Setting start address to execute raw binary file

Question

Bootloader is seperated into 2 stages. First stage is written in assembly and only loads second stage, second stage is in C. Stage1 loads code in C to address 0x0500:0, and jumps there. Stage2 have to write "hello message" and halt.

I tried different ways to set starting address to raw binary made by: (but nothing worked)

cc -nostartfiles -nostdlib -c stage2.c
ld -s -T scrptfile.ld stage2.o /* I'm using ld just to set starting address of executable */
objcopy -O binary stage2 stage2.bin /* delete all unuseful data  */

Linker script

SECTIONS
{
    . = 0x0500;
    .text : { *(.text)}
    .data : { *(.data)}
    .bss : { *(.bss)}
}

Maybe I delete with objcopy somethnig that shouldt be deleted.

How can I execute this stage2.bin then?

As I understand, written C code using 32-bits length instructions, when raw binary allows only 16?

P.S. Parameter -set-start (objcopy) returns an error: Invalid bfd target. It is because output file is binary?

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Thank you for answers.

Answer 1

. = 0x0500 does not correspond to 0x0500:0. 0x0500:0 is physical address 0x5000, not 0x500.

Also, if you're trying to compile C code as 32-bit and run it in real mode (which is 16-bit), it won't work. You need to either compile code as 16-bit or switch the CPU into 32-bit protected mode. There aren't that many C compilers still compiling 16-bit code. Turbo C++ is one, Open Watcom is another. AFAIK, gcc can't do that.

Finally, I'm guessing you expect the entry point to be at 0x500:0 (0x5000 physical). You need to either tell this to the linker (I don't remember how, if at all possible) or deal with an arbitrary location of the entry point (i.e. extract it from the binary somehow).