why does this bootloader only prints 'S' - c

Question

I am writing a simple x86 bootloader.
this is the c program that im having trouble with: test4.c
__asm__(".code16\n");
__asm__("jmpl $0x0, $main\n");
void prints ( char* str )
{
char* pStr = str;
while( *pStr )
{
__asm__ __volatile (
"int $0x10"
:
: "a"(0x0e00 | *pStr), "b"(7)
);
pStr++;
}
}
void main ( )
{
char* str = "\n\rHello World\n\r";
char* pStr = str;
while( *pStr )
{
__asm__ __volatile (
"int $0x10"
:
: "a"(0x0e00 | *pStr)
);
pStr++;
}
prints ( str );
}
when i try to print a string within main function, it works. But when i pass the string to another function which does carry out same instructions but still prints only S to the screen. So the final output looks something like this:
Hello World
S
Here is the linker file i used: test.ld
ENTRY(main);
SECTIONS
{
. = 0x7C00;
.text : AT(0x7C00)
{
*(.text);
}
.sig : AT(0x7DFE)
{
SHORT(0xaa55);
}
}
Here are the commands i used to compile the c program and to link it
$ gcc -c -g -Os -m32 -march=i686 -ffreestanding -Wall -Werror test4.c -o test4.o
$ ld -melf_i386 -static -Ttest.ld -nostdlib --nmagic -o test4.elf test4.o
$ objcopy -O binary test4.elf test4.bin
and i used bochs emulator to test out this bootloader

Answer 1

You can't do this with GCC. Ignore all the tutorials that say that you can -- they are wrong.
What's most important to keep in mind is that GCC is not a 16-bit compiler. The __asm__(".code16\n") directive does not turn it into one; it merely confuses the assembler into retargeting GCC's output from 32-bit x86 to 16-bit. This will cause strange and unexpected behavior, especially in any code using pointers.
If you want to write an x86 bootloader, you will need to:
Use a C compiler that can specifically target 16-bit x86 ("real mode"). Consider the OpenWatcom toolchain, for instance.
Become very familiar with the quirks of x86 real mode -- particularly segmentation.
Write some portions of the bootloader in assembly, particularly the startup code.