"If a computer handles data in 8-bit sizes and uses a 16-bit address to store and retrieve data in memory, its address space contains 2^16 (65536) bytes or 64k bytes"
My text book has this statement that I'm confused by. Where are they getting 2^16 from? If a computer uses a 16-bit address why isn't that just a 2 byte address space? The textbook hasn't explained how memory is stored in microcomputers and has this statement in the intro chapter. Am I missing something?
If an address is 16 bits, it means that you have 16 bits when referring to a location in memory. The address space is the range of valid addresses, not the physical size of an address.
These addresses start at address 0 (binary 0000 0000 0000 0000) and go up to address 216−1 (binary 1111 1111 1111 1111). That's total of 216 addresses that can be referenced. And if each address refers to 8 bits (i.e. a byte), the total amount of memory that you can refer to with those addresses is 216 × 8 bits, or 216 bytes.
As a smaller example, consider a system with 3-bit addresses, each referring to 4 bits (a nibble).
Address | 0 1 2 3 4 5 6 7
Memory | 0000 0000 0000 0000 0000 0000 0000 0000
Binary |
address | 000 001 010 011 100 101 110 111
The 3-bit addresses can have 23 values, from 0 to 7, and each refers to 4 bits of memory, so this system has a total of 23 = 8 nibbles of memory.
In this system, the only valid address are 0, 1, 2, 3, 4, 5, 6, and 7, so the address space is the set {0, 1, 2, 3, 4, 5, 6, 7}.
As an important point, don't forget that the address space is not necessarily the actual amount of memory available—computers use some handy tricks to use addresses spaces much larger than the memory they actual have available (for example, a 64-bit system can theoretically address 264 bytes of memory, but you don't even have a fraction of that in your computer).
Analogies for address spaces
Here are two analogies that might help you understand the difference between an address space, an address, and a pointer:
The web address space is the set of all URLs, basically the set of strings of the form https://[domain]/[path]. So https://example.com/page is an address, and this link is a pointer to that address.
The United States street address space is (approximately) the set of strings of this form:
[First name] [Last name]
[number] [Street name]
[Town], [STATE] [zip code]
In the same analogy, this is an address:
John Doe
10 Main St.
Faketown, NY 20164
Finally, a pointer is then analagous to the writing on the front of an envelope that the postal service uses to deliver letters.
Related
I've a question it is a kind of disagreement with a university professor, about Endianness, So I did not find any mean to solve this and find the right answer but asking and open a discussion in Stack Overflow community.
Let's say that we have this number (hex)11FF1 defined as an integer, for example in C++ it will be like : int num = 0x11FF1, and I say that the number will be presented in the memory in a little endian machine as :
addr[0] is f1 addr[1] is 1f addr[2] is 01 addr[3] is 00
in binary : 1111 0001 0001 1111 0000 0001 0000 0000
as the compiler considers 0x11ff1 as 0x0001ff1 and considers also 00 as the 1st byte and 01 as the 2nd byte and so on, and for the Big Endian I believe it will look like:
addr[0] is 00 addr[1] is 01 addr[2] is 1f addr[3] is f1
in binary : 0000 0000 0000 0001 0001 1111 1111 0001
but he has another opinion, he says :
Little Endian
Big Endian:
Actually I don't see anything logical in his representation, so I hope the developers Resolve this disagreement, Thanks in advance.
Your hex and binary numbers are correct.
Your (professor's?) French image for little-endian makes no sense at all, none of the 3 representations are consistent with either of the other 2.
73713 is 0x11ff1 in hex, so there aren't any 0xFF bytes (binary 11111111).
In 32-bit little-endian, the bytes are F1 1F 01 00 in order of increasing memory address.
You can get that by taking pairs of hex digits (bytes / octets) from the low end of the full hex value, then fill with zeros once you've consumed the value.
It looks like they maybe padded the wrong side of the hex value with 0s to zero-extend to 32 bits as 0x11ff1000, not 0x00011ff1. Note these are full hex values of the whole number, not an attempt to break it down into separate hex bytes in any order.
But the hex and binary don't match each other; their binary ends with an all-ones byte, so it has FF as the high byte, not the 3rd byte. I didn't check if that matches their hex in PDP (mixed) endian.
They broke up their hex column into 4 byte-sized groups, which would seem to indicate that it's showing bytes in memory order. But that column is the same between their big- and little-endian images, so apparently that's not what they're doing, and they really did just extend it to 32 bits by left shifting (padding with low instead of high zero).
Also, the binary field in the big vs. little endian aren't the reverse of each other. To flip from big to little endian, you reverse the order of the bytes within the integer, keeping each byte value the same. (like x86 bswap). Their 11111111 (FF) byte is 2nd in their big-endian version, but last in little-endian.
TL:DR: unfortunately, nothing about those images makes any sense that I can see.
I have a question about calculating memory addresses:
I am given 3 Memory blocks:
- 1x 1KByte (IC1) - 2^10 Byte
- 2x 4KByte (IC2 + IC3) 2^12 Byte
So far I calculated these memory addresses:
IC1:
0000 0000 0000 0000 (Starting adress)
0000 0011 1111 1111 (Ending adress, I got this from inverting the last 10 digits)
IC2:
0000 0100 0000 0000 (Starting adress)- Last ending adress +1
0000 1011 1111 1111 (Ending adress, I got this from inverting the last 12 digits)
However, at IC3 there has to be some method to get a carry bit into my first 0000-block, as I am running out of space when only using 3 the last 3 hex digits:
IC2:
0000 1100 0000 0000 (Starting adress)- Last ending adress +1
What is the ending address now? If I would invert the last 12 digits again, I would get a hex address which is already in use. It's pretty obvious that the next hex digit has to be increased to 1, but I can't find a rule to do this.
