signed two's complement arithmetic

I was thinking on data types ranges, a question arises. As we know signed char's range is from -128 to 127. I got the how 127 comes, i.e. 0111111 = +127

B开发者_开发知识库ut I could not get how -128 comes? if we just ON sign bit we get 11111111, how its is equal to -128 ?

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Most of the time, computers use what's called 2's complement to represent signed integers.

The way 2's complement works is that the possible values are in a huge loop, from 0, to MAX_VALUE, to MIN_VALUE, to zero, and so on.

So the minimum value is the maximum value +1 - 01111111 = 127, and 10000000 = -128.

This has the nice property of behaving exactly the same as unsigned arithmetic - if I want to do -2 + 1, I have 11111110 + 00000001 = 11111111 = -1, using all the same hardware as for unsigned addition.

The reason there's an extra value on the low end is that we choose to have all numbers with the high-bit set be negative, which means that 0 takes a value away from the positive side.

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In two's complement, -128 is 10000000.

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Negative numbers have the sign bit set to 1; -128 is the value with the sign bit set but no other bits (i.e., it is the smallest negative number). The binary representation of -128 is 10000000. For other data lengths, the smallest negative number in two's complement is always 1000... for the correct number of zeros.

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One simple way to think of this is to start at 01111111 and then keep subtracting 1 until it wraps around; the previous value is the smallest negative value. Subtracting 1 from 00000000 using the standard "borrow" technique yields 11111111, which is indeed the binary representation for -1. We can keep subtracting down to 10000000, which is -128, and subtracting one more yields 01111111 again, wrapping around.