Zack Downs - 1 year ago 170
C++ Question

# C++ integer overflow

I'm just starting to teach myself C++, and have begun learning about integer overflow. Out of curiosity I wrote some tests just to see what occurs with certain integer values.

Here's my program:

#include <iostream>

int main()
{
int x(0);
std::cout << x << std::endl;

x = x + 2147483647;
std::cout << x << std::endl;

x = x + 1;
std::cout << x << std::endl;
std::cout << std::endl;

unsigned int y(0);
std::cout << y << std::endl;

y = y + 4294967295;
std::cout << y << std::endl;

y = y + 1;
std::cout << y << std::endl;
}

And here's the output:

0
2147483647
-2147483648

0
4294967295
0

The output surprised me a bit, and I'm wondering if someone could explain why this happened, OR if the unexpected-ness of these results is expected; so this may just be the result of my specific machine.

Integers (generally) take a 32-bit representation. If you have 32 bits, you can address from 0 to 231-1. i.e.,

00000000000000000000000000000000
00000000000000000000000000000001
.
.
.
01111111111111111111111111111111
^-------------------------------
signed bit

0 indicates a positive number, 1 indicates a negative number.

If you add 1 to 01111111111111111111111111111111, you get 10000000000000000000000000000000, which is -2147483648 in decimal.

Using an unsigned integer, there's no signed bit and, ipso facto, can have a number twice as large as your largest signed integer. However, when the number rolls over again (i.e., 11111111111111111111111111111111 + 00000000000000000000000000000001), you simply roll back to 00000000000000000000000000000000.

For a more in depth understanding, you can look at two's complement, which is how integers are represented in computers.