I am using objcopy to remove some necessary scripting to embed a resource file (zip file) in flash memory (ARM embedded thingy).
I am using objcopy like this:
arm-none-eabi-objcopy.exe -I binary -O elf32-littlearm -B arm --rename-section .data=.rodata input.zip input.zip.o
arm-none-eabi-nm.exe -S -t d input.zip.o
00006301 R _binary_input_zip_end
00006301 A _binary_input_zip_size
00000000 R _binary_input_zip_start
What I need to know is what is the width of the _end and _size symbols. I can only guess that the _start is an address which can be accessed like a byte array: extern uint8_t _binary_input_zip_start[];. And I am assuming that the _end and _size are of 'native' int-size, and I suppose I can safely assume I can interpret these as uint32_t.
However I can't be certain. I can't find anything "size" related in the docs of objcopy: https://sourceware.org/binutils/docs/binutils/objcopy.html
I'm not %100 sure if this will work, but try adding the option --sort-size to arm-none-eabi-nm. This is supposed to sort the symbols by size, by comparing them to the next symbol above. In combination with the -S option, it should print a size. Hopefully, this will help you deduce their width.
What ARM micro are you using? 32-bits is a good guess, but there are exceptions. If you happen to be using a Texas Instruments part, I can help a lot more. I don't have an ARM project handy that I can test this on, but it's worth a shot. If that doesn't work, I'll keep digging.
Source: My knowledge, and double-checking via http://manned.org/arm-none-eabi-nm