Memory map 4GB or 4.096GB?

[JDirk.1]

Hi all,

1)

in STM32F446xC/E it is written that the memory consists out of 8 Blocks each 512-Mbytes. This leads to 4.096-GB. In PM0214 its written: The processor has a fixed memory map that provides up to 4 GB of addressable memory. Now my question: How much does it have in total, 4.096GB or 4.000GB? Or does it mean 0.096GB are not addressable?

2)

Is it correct, that some Addresses only refer to 4 Bit areas? Otherwise I cant explain the 4.096GB knowing that the last Address is 0xFFFFFFFF. In the beginning i thought it might be 8 bits per address but that would result into 4,29....GB.

Or is there something i got completly wrong?

[TDK]

The 32-bit addressable space is 2^32 bytes, or 4.294967296 GB, using the most commonly used definition of GB of 1000000000 bytes. That's just the number of unique values for a uint32_t.

The chip doesn't have anywhere near this amount of memory onboard. Only some portions of that memory space are mapped to peripherals, memories, and other stuff.

[JDirk.1]

Thanks. Your explanation for 4.29... is clear to me, its understandable.

But how can a address range from 0x0000 0000 to 0xFFFF FFFF (4294967296 different addresses) lead to 8 x 512-Mbytes (4.096-GB)? Thats fully confusing. 4294967296 x 8 Bit = 4.29... Giga-Byte. So the screenshot attached saying each block is 512-Mbytes with the mentioned 0xFFFF FFFF, something is wrong on how i read and calculate ...

[Piranha · Accepted solution]

https://en.wikipedia.org/wiki/Binary_prefix

In real life the numbering base depends on the context where it's used. For network speeds typically base 10 is used. For storage and addressing typically base 2 is used. And historically even when base 2 is used with IEC (capital) prefix, rarely anyone uses the "i" letter.

P.S. HDD manufacturers like to "cheat" and use base 10 also for storage capacity.

[TDK]

> But how can a address range from 0x0000 0000 to 0xFFFF FFFF (4294967296 different addresses) lead to 8 x 512-Mbytes (4.096-GB)? 

You seem to be okay with calling 1 MB = 1024^2 bytes, so why do you object to calling 1 GB = 1024^3 bytes.

As @Piranha mentions, the context for these numbers is important.

[JDirk.1]

many thanks to both of you! The binary prefix issue can be really confusing, its the first time i heard about ...

let me summarize:

The "512-Mbyte" given in the datasheet are equal to 536,871-Mega bytes assuming 8 Bits are one byte and Mega stands for 10^6 according to my calculation. Each Address refers to a 8-Bit register. 2^32 = 4294967296 possible addresses multiplied with 8 Bit result into 4,29GB memory (or expressed as 4GB if i use this binary prefix method).

is that correct?

for me 8 Bits are ony byte

1kilobyte are 1.000bytes

1 megabyte are 1.000.000bytes

in the PDF 1 MB is 1048576bytes

[JDirk.1]

i think i got it ...

[JDirk.1]

thanks!