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Yes, a byte is always 8 bits in modern computing. The book uses Words, not bytes. In the book, the word and the size of the word is explicitly mentioned, while there is not a word (haha) about bytes. Look at the phrase ..is represented in RAM by 32 consecutive 16-bit words.. The whole size is expressed in (16 bit) words rather than bytes.
For older architectures, "byte" indicated the size of the data bus, and as the original question states, a lot of different bus sizes existed (4, 5, 6, 8, 12 etc.). But since 1993 a byte has been defined as 8 bits, in order to have a standardized SI unit for data sizes. Hence the meaning of "byte" has changed from being an architecture ...
A word is a fixed sized group of bits that are handled as a unit by the instruction set and/or hardware of the processor. That means the size of a general purpose register ( which is generally more than a byte ) is a word. In the C, a word is most often called an integer => int. answered Oct 13, 2011 at 6:23.
The value is indeed the sum of all bits set to 1, but the place value of the eighth bit is 2 7 (128), not 256 as you suggest - the least significant bit is 2 0 (i.e. 1), so for eight bits the MSB is 2 7. You appear to have started from 2 1 (2) . For an unsigned integer: Bit 0 = 2 0 = 1 Bit 1 = 2 1 = 2 Bit 2 = 2 2 = 4 Bit 3 = 2 3 = 8 Bit 4 = 2 4 ...
6. A byte has only 8 bits. A bit is a binary digit. So a byte can hold 2 (binary) ^ 8 numbers ranging from 0 to 2^8-1 = 255. It's the same as asking why a 3 digit decimal number can represent values 0 through 999, which is answered in the same manner (10^3 - 1). Originally bytes weren't always 8 bits though.
Why not simply make the lower range -127 fot 8 bits instead of writing -128 as 10000000. 2^8 = 256. So whatever representation scheme you use, it should be able to represent 256 different values. And you can draw a circle to understand how good 2's complement system is. First look at this table :
A bit late post, but this works for me: byte[] ret = new byte[(bits.Length - 1) / 8 + 1]; bits.CopyTo(ret, 0); return ret; Works with: Instead of "bits.Length / 8", you should use " (bits.Length - 1) / 8 + 1", otherwise if the BitArray has a length of 7, your byte array will be empty.
The byte was originally the smallest number of bits that could hold a single character (I assume standard ASCII). We still use ASCII standard, so 8 bits per character is still relevant. This sentence, for instance, is 41 bytes. That's easily countable and practical for our purposes. If we had only 4 bits, there would only be 16 (2^4) possible ...
For example, a byte contains 8 bits; applying this operator to a value whose bit pattern is "00000000" would change its pattern to "11111111". A byte value IS integral, you can check bit state using masking operations.
NB: The >> 4 is because there are 8 bits in 1 byte, which is an unsigned char so we want to take the other half, and so on. We could easily apply this solution to 4 bytes with only two additional lines and following the same logic. Since both mask complement each other we can even use ~ in order to switch bits and saving some ink: