WAP to Check Integer OverFlow in Differnt Logical & Arithmetic Operation

Write a “C” function, int addOvf(int* result, int a, int b) If there is no overflow, the function places the resultant = sum a+b in “result” and returns 0. Otherwise it returns -1. The solution of casting to long and adding to find detecting the overflow is not allowed.

Method 1
There can be overflow only if signs of two numbers are same, and sign of sum is opposite to the signs of numbers.

1) Calculate sum
2) If both numbers are positive and sum is negative then return -1
Else
If both numbers are negative and sum is positive then return -1
Else return 0

?
#include
#include

/* Takes pointer to result and two numbers as
arguments. If there is no overflow, the function
places the resultant = sum a+b in “result” and
returns 0, otherwise it returns -1 */
int addOvf(int* result, int a, int b)
{
*result = a + b;
if(a > 0 && b > 0 && *result < 0)
return -1;
if(a < 0 && b < 0 && *result > 0)
return -1;
return 0;
}

int main()
{
int *res = (int *)malloc(sizeof(int));
int x = 2147483640;
int y = 10;

printf("%d", addOvf(res, x, y));

printf("\n %d", *res);
getchar();
return 0;
}

Time Complexity : O(1)
Space Complexity: O(1)


Method 2

#include
#include
#include

int addOvf(int* result, int a, int b)
{
if( a > INT_MAX - b)
return -1;
else
{
*result = a + b;
return 0;
}
}

int main()
{
int *res = (int *)malloc(sizeof(int));
int x = 2147483640;
int y = 10;

printf("%d", addOvf(res, x, y));
printf("\n %d", *res);
getchar();
return 0;
}

Time Complexity : O(1)
Space Complexity: O(1)


Source http://joyofprogramming.com/Docs_ColumnArticles/27-JoP-Mar-09.pdf