How many memory addresses can we get with a 32-bit processor and 1gb ram?

Question

How many memory addresses can we get with a 32-bit processor and 1gb ram and how many with a 64-bit processor?

I think that it's something like this:

1GB of ram divided by 32bits or divided by 4? to get the number of memory addresses?

But i'm not sure. That's why i'm asking.

I red on wikipedia that 1 memory addresses is 32bits wide or 4 octets (1 octet = 8 bits), compared to 64bit a processor where 1 memory addresses or 1 integer is 64bits wide or 8 octets. But don't know if i understood it correctly either.

Solution 1

Short answer: The number of available addresses is equal to the smaller of those:

• Memory size in bytes
• Greatest unsigned integer that can be saved in CPU's machine word

---

Long answer and explaination of the above:

Memory consists of bytes (B). Each byte consists of 8 bits (b).

1 B = 8 b

1 GB of RAM is actually 1 GiB (gibibyte, not gigabyte). The difference is:

1 GB  = 10^9 B = 1 000 000 000 B
1 GiB = 2^30 B = 1 073 741 824 B

Every byte of memory has its own address, no matter how big the CPU machine word is. Eg. Intel 8086 CPU was 16-bit and it was addressing memory by bytes, so do modern 32-bit and 64-bit CPUs. That's the cause of the first limit - you can't have more addresses than memory bytes.

Memory address is just a number of bytes the CPU has to skip from the beginning of the memory to get to the one it's looking for.

• To access the first byte it has to skip 0 bytes, so first byte's address is 0.
• To access the second byte it has to skip 1 byte, so its address is 1.
• (and so forth...)
• To access the last byte, CPU skips 1073741823 bytes, so its address is 1073741823.

Now you have to know what 32-bit actually means. As I mentioned before, it's the size of a machine word.

Machine word is the amount of memory CPU uses to hold numbers (in RAM, cache or internal registers). 32-bit CPU uses 32 bits (4 bytes) to hold numbers. Memory addresses are numbers too, so on a 32-bit CPU the memory address consists of 32 bits.

Now think about this: if you have one bit, you can save two values on it: 0 or 1. Add one more bit and you have four values: 0, 1, 2, 3. On three bits, you can save eight values: 0, 1, 2... 6, 7. This is actually a binary system and it works like that:

Decimal Binary
0       0000
1       0001
2       0010
3       0011
4       0100
5       0101
6       0110
7       0111
8       1000
9       1001
10      1010
11      1011
12      1100
13      1101
14      1110
15      1111

It works exactly like usual addition, but the maximum digit is 1, not 9. Decimal 0 is 0000, then you add 1 and get 0001, add one once again and you have 0010. What happend here is like with having decimal 09 and adding one: you change 9 to 0 and increment next digit.

From the example above you can see that there's always a maximum value you can keep in a number with constant number of bits - because when all bits are 1 and you try to increase the value by 1, all bits will become 0, thus breaking the number. It's called integer overflow and causes many unpleasant problems, both for users and developers.

   11111111    = 255
+         1
-----------
  100000000    = 0   (9 bits here, so 1 is trimmed)

• For 1 bit the greatest value is 1,
• 2 bits - 3,
• 3 bits - 7,
• 4 bits - 15

The greatest possible number is always 2^N-1, where N is the number of bits. As I said before, a memory address is a number and it also has a maximum value. That's why machine word's size is also a limit for the number of available memory addresses - sometimes your CPU just can't process numbers big enough to address more memory.

So on 32 bits you can keep numbers from 0 to 2^32-1, and that's 4 294 967 295. It's more than the greatest address in 1 GB RAM, so in your specific case amount of RAM will be the limiting factor.

The RAM limit for 32-bit CPU is theoretically 4 GB (2^32) and for 64-bit CPU it's 16 EB (exabytes, 1 EB = 2^30 GB). In other words, 64-bit CPU could address entire Internet... 200 times ;) (estimated by WolframAlpha).