Any advice?
I'm not sure why you're using bit flipping for this, it looks like it should be a very efficient implementation if it works, but it doesn't seem to:
Your IC2 block starting address (in Hex) is 400 (which is 1K from the start of memory, all good so far), but the ending address in hex is BFF when it should be 13FF (1k+4k = 5k) in binary that is 0001 0011 1111 1111
Is there a reason why you cannot calculate these addresses using addition instead of bit-flipping?
Given this data,
.data
Alpha WORD 0022h, 45h
Beta BYTE 56h
Gamma DWORD 4567h
Delta BYTE 23h
Assuming that the data segment begins at 0x00404000, can anyone verify how correct this table is?
Address Variable Data
00404000 Alpha 22
00404001 Alpha + 1 00
00404002 Alpha + 2 45
00404003 Beta 56
00404004 Gamma 67
00404005 Gamma+1 45
00404006 Delta 23
Impossible to answer without knowing the addressing of the processor in question (and how the assembler views the addressing). Nonetheless, you'd need a pretty unusual system for it to be correct.
Alpha is defined has having the type "word". You're showing the first word as allocating two bytes (fairly reasonable), but the second only one byte. This is much less reasonable--a word might be one byte or it might be two, but its size is normally going to at least be consistent.
For the moment, let's assume a word is two bytes, and a dword is four bytes. In that case, I'd expect something more like:
Alpha 22h
alpha+1 00h
alpha+2 45h
Alpha+3 00h
Beta 56h
Gamma 67h
Gamma+1 45h
Gamma+2 00h
Gamma+3 00h
Delta 23h
I'm having a bit of an issue understanding what is going on here, and can't seem to wrap my head about it.
Notes:
Course notes about topic
Example:
Memory location 0x1f6
What is the binary format of this address? 1 1111 0110
What are tag, block index, and block offset? 3, 7, 6
My own work:
Memory location 0x033
What is the binary format of this address? 0 0011 0011
What are tag, block index, and block offset? 0 6, 3
Memory location 0x009
What is the binary format of this address? 0 0000 1001
What are tag, block index, and block offset? 0, 1, 1
Memory location 0x652
What is the binary format of this address? 0110 0101 0010
What are tag, block index, and block offset? 12, 10, 2
These are my attempts, but I have not a clue if I'm doing it right, and I have a feeling that I am not, as least for the last one, which I believe is wrong. Can anyone point me in the right direction?
I ended up figuring it out. The block offset is dependent upon the block size, in this case 16bytes, so it requires 4 binary digits to represent it. Next, the block index is dependent upon the number of blocks, in this case 8 (0-7), which requires 3 binary digits. Finally, the tag is made up of the remaining binary digits after you convert the hex memory location to binary.
Example
Memory location 0x652
What is the binary format of this address? 0110 0101 0010
What is the binary representation of tag, block index, and block offset? 1100 101 0010
What are tag, block index, and block offset? 12, 7, 2
I am confused as to how memory is stored when declaring variables in assembly language. I have this block of sample code:
val1 db 1,2
val2 dw 1,2
val3 db '12'
From my study guide, it says that the total number of bytes required in memory to store the data declared by these three data definitions is 8 bytes (in decimal). How do I go about calculating this?
It also says that the offset into the data segment of val3 is 6 bytes and the hex byte at offset 5 is 00. I'm lost as to how to calculate these bytes and offsets.
Also, reading val1 into memory will produce 0102 but reading val3 into memory produces 3132. Are apostrophes represented by the 3 or where does it come from? How would val2 be read into memory?
You have two bytes, 0x01 and 0x02. That's two bytes so far.
Then you have two words, 0x0001 and 0x0002. That's another four bytes, making six to date.
The you have two more bytes making up the characters of the string '12', which are 0x31 and 0x32 in ASCII (a). That's another two bytes bringing the grand total to eight.
In little-endian format (which is what you're looking at here based on the memory values your question states), they're stored as:
offset value
------ -----
0 0x01
1 0x02
2 0x01
3 0x00
4 0x02
5 0x00
6 0x31
7 0x32
(a) The character set you're using in this case is the ASCII one (you can follow that link for a table describing all the characters in that set).
The byte values 0x30 thru 0x39 are the digits 0 thru 9, just as the bytes 0x41 thru 0x5A represent the upper-case alpha characters. The pseudo-op:
db '12'
is saying to insert the bytes for the characters '1' and '2'.
Similarly:
db 'Pax is a really cool guy',0
would give you the hex-dump representation:
addr +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +A +B +C +D +E +F +0123456789ABCDEF
0000 50 61 78 20 69 73 20 61 20 72 65 61 6C 6C 79 20 Pax is a really
0010 63 6F 6F 6C 20 67 75 79 00 cool guy.
val1 is two consecutive bytes, 1 and 2. db means "direct byte". val2 is two consecutive words, i.e. 4 bytes, again 1 and 2. in memory they will be 1, 0, 2, 0, assuming you're on a big endian machine. val3 is a two bytes string. 31 and 32 in are 49 and 50 in hexadecimal notation, they are ASCII codes for the characters "1" and "2".