Simple C Kernel char Pointers Aren't Working

问题

I am trying to make a simple kernel using C. Everything loads and works fine, and I can access the video memory and display characters, but when i try to implement a simple puts function for some reason it doesn't work. I've tried my own code and other's. Also, when I try to use a variable which is declared outside a function it doesn't seem to work. This is my own code:

#define PUTCH(C, X) pos = putc(C, X, pos)
#define PUTSTR(C, X) pos = puts(C, X, pos)

int putc(char c, char color, int spos) {
    volatile char *vidmem = (volatile char*)(0xB8000);
    if (c == '\n') {
        spos += (160-(spos % 160));
    } else {
        vidmem[spos] = c;
        vidmem[spos+1] = color;
        spos += 2;
    }
    return spos;
}
int puts(char* str, char color, int spos) {
    while (*str != '\0') {
        spos = putc(*str, color, spos);
        str++;
    }
    return spos;
}
int kmain(void) {
    int pos = 0;
    PUTSTR("Hello, world!", 6);
    return 0;
}

The spos (starting position) stuff is because I can't make a global position variable. putc works fine, but puts doesn't. I also tried this:

unsigned int k_printf(char *message, unsigned int line) // the message and then the line #
{
    char *vidmem = (char *) 0xb8000;
    unsigned int i=0;

    i=(line*80*2);

    while(*message!=0)
    {
        if(*message=='\n') // check for a new line
        {
            line++;
            i=(line*80*2);
            *message++;
        } else {
            vidmem[i]=*message;
            *message++;
            i++;
            vidmem[i]=7;
            i++;
        };
    };

    return(1);
};

int kmain(void) {
    k_printf("Hello, world!", 0);
    return 0;
}

Why doesn't this work? I tried using my puts implementation with my native GCC (without the color and spos data and using printf("%c")) and it worked fine.

回答1:

Since you're having an issue with global variables in general, the problem most likely has to-do with where the linker is placing your "Hello World" string literal in memory. This is due to the fact that string literals are typically stored in a read-only portion of global memory by the linker ... You have not detailed exactly how you are compiling and linking your kernel, so I would attempt something like the following and see if that works:

int kmain(void) 
{
    char array[] = "Hello World\n";
    int pos = 0;
    puts(array, 0, pos);
    return 0;
}

This will allocate the character array on the stack rather than global memory, and avoid any issues with where the linker decides to place global variables.

In general, when creating a simple kernel, you want to compile and link it as a flat binary with no dependencies on external OS libraries. If you're working with a multiboot compliant boot-loader like GRUB, you may want to look at the bare-bones sample code from the multiboot specification pages.

回答2:

Since this got references outside of SO, I'll add a universal answer

There are several kernel examples around the internet, and many are in various states of degradation - the Multiboot sample code for instance lacks compilation instructions. If you're looking for a working start, a known good example can be found at http://wiki.osdev.org/Bare_Bones

In the end there are three things that should be properly dealt with:

The bootloader will need to properly load the kernel, and as such they must agree on a certain format. GRUB defines the fairly common standard that is Multiboot, but you can roll your own. It boils down that you need to choose a file format and locations where all the parts of your kernel and related metadata end up in memory before the kernel code will ever get executed. One would typically use the ELF format with multiboot which contains that information in its headers
The compiler must be able to create binary code that is relevant to the platform. A typical PC boots in 16-bit mode after which the BIOS or bootloader might often decide to change it. Typically, if you use GRUB legacy, the Multiboot standard puts you in 32-bit mode by its contract. If you used the default compiler settings on a 64-bit linux, you end up with code for the wrong architecture (which happens to be sufficiently similar that you might get something that looks like the result you want). Compilers also like to rename sections or include platform-specific mechanisms and security features such as stack probing or canaries. Especially compilers on windows tend to inject host-specific code that of course breaks when run without the presence of Windows. The example provided deliberately uses a separate compiler to prevent all sorts of problems in this category.these
The linker must be able to combine the code in ways needed to create output that adheres to the bootloader's contract. A linker has a default way of generating a binary, and typically it's not at all what you want. In pretty much all cases, choosing gnu ld for this task means that you're required to write a linker script that puts all the sections in the places where you want. Omitted sections will result in data going missing, sections at the wrong location might make an image unbootable. Assuming you have gnu ld, you can also use the bundled nm and objdump tools besides your hex editor of choice to tell you where things have appeared in your output binary, and with it, check if you've been following the contract that has been set for you.

Problems of this fundamental type are eventually tracked back to not following one or more of the steps above. Use the reference at the top of this answer and go find the differences.

来源：https://stackoverflow.com/questions/11305075/simple-c-kernel-char-pointers-arent-working

标签

pointers

x86

kernel

osdev