I have this confusion regarding binary representation of decimal value 129 (or even 128). If 8 bits are used to represent numbers when doing the two's complement, then we know that '00000000' to '01111111' are used for 0 to 127 and leading 1 is used for negative numbers i.e. 1xxxxxxx where x is any combination of 0's and 1's to represent the value of that negative number. But then how is 128 represented in 8 bits because the first bit is reserved for telling the number is negative (basically a signed number)?
1 Answer
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The range of integers representable in 8 bits using two's complement is –128 to +127.
If instead the numbers are unsigned, then the range becomes 0 to 255.
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$\begingroup$ So does this mean that 128 is 10000000 when we are using 8 bits and it will not be confused for a negative number? Then how can we tell if we are looking at a negative 8-bit number or an unsigned 8-bit number (in cases where the first bit is 1)? $\endgroup$bee5911– bee59112019-08-03 20:51:28 +00:00Commented Aug 3, 2019 at 20:51
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$\begingroup$ 128 is not between -128 and +127. As for how to tell whether a number is signed or unsigned, it’s impossible. It’s up to the programmer to decide whether she interprets a byte as a signed 8-bit integer, an unsigned 8-bit integer, or anything else. $\endgroup$Yuval Filmus– Yuval Filmus2019-08-03 22:03:28 +00:00Commented Aug 3, 2019 at 22:03