However, in real-life operating systems 32-bit CPUs can address about 3 GiB of RAM. That's because of operating system's internal architecture - some addresses are reserved for other purposes. You can read more about this so-called 3 GB barrier on Wikipedia. You can lift this limit with Physical Address Extension.

---

Speaking about memory addressing, there are few things I should mention: virtual memory, segmentation and paging.

Virtual memory

As @Daniel R Hicks pointed out in another answer, OSes use virtual memory. What it means is that applications actually don't operate on real memory addresses, but ones provided by OS.

This technique allows operating system to move some data from RAM to a so-called Pagefile (Windows) or Swap (*NIX). HDD is few magnitudes slower than RAM, but it's not a serious problem for rarely accessed data and it allows OS to provide applications more RAM than you actually have installed.

Paging

What we were talking about so far is called flat addressing scheme.

Paging is an alternative addressing scheme that allows to address more memory that you normally could with one machine word in flat model.

Imagine a book filled with 4-letter words. Let's say there are 1024 numbers on each page. To address a number, you have to know two things:

• The number of page on which that word is printed.
• Which word on that page is the one you're looking for.

Now that's exactly how modern x86 CPUs handle memory. It's divided into 4 KiB pages (1024 machine words each) and those pages have numbers. (actually pages can also be 4 MiB big or 2 MiB with PAE). When you want to address memory cell, you need the page number and address in that page. Note that each memory cell is referenced by exactly one pair of numbers, that won't be the case for segmentation.

Segmentation

Well, this one is quite similar to paging. It was used in Intel 8086, just to name one example. Groups of addresses are now called memory segments, not pages. The difference is segments can overlap, and they do overlap a lot. For example on 8086 most memory cells were available from 4096 different segments.

---

An example:

Let's say we have 8 bytes of memory, all holding zeros except for 4th byte which is equal to 255.

Illustration for flat memory model:

 _____
|  0  |
|  0  |
|  0  |
| 255 |
|  0  |
|  0  |
|  0  |
|  0  |
 -----

Illustration for paged memory with 4-byte pages:

 PAGE0
 _____
|  0  |
|  0  |
|  0  |  PAGE1
| 255 |  _____
 -----  |  0  |
        |  0  |
        |  0  |
        |  0  |
         -----

Illustration for segmented memory with 4-byte segments shifted by 1:

 SEG 0
 _____   SEG 1
|  0  |  _____   SEG 2
|  0  | |  0  |  _____   SEG 3
|  0  | |  0  | |  0  |  _____   SEG 4
| 255 | | 255 | | 255 | | 255 |  _____   SEG 5
 -----  |  0  | |  0  | |  0  | |  0  |  _____   SEG 6
         -----  |  0  | |  0  | |  0  | |  0  |  _____   SEG 7
                 -----  |  0  | |  0  | |  0  | |  0  |  _____
                         -----  |  0  | |  0  | |  0  | |  0  |
                                 -----   -----   -----   -----

As you can see, 4th byte can be addressed in four ways: (addressing from 0)

• Segment 0, offset 3
• Segment 1, offset 2
• Segment 2, offset 1
• Segment 3, offset 0

It's always the same memory cell.

In real-life implementations segments are shifted by more than 1 byte (for 8086 it was 16 bytes).

What's bad about segmentation is that it's complicated (but I think you already know that ;) What's good, is that you can use some clever techniques to create modular programs.

For example you can load some module into a segment, then pretend the segment is smaller than it really is (just small enough to hold the module), then choose first segment that doesn't overlap with that pseudo-smaller one and load next module, and so on. Basically what you get this way is pages of variable size.

Solution 2

In addition to the above, note that virtual addressing is used, along with multiple address spaces. So, even though you only have 1GB of RAM, a program could conceptually use up to 4GB of virtual memory (though most operating system will limit it to less than this). And you can conceptually have a (nearly) infinite number of such 4GB address spaces.

