memory addresses increasing

[tawny hare]

Does anyone know why memory addresses in microcontrollers increase by 4? Like starts at 0x00000000 and the next is 0x00000004, the registers are 4 bytes, does this have a correlation or is it just how it is? I thought that address are different from what they contain so if this is the case why would the register size affect the address.

[bleak vigil]

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[vale idol]

if your register stores 4 bytes of data, you're gonna need 4 bytes in memory to store the same data. if I put 0xDEADBEEF at address 0x00000000, then put 0xCAFEBABE at address 0x00000001, then memory may look like

0 1 2 3 4
DECAFEBABE

but if we put the second word at address 0x00000004, it may look like

0 1 2 3 4 5 6 7
DEADBEEFCAFEBABE

[tawny hare]

So

Does this mean that every byte of memory takes up 1 address ?

[vale idol]

yes. memory in modern computers is what's called "byte-addressable"

[tawny hare]

I understand now, so if I have data ie, 0x87654321 at address 0x00000000

How do I know which byte is at 0x00000000

And which is at 0x00000003

[vale idol]

depends on if it's a big-endian or little-endian system

big-endian stores with the most significant byte first, so if storing 0xCAFEBABE we'd get

0 1 2 3
CAFEBABE

but little-endian stores with the least-significany byte first, so we'd get

0 1 2 3
BEBAFECA

[graceful mulch]

you can bitwise and 0x87654321 for the first byte in bigendian if you do
0x87654321 & 0xff000000

but probably there is a better way ?

[vale idol]

endianness only exists for storing in memory, it's always the "as expected" value when you're interacting with it

[graceful mulch]

i suppose one must bitshift it as well

[vale idol]

0x12345678 & 0xff will be 0x78 on both little- and big-endian

[graceful mulch]

oh ok

(0x87654321 & 0xff000000) >> 6
so this would yield what jesus asked then?

[vale idol]

on a big-endian system specifically, sure

could also just e.g. ((unsigned char*)&value)[3]

[graceful mulch]

is that faster, or is casting+accessing it doing something similar to what i did with bitoperations?

[vale idol]

depends™️

compilers can do lots of neato optimizations, and not all hardware is equal. try it out

[tawny hare]

So like for this example how would I know?

[vale idol]

which do you think it is?

[tawny hare]

Well

Using process of elimination it would be E3

But this example doesn’t do well.

How am I supposed to know the order?

[vale idol]

B is after A

[tawny hare]

Yes

But how do I know it’s E3 and not 95

[vale idol]

because A + 2 is C, not B

[tawny hare]

Yea but I’m confused on the ordering

So this means at FA the value is A2

How do I know it’s A2 and not 74

[vale idol]

because the 74 is three bytes after FA

[tawny hare]

So the most significant is always on the left?

[vale idol]

no

[tawny hare]

And most significant is the lowest address?

[vale idol]

the byte with the smallest address is

endian-ness doesn't effect anything here

[tawny hare]

The byte with the smallest address is on the left

[vale idol]

which has nothing to do with the value of the larger word

[tawny hare]

How come in this example it becomes flipped?

[vale idol]

the 4-byte word at 0.04FA could be 0xA2E39574 if it's a big-endian system, or it could be the storage of 0x7495E3A2 if it's little-endian

[tawny hare]

4B 57 turned to 574B

Due to little endian

[tawny hare]

Oh yes I understand

But why does endian not affect this

[vale idol]

endianness just effects the order of the bytes for each value, not the order of bytes as a whole

effect what

[tawny hare]

Hmm

I don’t understand

So what determines the order of the bytes?

[vale idol]

regardless of the byte order, it's still 0x05, then 0x06, then 0x07, etc

[tawny hare]

Yes I understand this

But

Regardless of byte order we still have the same address order

[vale idol]

the top-left cell here will always be 0x0.04FA. What endianness effects is if that's the first or the last byte of the data in those four bytes

[tawny hare]

But how do we know the byte order

Oh

Okay

I see now

So the left cell is always the lowest address

[vale idol]

same as a number line

[tawny hare]

But how come in this the left cell isn’t the lowest addresss

Any of the data

[vale idol]

because it's little-endian, so the least-significant byte is stored first

[tawny hare]

How do we know which is the LSB

57?

[vale idol]

well if you look at the right side of the image, where the values are being shown instead of the memory representation, we can see that 4B is the least significant byte

[tawny hare]

yes

but how do I know

[vale idol]

because it's the least significant byte

[tawny hare]

my misunderstanding is how do we know if 4B or 57 is the LSB

from this

[vale idol]

do you know what LSB means

[tawny hare]

the byte on the right

yes

but given 4B and 57 howdo weknow which is LSB

[vale idol]

by looking

[tawny hare]

wait

im looking and the order is different

in this

the firstsquare on the left (4B) is 03FC?

[tawny hare]

?

[vale idol]

yes

[tawny hare]

so this means that in little endian 4B is LSB

thats all I wanted to figure out

[vale idol]

yes

[tawny hare]

okay thanks

i understand now

[vale idol]

you can also just look at the number and see that it's the LSB

[tawny hare]

yea I got that from the start

i was just trying to figure out the otherimage

i just didnt know how the address and the squares in the other image correlated

[bleak vigil]

@tawny hare Has your question been resolved? If so, type !solved :)