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Why does integer overflow result in different behavior for signed and unsigned integers?

Susan Sarandon
Release: 2024-11-17 08:15:03
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Why does integer overflow result in different behavior for signed and unsigned integers?

Understanding Signed/Unsigned Integer Overflow

In the process of learning about integer overflow in C , the following questions arose:

Why do I obtain these results with signed/unsigned integer overflow?"

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;
}
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Output:

0
2147483647
-2147483648

0
4294967295
0
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Explanation:

Understanding integer overflow in C requires an understanding of signed and unsigned integer types.

  • Signed Integer Override:

Signed integer overflow is considered undefined behavior, meaning the compiler is not required to specify the exact behavior when a signed integer overflows. In practice, most implementations store signed integers using 2's complement representation. This means that when a signed integer overflows, its value "wraps around" to the negative range. In the example, adding 1 to the maximum positive integer (2147483647) using a signed integer results in -2147483648, the minimum negative integer.

  • Unsigned Integer Overflow:

In contrast, unsigned integer overflow is well-defined, with the value wrapping around to zero when it exceeds the maximum value. In the example, adding 1 to the maximum unsigned integer (4294967295) results in 0. This is because unsigned integers do not have a concept of negative values.

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