p + b.