RAM size doesn't constrain (so much) the maximum size of a program or the number of programs you can run, but rather constrains performance. When real memory becomes "over-committed" and the system begins to "thrash" as it "swaps" "pages" of memory back and forth between RAM and disk, performance plummets.

Solution 3

The 1GByte of RAM would occupy 1024*1024*1024 bytes, or 1,073,741,824 bytes.

A 32-bit processor always has 4*1024*1024*1024 bytes, or 4,294,967,296 bytes of address space The 1Gbyte of RAM appears within this space. On Intel processors, some RAM needs to appear at address 0 for the interrupt vectors, so physical RAM starts at address 0 and goes up.

Other things appear in that address space, such as BIOS and option ROMs (in the upper 384Kbytes within the first 1Mbyte), I/O devices (like the APIC), and the video RAM. Some weird things also go on with system management mode "SMRAM" that I don't completely understand yet.

Note this is physical address space, from the kernel's point of view. The MMU can rearrange all of this in any manner to a userspace process.

Comments

[Daniel R Hicks]

Virtually all modern systems are byte-addressed, meaning that 32 bits can address about 4 gigabytes (if that much RAM is installed). Historically, there have been word-addressed schemes, with a "word" being 12, 15, 16, 17, 24, 32, 36, or 48 bits, and likely some others, plus decimal machines which addressed a 4 or 6-bit unit. But also keep in mind that most modern systems employ virtual memory, meaning that the processor can address even more memory than what is installed.

[Jamie Hanrahan]

@DanielRHicks Virtual memory doesn't affect how much RAM can be addressed.

[Daniel R Hicks]

@JamieHanrahan - In order for virtual memory to work there must be some addressing mechanism that spans the virtual address range (of a single process). This can be done with software simulation, but that's quite inefficient, so in most systems the processor's addressing range is made large enough to incorporate the max address space of a process.

[Jamie Hanrahan]

@DanielRHicks Sure, but that doesn't affect how much RAM can be addressed. RAM is physical memory, not virtual. Virtual addresses are not RAM addresses and virtual memory is not RAM.

[Daniel R Hicks]

@JamieHanrahan - You're not making any sense. If you cannot address pages of RAM in the virtual address space then the RAM is useless. (I know very well how virtual memory works, having worked on virtual memory designs since 1972.)

[Jamie Hanrahan]

@DanielRHicks I never said RAM was not reachable via virtual addresses. Only that virtual and physical addresses are not the same things. Once virtual addresses are as wide as physical, increases in VA width do not mean you can address more RAM than before. Using x86/x64 (and VAX) terms: the Virtual Page Number from a VA does not address RAM, it is merely used to find a PTE. The number of bits in the PTE's PFN field are what determine the max addressable RAM. Re your previous, the "max address space of a process" is virtual address space, not physical (once address translation is enabled).

[Daniel R Hicks]

I never said they were the same thing.

[Jamie Hanrahan]

Not exactly, no. But the OP was clearly asking about RAM, and you said that virtual memory means "the processor can address even more memory than what is installed." I feel this is very misleading, What the program code addresses when VM is enabled is virtual memory, not RAM; but when you speak of "memory that is installed" you're clearly talking about RAM (as v.m. is not something that is "installed"). I'm sure you know what you're talking about, I just think you're expressing it unclearly. I've found that the unqualified term "memory" is a frequent contributor to misunderstandings.

[Paul A. Clayton]

An OS can give the full 4GiB virtual address space (possibly minus one page--4KiB for x86) to the user process, but it makes system calls more expensive since the address space must be changed. With PAE and similar mechanisms, the amount of physical memory that is addressable by the system can be increased though addresses are still limited to 32 bits.

[johan smohan]

I red on wikipedia that 1 memory addresses is 32bits wide or 4 octets (1 octet = 8 bits), compared to 64bit a processor where 1 memory addresses or 1 integer is 64bits wide or 8 octets. You are correct about the 4*1024*1024*1024 bytes address space, but i was searching for the memory address space, which i think is 1GB/32bits, but i still don't know if i'm correct or not. :) Thank you for your answer!

