为什么符号字节的范围是从-128到127（2的补数），而不是从-127到127？

问题描述

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它说：

so it concludes -128 is 10000000

so +128 cannot be represented in 2's complement in 8 bits, but that means we can represent it in 9 bits, so 128 is 010000000 and so taking its 2's complement -128 is 110000000,

so is representation of -128 10000000 or 110000000 ?Is the representaion bit dependent ?

Why not simply make the lower range -127 fot 8 bits instead of writing -128 as 10000000 ?

解决方案

It's not. An unsigned byte (assuming 8-bit) is from 0 to 255.

The range of a signed byte using 2's complement is from -128 to 127, directly from the definition of 2's complement:

01111111 = +127
01111110 = +126
01111101 = +125
...
00000001 = +1
00000000 =  0
11111111 = -1
...
10000010 = -126
10000001 = -127
10000000 = -128

In 8-bit, it's 10000000, in a hypothetical 9-bit representation it's 110000000.

Artificially restricting the range to -127 wouldn't achieve very much; you'd be disallowing a perfectly valid value, and generally making code more complex (what else would you do with the bit pattern 10000000?).

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