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.

  • 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. Aug 8, 2014 at 11:36
  • @DanielRHicks Virtual memory doesn't affect how much RAM can be addressed. Jun 7, 2017 at 19:59
  • @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. Jun 7, 2017 at 20:13
  • @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. Jun 7, 2017 at 23:31
  • @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.) Jun 8, 2017 at 1:57

6 Answers 6


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.


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.


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:

|  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.

  • 1
    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. May 10, 2013 at 12:26
  • 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! May 10, 2013 at 13:09
  • 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? May 10, 2013 at 13:22
  • 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.
    – gronostaj
    May 10, 2013 at 14:16
  • 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? May 10, 2013 at 14:31

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.


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.

  • 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! May 10, 2013 at 13:06
  • Np. Intel CPU's have two address spaces: "Memory" and "I/O". Things other than RAM appear in the "Memory" space. Other things such as I/O devices or ROM may be at spots unoccupied by RAM. Generally only I/O devices appear in the I/O address space.
    – LawrenceC
    May 10, 2013 at 17:23
  • @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. Jun 8, 2017 at 21:19

A 32bit processor can address at most 2^32 individual bytes of memory (about 4GB), but having 1GB of memory would make 1*1024*1024*1024 addressable bytesof memory (though you would probably still have a 2^32 virtual address space). A 64bit CPU could address 2^64 individual bytes, but I think most systems use only 48 bits for memory addresses making the upper bound. addressable bytes 2^48.

  • You meant 1024*1024*1024 and not 1*1024*1024 right? May 10, 2013 at 13:26
  • A 32bit processor can address at most 2^32 octets or bits? Just checking, because i must know it for sure. May 10, 2013 at 13:29
  • @johan smohan Correct, it should have been 1*1024*1024*1024
    – AcId
    May 10, 2013 at 14:17
  • @johan smohan A 32bit processor can address at most 2^32 bytes, with one byte being 8 bits (one octet t of bits)
    – AcId
    May 10, 2013 at 14:19

The accepted answer gives a good explanation. But I don't think that it is the answer. It doesn't contain anything about address bus. And its size is actually the main reason of memory constrains. For example 8080 is 8-bit processor (size of its data bus is 8 bits), but it has a 16-bit address bus. It can address 2^16=(2^6)*(2^10)=64*1024 bytes=64KB.

You can find more here (32-bit) in "Technical history" section.

  • 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). Jun 8, 2017 at 20:27

I belive the most basic info is lost in this conversation, so here is my answer:

Saying "This is a a 32 bit processor" means that the instruction size, or command size, that cpu can understand and work with at one time is 32 bits. Likewise with 64 bit processors: they can handle instructions of at most 64 bits.

Think of this like an old mechanical calculator: you only have so many digits, so simply cannot input any longer number.

Now, an address a CPU can use also has to fit into that same space, so for a 32 bit processor, the address it uses can also only be 32 bit at most. So, from here we can simply calculate the maximum number of addresses (ie, maximum amount of RAM usable by the CPU):

2^32 = 4294967296 (= 4 GB)


2^64 = 18446744073709551616 (= way more ;)

Or, as a fun example, my old Commodore 64 had a 16 bit CPU, so it was capable of managing a memory of:

2^16 = 65536 bytes (= 64 KB)

This is the basic logic, but, as stated before, there are ways around this limitation, like virtual address spaces, memory mapping etc.

  • 2
    The significant difference between 32 and 64 bit processors is address range (which affects virtual address space more than "real"). Processors use so many tricks that it's hard to say what the actual data path width is, in many cases. And the instruction length is largely unrelated to the processor "width". May 10, 2013 at 21:30
  • 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. Aug 8, 2014 at 10:45
  • 2
    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. Aug 8, 2014 at 11:30
  • 1
    @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. Jun 7, 2017 at 20:07

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