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Davis Apanasova
Davis Apanasova

Reputation: 51

Little Endian - Memory content/address

Consider a system that has a byte-addressable memory organized in 32-bit words according to the big-endian scheme. A program reads ASCII characters entered at a keyboard and stores them in successive byte locations starting at location 1000.

Show the contents of the two memory words at locations 1000 and 1004 after the name johnson has been entered. Write this in the little-endian scheme.

What I got was:

[NULL, n], [o, s], [n,h], [o,j] 00, 6E 6F, 73 6E, 68 6F, 6A

I just want to know if this is correct and if not, what I did wrong.

Thank you all!

Upvotes: 1

Views: 7276

Answers (2)

6502
6502

Reputation: 114461

I find this question quite confusing.

byte is normally defined as the smallest addressable unit so saying that a machine has byte-addressable memory just tells nothing: every machine has byte-addressable memory because that's the definition of what a byte is, what can change is how many bits is a byte.

If the question is talking about a 32-bit byte machine (I know they exists, but I personally used only machines with 8-bit and 16-bit bytes) then it's not clear what role is playing endian-ness given that no multibyte processing is needed for storing ASCII. What is often done in large-byte machines is however storing multiple characters per byte to save space (not necessarily a 16-bit byte machine is "big": the one I know is a DSP with a very limited amount of memory) but this seems unrelated to the question and there so "standard" way to do so anyway.

If instead the question assumes that a byte is always 8 bit by definition and talks about storing ASCII chars then once again endian-ness plays no role; chars are just store in memory one after another in consecutive locations. For example if the string "johnson" has been stored (assuming a C string convention) the content of memory would be:

0x6A 0x6F 0x68 0x6E 0x73 0x6F 0x6E 0x00

Reading this memory content as two 32-bit words would be affected by endian-ness of course, but saying that the machine uses big-endian and asking to display the result in little-endian scheme is nonsense.

In a big endian scheme (e.g. 68k) the two 32-bit words would be 0x6A6F686E and 0x736F6E00, in a little-endian scheme (e.g. x86) they would be 0x6E686F6A and 0x006E6F73.

Upvotes: 0

Bishop
Bishop

Reputation: 188

There is no such thing as endianes for storing a single byte (such as an ASCII character). Endianes only comes into play when a value is represented as multiple bytes. So for example, storing a sequence of bytes is the same in little- and big-endian, only the representation of the bytes are different. For example, take the number 3 735 928 559 (or 0xdeadbeef in hex notation) and store that as a 32-bit word (e.g., an int) at memory location 1000 will give:

ADR: 1000 1001 1002 1004  
BE:   de   ad   be   ef  
LE:   ef   be   ad   de

So, if you were to actually represent your ASCII character as a 32-bit word you would get:

[0, 0, 0, 6a], [0, 0, 0, 6f], ... or,  
[6a, 0, 0, 0], [6f, 0, 0, 0], ...

for BE and LE respectively.

Upvotes: 1

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