Why is this kernel causing problems?

Question

I decided to try and write a C Hello world protected mode Kernel, and although directly accessing the video memory works to print characters manually, I decided to try to write a string, and there are no warnings or errors but it boots to a blinking cursor, but after about half a millisecond it black screens. Here is the code:

#include <stdint.h>

void kmain() {
    // Startup
    int i = 0;
    
    uint16_t* vidmem = (uint16_t*) 0xB8000;
    
    char string[] = "Hello, kernel world!";
    while (string[i]) {
        vidmem[i] = string[i];
        vidmem[i+1] = 0;
        i+=2;
    }

    // Hang the system
    for (;;) {
        
    }
}

Compiled with: gcc -m32 -c kernel.c -o kernel.o -std=gnu99 -ffreestanding -O1 -Wall -Wextra

And then linked with a generic boot.o and then made into an ISO with grub

I tested it in qemu on linux, any help would be appreciated!

Before I attempted the printstring funciton, this was able to print characters:

vidmem[0] = "C";
vidmem[1] = 1;

Which would print the letter C. Nothing fancy though.

After trying to fix some erros I came up with this:

#include <stdint.h>

void kmain() {
    // Startup
    int i = 0;
    int j = 0;
    
    uint16_t* vidmem = (uint16_t*) 0xB8000;
    
    char string[] = "Hello, kernel world!";
    while (string[i]) {
        vidmem[i] = string[i];
        vidmem[j] = 0;
        i+=1;
        j=i*2;
    }

    // Hang the system
    for (;;) {
        
    }
}

Which still blackscreens.

If you defined `uint16_t* vidmem` how many bytes are written by `vidmem[i] = string[i];` and how many bytes are added by `vidmem[i+1]`, and how many bytes advanced by `i+=2;`? — Weather Vane, Jul 28 '20 at 19:05
`uint16_t` is 2 bytes and so `vidmem[i]` writes to 2 bytes, and `vidmem[i+1]` is 2 bytes further on. `i+=2` used as an index advances by 4 bytes in the video memory and 2 bytes in your source string. — Weather Vane, Jul 28 '20 at 19:09
You know you are skipping characters of `string`? You can easily miss the `\0` this way. — Eugene Sh., Jul 28 '20 at 19:12
Thanks I didn't realize that, but making the adjustments still does not fix the black screen problem... — A user, Jul 28 '20 at 19:14
Please post your revised code. If you can also post the earlier version that is able to print characters, that would help too. — Nate Eldredge, Jul 28 '20 at 19:29
try using 'uint8_t*' as the pointer to the video memory , instead of 'uint16_t*' . — Amine Bensalem, Jul 31 '20 at 15:25

score 3 · Accepted Answer · answered Jul 28 '20 at 20:20

You are trying too hard to do arithmetic that isn't needed. Remember that pointer arithmetic and array indexing in C are always in units of the size of the type pointed to. If you want to copy the bytes of string into vidmem with a zero byte after each one, then thanks to the fact that integers on x86 are little-endian, it should be sufficient to do

int i = 0;
uint16_t* vidmem = (uint16_t*) 0xB8000;

unsigned char string[] = "Hello, kernel world!";
while (string[i]) {
    vidmem[i] = string[i];
    i++;
}

Or if you prefer

for (i = 0; string[i]; i++)
    vidmem[i] = string[i];

That's it.

If you want a different color that is not black-on-black, you could change this to vidmem[i] = string[i] | 0x7000;.

0___________ · Answer 2 · 2020-07-28T20:39:27.290

0

Both versions are incorrect;

    uint8_t* vidmem = (uint8_t*) 0xB8000;
    
    char string[] = "Hello, kernel world!";
    for(size_t i = 0; string[i]; i++) {
        vidmem[i*2] = string[i];
        vidmem[i*2 + 1] = 0;
    }

edited Jul 28 '20 at 20:39

answered Jul 28 '20 at 20:11

0___________

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Shouldn't that be vidmem[i*2 + 1] = 0x70; or somesuch. 0 is "black on black", which is a bit tough on the eyes. – mevets Jul 28 '20 at 20:17
@mevets it can be. Who knows. – 0___________ Jul 28 '20 at 20:20
Your cast on the first line is incorrect, though it will probably just give you a warning -- should cast to `uint8_t *` to match the pointer type... – Chris Dodd Jul 28 '20 at 20:36

Why is this kernel causing problems?

2 Answers2