问题
Just reading some notes in a purdue lecture about OSs, and it says:
A program sees memory as an array of bytes that goes from address 0 to 2^32-1 (0 to 4GB-1)
Why 4 GB?
回答1:
Because 32 bits are able to represent numbers up to 232 − 1 = 4294967295 = 4 GiB − 1 and therefore address up to 232 individual bytes which would be 4 GiB then.
There are ways to circumvent that, though. For example using PAE even a 32-bit operating system can support more memory. Historically this has most commonly been used on servers, though. Also, the non-server Windows SKUs don't support it. By now all that is moot, though, given that 64-bit CPUs, OSes and driver support are commonplace.
回答2:
Because each byte of memory has to have an address. In a 32-bit operating system, an address is 32 bits long; thus, there are 2^32 possible addresses, which means there are 2^32 bytes = 4 GB.
回答3:
Everybody is saying 2^32 = 4GiB, which is right. Just in case, here is how we got there:
A 32-bit machine uses 32 bits to address memory. Each bit has a value of 0 or 1. If you have 1 bit, you have two possible addresses: 0 or 1. A two-bit system ( pun aside ) has four possible address: 00 =0, 01=1, 10=2, 11=3. 2^2=4. Three bits have 8 possble addresses: 000=0, 001=1, 010=2, 011=3, 100=4, 101=5, 110=6, and 111=7.
Each bit doubles the potential address space, which is why 2^n tells you how many addresses you use for a given number of bits. 2^1 = 2, 2^2 = 2*2 = 4, 2^3 = 2*2*2 = 8, etc.
By the time you get to 32 bits, you are at 4GiB.
回答4:
4 GB = 2^32 bytes.
回答5:
If you have a 4-bit system, this means the address for each byte is 4 binary digits, the probability of all your address will range from 0000 through 1111 which is 2^4 = 16 (2 because there is either 0 or 1), with four bits it's possible to create 16 different values of zeros and ones, If you have 16 different addr. each represent a byte then you can have a max of 16 bytes
4-bit system will look like this:
For a 32-bit system, your max is 2^32 = 4294967292 bytes
回答6:
2 ^ 32 = 4 * 1024 * 1024 * 1024
That, in bytes, is the definition of 4 GB. In other words a 32-bit register as a memory pointer can address 4 GB of memory and no more.
回答7:
Actually, it's not as simple as 2^32 = 4294967296 bytes. You see in x86 protected mode, with paging enabled (that is, what you get when you use any modern OS), you don't address memory locations directly, even though the paging translation mechanism is transparent for client applications.
Of a logical 32 bit memory address, when using 4K pages:
- bits 22-31 refer to a page directory
- bits 12-21 refer to a page table
- bits 11-0 refer to an offset in the 4096 byte page
As you can see, you have 2^10 (1024) page directories, in each page directory, you have 2^10 page tables and each page is 2^12 (4096) bytes long, hence 2^32 = 4294967296 bytes. The width of the memory bus is conveniently the same as the word length of the CPU but it's not necessary to be like this at all. In fact, more modern x86 CPUs support PAE which enables addressing more than 4GB (or GiB) even in 32-bit mode.
回答8:
Because is the amount of different memory addresses (in Bytes) that can be stored in a Word.
But, in fact, that's not always true (in most of cases it isn't), the OS can handle more physical memory than that (with PAE) and the applications can use less than 4GB of virtual memory (because part of that virtual memory is mapped to the OS, 1GB in Linux and 2GB in Windows, for example).
Another scenario where that doesn't apply is if the memory was addressed by Words instead of Bytes, then the total memory addressable would be 16GB, for example.
回答9:
Mainly due to 32bit OS chosing to support only 2^32-1 addresses.
If the CPU has more than 32 address lines on the FSB, then the 32bit OS can choose to use a paging mechanism to access more than 4GiB. (For example Windows 2000 Advanced Server/Data Center editions on PAE supported Intel/AMD chips)
回答10:
- 32bits can represent numbers
0..2^32=0..4,294,967,296 - 32bits can address up to 2^32Bytes (assuming Byte-size blocks)
- 2^32Bytes is the max size
2^32B = 4,194,304KiB = 4,194MiB = 4GiB
回答11:
4 GB = 2^32 bytes. But remember its max 4gb allocated by a 32 bit OS. In reality, the OS will see less e.g. after VRAM allocation.
回答12:
As previously stated by other users, 32-bit Windows OSes use 32-bit words to store memory addresses.
Actually, most 32-bit chips these days use 36-bit addressing, using Intel's Physical Address Extension (PAE) model. Some operating systems support this directly (Linux, for example).
As Raymond Chen points out, in Windows a 32-bit application can allocate more than 4GB of memory, and you don't need 64-bit Windows to do it. Or even PAE.
For that matter, 64-bit chips don't support the entire 64-bit memory space. I believe they are currently limited to 42-bit space... the 36-bit space that PAE uses, plus the top 8-bit addresses,
来源：https://stackoverflow.com/questions/1119278/why-does-a-32-bit-os-support-4-gb-of-ram