/*
    memblock.c

    Written by Tanya Amert for Fall 2024

    This program generates a specific block of 32 bytes in memory.
    These correspond to a table that, on paper, looks like this:
    
       0x60    3F  24  07  00  12  00  00  00
       0x68    75  00  00  00  00  00  00  00   <-- different in this program
       0x70    59  41  31  00  00  43  61  54
       0x78    20  3B  29  00  00  00  00  00

    Compile as usual:

        gcc -Wall -Werror -g -o memblock memblock.c

    Run with no command-line arguments:

        ./memblock
*/

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
    // We'll build a 4x8 table (four rows, each depicting eight bytes)
    int num_bytes = 32;
    char *mem = (char *)malloc(sizeof(char) * num_bytes);

    // Row 1 (0x60)
    *((int *)mem) = 0x7243F;
    *((short *)mem + 2) = 0x12;
    *((short *)mem + 3) = 0xCA;

    // Row 2 (0x68)
    *((char**)mem + 1) = mem+0x15; // 0x75

    // Row 3 (0x70)
    *((int *)mem + 4) = 0x314159;
    mem[20] = 0x00;
    mem[21] = 0x43;
    mem[22] = 0x61;
    mem[23] = 0x54;

    // Row 4 (0x78)
    mem[24] = 0x20;
    mem[25] = 0x3B;
    mem[26] = 0x29;
    mem[27] = 0x00;
    mem[28] = 0x00;
    mem[29] = 0x00;
    mem[30] = 0x00;
    mem[31] = 0x00;

    // Print out the memory
    for (int i = 0; i < num_bytes; i++)
    {
        if (i % 8 == 0)
        {
            printf("%p ", mem + i);
        }

        printf("%.2x ", mem[i] & 0xFF);
        if ((i+1) % 8 == 0) printf("\n");
    }

    printf("\n");

    // Now onto the questions!
    int *num_addr = (int *)mem; // we can't actually hard-code the address, but we know it relative to mem
    printf("a) 0x%x\n", *num_addr); // a

    short two_byte_num = (short) *num_addr;
    printf("b) 0x%x\n", two_byte_num & 0xFFFF); // b

    char c = mem[1];
    printf("c) %c %d\n", c, c); // c

    char **str_addr = (char **)(mem + 0x8);
    printf("d) %p\n", str_addr); // d

    char *str = *str_addr;
    printf("e) '%s' at %p\n", str, str); // e

    printf("f) %p\n", str + 4); // f
    printf("g) %p\n", num_addr + 3); // g
    printf("h) %p\n", str_addr + 1); // h

    char *ch_addr = &(str[2]);
    printf("i) %c %p\n", *ch_addr, ch_addr); // i

    // Don't forget to free the memory!
    free(mem);

    return 0;
}