CODEWITHSUNDEEP
cpointers
Nishit Shah
Nishit Shah

Reputation: 192

adding two number using pointers

I found this code in the internet for adding two numbers using pointers. couldn't understand how it is working? Any help would be appreciated.

#include <stdio.h>
#include <conio.h>
int main()
{
      int  a,b,sum;
      char *p;
      printf("Enter 2 values : ");
      scanf("%d%d",&a,&b);
      p = (char *)a; // Using pointers
      sum = (int)&p[b];
      printf("sum = %d",sum);
      getch(); 
      return 0;
}

Upvotes: 1

Views: 1381

Answers (3)

Christophe
Christophe

Reputation: 73587

The following line interprets the value in a as an address:

p = (char *)a;

&p[b] is the address of the b th element of the array starting at p. So, as each element of the array has a size of 1, it's a char pointer pointing at address p+b. As p contains a, it's the address at p+a.

Finally, the following line converts back the pointer to an int:

 sum = (int)&p[b];

But needless to say: it's a weird construct.

Additional remarks:

Please note that there are limitations, according to the C++ standard:

5.2.10/5: A value of integral type (...) can be explicitly converted to a pointer.

5.2.10/4: A pointer can be explicitly converted to any integral type large enough to hold it.

So better verify that sizeof(int) >= sizeof(char*).

Finally, although this addition will work on most implementations, this is not a guaranteed behaviour on all CPU architectures, because the mapping function between integers and pointers is implementation-defined:

A pointer converted to an integer of sufficient size (if any such exists on the implementation) and back to the same pointer type will have its original value; mappings between pointers and integers are otherwise implementation-defined.

Upvotes: 5

Weather Vane
Weather Vane

Reputation: 34583

As commented, it is valid, but horrible code - just a party trick.

p = (char *)a;

p takes the value of a entered as a supposed address.

sum = (int)&p[b];

the address of the bth element of a char array is at p + b.

Since p == a (numerically), the correct sum is obtained.

To take a worked example, enter 46 and 11.

p = (char *)a;            // p = 46
sum = (int)&p[b];         // the address of p[b] = 46 + 11 = 57

Note: nowhere is *p or p[b] written or read, and size does not matter - except for the char array, where pointer arithmetic is in units of 1.

Upvotes: 2

Khouri Giordano
Khouri Giordano

Reputation: 1461

First a is converted to a pointer with the same value. It doesn't point to anything really, it's just the same value.

The expression p[b] will add b to p and refer to the value at that position.

Then the address of the p[b] element is taken and convert to an integer.

Upvotes: 2

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