RAM Mapping

0x4000 is the size, 0x7FE00000 is the start address:

  RW_IRAM3 0x7FE00000 0x00004000  {
        LPC24_EMAC.o (+ZI +RW)
  }

refer to the RealView linker and utilities documentation for details.

Thanks for your help.

But I think is not exactly that.
Look a simple example, a have this three areas

long Bank0[0x80000];
long Bank1[0x80000];
long Bank2[0x80000];

But when I compile the folowing messages is shwon:

"No space in execution regions with .ANY selector matching Section .bss(banck.c)"

"Sections of aggregate size 0x600000 bytes could not fit into .ANY selector(s)."

But in the memory window all parameter are configured with the write address and size. I have area enough to alocate those banks. How do I write the all scatter file?
What should I do? I'm getting crazy!

I assume you don't have enough RAM (at need 1536 KB for these arrays only!).
don't go mad. that will disturb you reading the manual :-)

Hello,

I have RAM enough, that I'm sure.

A long use 4 bytes, so an array with 0x80000 longs will use

0x80000 x 4 bytes = 2097152 2Mb each array

So if I have 3 external memorys with 2097152bytes and I want to alocate 3 arrays with 2097152bytes I dont understand how is not memory enough.

Excuse me if I was not clear enough.

Thank you

the best things to do are:
* place every buffer in a separate C module, and treat it like in the previous post - namely, place the generate .o file in a particular section in memory as demonstrated above.
* read the document to which I refer above. it will answer all of your questions.

or you can use the __at directive, like this:

const unsigned long long sys_software_programming_success_high __attribute__( (at(FIRMWARE_ID_ADDR+0xFF8) ) ) = 0xFFFFFFFFFFFFFFFF ;