The following site "Writing Boot Sector Code" provides a sample of code that prints 'A' to the screen when the system boots. From what I have been reading don't you have to use INT opcode to get BIOS to do certain things? How does the code below, from the site referenced above work without using interrupts? What portion of code actually tells the hardware to print 'A' to the screen?
Code in question:
.code16
.section .text
.globl _start
_start:
mov $0xb800, %ax
mov %ax, %ds
movb $'A', 0
movb $0x1e, 1
idle:
jmp idle
APPENDING TO ORIGINAL QUESTION
If I use the following code does the BIOS call write to the text buffer for me? The buffer starting at address 0xb800?
# Author: Matthew Hoggan
# Date Created: Tuesday, Mar 6, 2012
.code16 # Tell assembler to work in 16 bit mode (directive)
.section .text
.globl _start # Help linker find start of program
_start:
movb $0x0e, %ah # Function to print a character to the screen
movb $0x00, %bh # Indicate the page number
movb $0x07, %bl # Text attribute
mov $'A', %al # Move data into low nibble
int $0x10 # Video Service Request to Bios
_hang:
jmp _hang
.end
Direct answer to your question: The line "movb $'A', 0" effectively completes the print to the screen (and the following line, "movb $0x1e, 1" specifies what color it should be).
Longer answer: The video hardware draws the screen based on the contents of memory. When in text mode, the video hardware starts drawing based on memory segment 0xB800. Whatever is at byte 0 defines the character to be drawn at the first text cell on the screen. The next byte defines the attributes (foreground color, background color, and blink status). This pattern repeats (char - attr - char - attr) throughout the entire screen.
So, technically, my direct answer wasn't true. The 2 'movb' statements simply stage the letter 'A' to be printed. 'A' is not printed until the next time hardware refreshes the display based on the memory.