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
C++
#include <bits/stdc++.h>
using namespace std;
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 = new int[(sizeof(int))];
int x = 2147483640;
int y = 10;
cout<<addOvf(res, x, y);
cout<<"\n"<<*res;
return 0;
}
|
C
#include<stdio.h>
#include<stdlib.h>
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;
}
|
Java
class GFG
{
static 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;
}
public static void main(String args[]) {
int res = -2147483646;
int x = 2147483640;
int y = 10;
System.out.println(addOvf(res, x, y));
System.out.print(res);
}
}
|
C#
using System;
class GFG {
static 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;
}
public static void Main()
{
int res = -2147483646;
int x = 2147483640;
int y = 10;
Console.WriteLine(addOvf(res, x, y));
Console.Write(res);
}
}
|
Javascript
<script>
function addOvf(result, a, b)
{
let resultt = a + b;
if(a > 0 && b > 0 && result < 0)
return -1;
if(a < 0 && b < 0 && result > 0)
return -1;
return 0;
}
let res = -2147483646;
let x = 2147483640;
let y = 11;
document.write(addOvf(res, x, y) + "<br/>");
document.write(res);
</script>
|
Python3
def addOvf(result, a, b):
resultt = a + b;
if(a > 0 and b > 0 and result < 0):
return -1;
if(a < 0 and b < 0 and result > 0):
return -1;
return 0;
res = -2147483646
x = 2147483640
y = 11
print(addOvf(res, x, y))
print(res)
|
Time Complexity: O(1)
Space Complexity: O(1)
Method 2
Thanks to Himanshu Aggarwal for adding this method. This method doesn’t modify *result if there us an overflow.
C++
#include <bits/stdc++.h>
using namespace std;
int addOvf(int* result, int a, int b)
{
if (a >= 0 && b >= 0 && (a > INT_MAX - b)) {
return -1;
}
else if (a < 0 && b < 0 && (a < INT_MIN - b)) {
return -1;
}
else {
*result = a + b;
return 0;
}
}
int main()
{
int a, b;
int* res;
a = INT_MAX;
b = 8192;
cout << addOvf(res, a, b);
return 0;
}
|
C
#include <limits.h>
#include <stdio.h>
#include <stdlib.h>
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;
}
|
Java
class GFG {
public static int addOvf(int[] result, int a, int b)
{
if (a >= 0 && b >= 0
&& (a > Integer.MAX_VALUE - b)) {
return -1;
}
else if (a < 0 && b < 0
&& (a < Integer.MIN_VALUE - b)) {
return -1;
}
else {
result[0] = a + b;
return 0;
}
}
public static void main(String[] args)
{
int a, b;
int[] res = new int[1];
a = Integer.MAX_VALUE;
b = 8192;
System.out.println(addOvf(res, a, b));
}
}
|
Python3
import sys
def addOvf(result, a, b):
if (a >= 0 and b >= 0 and (a > sys.maxsize - b)) or (a < 0 and b < 0 and (a < -sys.maxsize - b)):
return -1
else:
result[0] = a + b
return 0
if __name__ == "__main__":
a = sys.maxsize
b = 8192
res = [0]
print(addOvf(res, a, b))
|
C#
using System;
class GFG {
public static int addOvf(int[] result, int a, int b)
{
if (a >= 0 && b >= 0 && (a > int.MaxValue - b)) {
return -1;
}
else if (a < 0 && b < 0 && (a < int.MinValue - b)) {
return -1;
}
else {
result[0] = a + b;
return 0;
}
}
public static void Main(string[] args)
{
int a, b;
int[] res = new int[1];
a = int.MaxValue;
b = 8192;
Console.WriteLine(addOvf(res, a, b));
}
}
|
Javascript
function addOvf(result, a, b) {
if (a >= 0 && b >= 0 && a > Number.MAX_SAFE_INTEGER - b) {
return -1;
} else if (a < 0 && b < 0 && a < Number.MIN_SAFE_INTEGER - b) {
return -1;
} else {
result[0] = a + b;
return 0;
}
}
let a = 0, b = 0;
let res = [0];
a = Number.MAX_SAFE_INTEGER;
b = 8192;
console.log(addOvf(res, a, b));
|
Time Complexity: O(1)
Space Complexity: O(1)
Please write comments if you find any bug in above codes/algorithms, or find other ways to solve the same problem
