Question

#include<stdio.h>
int main(void)
{
    unsigned short a,e,f ;    // 2 bytes data type
    unsigned int temp1,temp2,temp4; // 4 bytes data type
    unsigned long temp3; // 8 bytes data type
    a=0xFFFF;
    e=((a*a)+(a*a))/(2*a); // Line 8
    //e=(((unsigned long)(a*a)+(unsigned long)(a*a)))/(unsigned int)(2*a);    

    temp1=a*a;
    temp2=a*a;
    temp3=(unsigned long)temp1+(unsigned long)temp2; // Line 14
    temp4=2*a;

    f=temp3/temp4;

    printf("%u,%u,%lu,%u,%u,%u,%u\n",temp1,temp2,temp3,temp4,e,f,a);
    return(1);
}

How do I fix the arithmetic (At Line 8 by appropriate typecasting of intermediate results) so that overflows are taken care of ? Currently it prints 65534 instead of expected 65535.

Why is the typecast necessary for Line 14 ?

Answer 1

You have to promote the types before performing the overflowing operation. In line 8 that would be multiplication, so

e = ((unsigned) a * a + (unsigned) a * a) / (2 * (unsigned) a); 

Note that promoting only one operand of such symmetrical operations as * is enough. You can use (unsigned) a * (unsigned) a if you wish, but (unsigned) a * a will work as well.

This will take care of multiplication, which will no longer overflow. However, now the addition will overflow. While 32-bit unsigned is enough for a * a, it is not enough for a * a + a * a. For that you’ll need unsigned long (assuming it is larger). You can formally promote the first operand of + to unsigned long

e = ((unsigned long) ((unsigned) a * a) + (unsigned) a * a) / (2 * (unsigned) a); 

(again, promoting only the first operand of + is enough, meaning that the second multiplication can be left in unsigned).

The above looks a bit too convoluted, and to make it look cleaner you can use unsigned long in the first multiplication from the very beginning

e = ((unsigned long) a * a + (unsigned) a * a) / (2 * (unsigned) a); 

or you can just use unsigned long everywhere to make it look even cleaner

e = ((unsigned long) a * a + (unsigned long) a * a) / (2 * (unsigned long) a); 

The very same problem appears in your temp1 = a * a; lines. They will overflow for the very same reason. You have to do

temp1 = (unsigned) a * a;
temp2 = (unsigned) a * a;

to avoid overflow, i.e. promote a before multiplication.

And that is exactly what you correctly do in line 14, i.e. promote operands of + before addition, although promoting only one operand is perfectly sufficient

temp3 = (unsigned long) temp1 + temp2;