Integer Overflow

Integer Overflow occurs when an arithmetic operation results in a value that exceeds the range of representable values in the given data type. In C, this behavior is undefined.

For example, consider the following code snippet:

#include <stdio.h>
 
int main(void) {
    int x = 2147483647;  // INT_MAX
    x = x + 1;           // Overflow
    printf("%d\n", x);   // -2147483648 (INT_MIN)
 
    int j = 2147483647;  // INT_MAX
    j = j + 5;           // Overflow
    printf("%d\n", j);   // -2147483644
    return(0);
}

Here we can see that adding to a INT_MAX results in a wraparound to INT_MIN. Thats why the output is -2147483648 instead of 2147483648. In general, that’s what integer overflow in C is.

Detailed Review

Let’s understand this behavior in more detail.

int a = 2147483647;
int b = 2147483646;
int c = a + b;
printf("%d\n", c);

In the above code snippet, make your guess about the value of c before running the code. After running the code, you will see that the output is -3. This is because the sum of 2147483647 and 2147483646 is 4294967293, which is out of the range of int - but wait, why is the output -3? It’s a bit more complicated than you might think.

When adding 2147483647 and 2147483646, the binary addition would go as follows:

2147483647 = 01111111 11111111 11111111 11111111
2147483646 = 01111111 11111111 11111111 11111110

Adding them gives:

01111111 11111111 11111111 11111111 +
01111111 11111111 11111111 11111110 =
11111111 11111111 11111111 11111101

So our result is 11111111 11111111 11111111 11111101.

Warning

If you try to use any Binary Calculator to validate this addition, they won’t probably show the correct result because they are using 32-bit integers, therefore this operation will result in an error

The result is 4294967293 in unsigned binary. However, since the result is stored in a 32-bit int (signed), it wraps around modulo 2^32, because the range of int is limited to [−2^31,2^31−1].

Because the MSB of the result is 1, the binary value 11111111 11111111 11111111 11111101 is negative. Two’s Complement is used to determine its value.

1. Invert all bits: 11111111 11111111 11111111 11111101 → 00000000 00000000 00000000 00000010
2. Add 1: 00000000 00000000 00000000 00000010 → 00000000 00000000 00000000 00000011

All what’s left is to convert 00000000 00000000 00000000 00000011 to decimal, which is -3.

This is why 2147483647 + 2147483646 = -3 in C.

Congrats!