How many bits of address is required (for the program counter for example) in a byte-addressed computer with 512 Mbyte RAM?
What does the formula look like?
How is this connected with the fact that 32 bits can address no more than 4 GB RAM?
Computer Science – How Many Bits Address Required for n Bytes of Memory?
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Related Solutions
A lot of really early work was done with 5-bit baudot codes, but those quickly became quite limiting (only 32 possible characters, so basically only upper-case letters, and a few punctuation marks, but not enough "space" for digits).
From there, quite a few machines went to 6-bit characters. This was still pretty inadequate though -- if you wanted upper- and lower-case (English) letters and digits, that left only two more characters for punctuation, so most still had only one case of letters in a character set.
ASCII defined a 7-bit character set. That was "good enough" for a lot of uses for a long time, and has formed the basis of most newer character sets as well (ISO 646, ISO 8859, Unicode, ISO 10646, etc.)
Binary computers motivate designers to making sizes powers of two. Since the "standard" character set required 7 bits anyway, it wasn't much of a stretch to add one more bit to get a power of 2 (and by then, storage was becoming enough cheaper that "wasting" a bit for most characters was more acceptable as well).
Since then, character sets have moved to 16 and 32 bits, but most mainstream computers are largely based on the original IBM PC. Then again, enough of the market is sufficiently satisfied with 8-bit characters that even if the PC hadn't come to its current level of dominance, I'm not sure everybody would do everything with larger characters anyway.
I should also add that the market has changed quite a bit. In the current market, the character size is defined less by the hardware than the software. Windows, Java, etc., moved to 16-bit characters long ago.
Now, the hindrance in supporting 16- or 32-bit characters is only minimally from the difficulties inherent in 16- or 32-bit characters themselves, and largely from the difficulty of supporting i18n in general. In ASCII (for example) detecting whether a letter is upper or lower case, or converting between the two, is incredibly trivial. In full Unicode/ISO 10646, it's basically indescribably complex (to the point that the standards don't even try -- they give tables, not descriptions). Then you add in the fact that for some languages/character sets, even the basic idea of upper/lower case doesn't apply. Then you add in the fact that even displaying characters in some of those is much more complex still.
That's all sufficiently complex that the vast majority of software doesn't even try. The situation is slowly improving, but slowly is the operative word.
Even nowadays you can find examples of such processors, for example in complex interlocking systems.
However, these processors are not off-the shelf, and typically the production numbers are so low that in the end these end up being implemented in programmable logic (such as FPGA).
Best Answer
You need log2(n) bits to address n bytes. For example, you can store 256 different values in an 8 bit number, so 8 bits can address 256 bytes. 210 = 1024, so you need 10 bits to address every byte in a kilobyte. Likewise, you need 20 bits to address every byte in a megabyte, and 30 bits to address every byte in a gigabyte. 232 = 4294967296, which is the number of bytes in 4 gigabytes, so you need a 32 bit address for 4 GB of memory.