# Check for Integer Overflow

• Difficulty Level : Easy
• Last Updated : 09 Nov, 2020

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.

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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 ``using` `namespace` `std;`` ` `/* 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; ``} `` ` `// Driver code``int` `main() ``{ ``    ``int` `*res = ``new` `int``[(``sizeof``(``int``))]; ``    ``int` `x = 2147483640; ``    ``int` `y = 10; `` ` `    ``cout<

## C

 `#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;``}`
Output:
```-1
-2147483646
```

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 ``using` `namespace` `std;`` ` `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 = ``new` `int``[(``sizeof``(``int``))]; ``    ``int` `x = 2147483640; ``    ``int` `y = 10; ``     ` `    ``cout<

## C

 `#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;``}`
Output:
```-1
0
```

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

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