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; }
Output:
0 2147483647 -2147483648 0 4294967295 0
Explanation:
Understanding integer overflow in C requires an understanding of signed and unsigned integer types.
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.
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.
The above is the detailed content of Why does integer overflow result in different behavior for signed and unsigned integers?. For more information, please follow other related articles on the PHP Chinese website!