I have a somewhat niche question. A variable stored as a byte uses 1 byte to store a number. 1 byte is 8 bits, which can take \(2^8=256\) combinations. 26 (.a to .z) + 1 (.) are reserved for missing values, leaving 229 combinations. A byte variable can store values from -127 to 100, requiring 228 combinations. 'That leaves 1 combination unaccounted for. Did I make an error or is that 1 remaining combination used for something else?
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