[johan smohan]

It's not what i was searching for, but it's good information anyways! Thank you for the good explanation of the max. digit thing (1=max. as in binary 9 is max.) for the binary and decimal table comparison. Really a good way of thinking about this. I learned something new. :) Thanks!

[johan smohan]

This Gibi and Giga is so confusing... Because on some sites i read that the memory is measured in Gibi and on others that it's in Giga... Do you have some good / reliable source?

[johan smohan]

You meant 1024*1024*1024 and not 1*1024*1024 right?

[johan smohan]

A 32bit processor can address at most 2^32 octets or bits? Just checking, because i must know it for sure.

[gronostaj]

Wikipedia has an article on binary prefixes, including some historic background. Most hardware-related numbers use decimal prefixes, most notable exceptions are probably RAM and maybe color scale - eg. LCDs with 16M colors have three 8-bit color channels (2^24). Answering your question directly: In this case number of available addresses is equal to the number of memory bytes, because RAM is addressed on per-byte basis. 32-bit CPU could handle up to 2^32 B, 64-bit one - 2^64.

[AcId]

@johan smohan Correct, it should have been 1*1024*1024*1024

[AcId]

@johan smohan A 32bit processor can address at most 2^32 bytes, with one byte being 8 bits (one octet t of bits)

[johan smohan]

Thanks! I need this for school exams. :) I think that i understand the most things now. The only thing which still bothers me is why 2^32 B, if it's a 32bit processor and not a 32byte?

[gronostaj]

Because CPU is addressing bytes, not bits. You have 2^32 different addresses, so you can handle up to 2^32 bytes.

[GalacticCowboy]

"Now think about this: if you have one bit, you can save two values on it: 0 or 1.", etc. throughout the paragraph.

[Scott Chamberlain]

@GalacticCowboy corrected, I think I got them all.

[johan smohan]

Oh... i didn't knew that CPU addresses bytes and not bites. Thanks a lot for the whole answer!

[Tuncay Göncüoğlu]

The explanation is perfectly correct. Assuming you do not mean the actual chip dimension in centimeters when saying "processor width", in which case you would be correct in saying its unrelated, you are mistaking memory mapping techniques / virtual address spaces with physical memory addressing. Moreover what you say is more related to kernel implementations, you might want to check on PAE linux kernels.

[Daniel R Hicks]

The length of an instruction has no relationship at all to the "width" of the processor in modern systems. The most relevant values are the width of registers (though that can be deceptive), the width off the transfer path between processor and memory, and the size in bits of a memory address. But these 3 values can very easily be different from each other.

[Jamie Hanrahan]

@DanielRHicks Daniel Hicks is correct. The "bit width" of the CPU does not necessarily have anything to do with the "instruction size, or command size". There have been CPUs that were built that way but today's commodity processors (x86/x64) are not among them.

[Jamie Hanrahan]

I agree. Note that the later models of the 16-bit PDP-11 had a 22-bit address bus (and so could address 4 MB of RAM), the HP 1000MX, also "16-bit", eventually reached 16 MB of RAM (24-bit addresses); the VAX was a 32-bit CPU but had a 30-bit physical address space, but half was reserved for I/O space, for a RAM limit of 512 MB; the "16-bit" 8086, 1 MB; the "16-bit" 80286, 16 MB; etc. And when PAE was introduced with the Pentium Pro, 32-bit x86 could address up to 64 GB RAM (24-bit physical address space, although the low-order three bits never actually make it out of the CPU).

[Jamie Hanrahan]

@johansmohan Neither your numbers nor Lawrence's answer are correct. There is no fixed relationship between the "bit width" of a processor and the width of the RAM addresses it can use. 32-bit-only x86 processors can address 64 GB of RAM. x64 processors started at 40-bit physical address space and are now at 52 bits. As for virtual address space, that can be different too. On x64, although the virtual addresses do take 64 bits to store, only 48 bits are implemented, for a VAS of 256 TiB instead of the 16 EiB you'd expect from 64 bits.