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What is the range of an 8-bit signed (two's complement) integer?
-128 to 127. There is one more negative number than positive.
- -127 to 127
- -128 to 127 ✔
- -128 to 128
Why: With n bits the range is minus 2 to the (n-1) up to 2 to the (n-1) minus 1. For 8 bits that is -128 to 127. Zero takes one of the non-negative slots, so the positive side is one short.
Huh, didn't know that: In 8-bit two's complement, 01111111 represents 127 and 10000000 represents -128.
Why did 127 refuse to add one more? It knew it would end up deeply negative.
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References
Answer: -128 to 127. With n bits the range is minus 2 to the (n-1) up to 2 to the (n-1) minus 1. For 8 bits that is -128 to 127. Zero takes one of the non-negative slots, so the positive side is one short.
- OpenStax Introduction to Computer Science 5.3: Machine-Level Information Representation openstax.org/books/introduction-computer-science/pages/5-3-machine-level-information-representation
with three bits, the two's complement method can present up to -4, which we did not find in the other two methods
Checked 2026-10-10.
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