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Question: large integer implicitly truncated to unsigned type
Time: Asked 51 days ago
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large integer implicitly truncated to unsigned type

When a large integer is implicitly truncated to an unsigned type, the result depends on the size of the unsigned type and the value of the original integer. If the original integer is too large to fit in the unsigned type, the result will be reduced modulo the maximum value of the unsigned type, which is 2^n, where n is the number of bits used to represent the unsigned type.

For example, suppose we have an 8-bit unsigned integer type and an integer value of 300. In binary, 300 is represented as 0b100101100, which is 9 bits long. When this value is implicitly converted to an 8-bit unsigned type, the result is truncated to the least significant 8 bits, which gives us 0b0101100 or 44 in decimal. This is because 300 modulo 256 (which is 2^8) is equal to 44.

It’s important to be aware of these potential truncation issues when working with large integers and unsigned types in programming languages, as they can lead to unexpected behavior and bugs in your code.

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