How to print a pointer address without printf

Question

I\'m doing an exercise in which I need to print the memory (address) of a pointer. It would be easy to do it with printf(\"%p\", ..) but I\'m not allowed to use

Answer 1

You don't have to print the memory, you have to transform each byte from addr in hexadecimal and print it, the second part with dots and asterisk is transform the value of each byte in char, if it isn't printable it will print dot.

#include 

#include 

void    ft_putchar(char c)
{
    write(1, &c, 1);
}

void    print_ascii(const char *addr, int i)
{
    int j;
    int len;

    j = 0;
    if ((i + 1) % 16 == 0)
        len = 16;
    else
        len = (i + 1) % 16;
    while (j < 16 - len)
    {
        ft_putchar(' ');
        ft_putchar(' ');
        if (j % 2)
            ft_putchar(' ');
        j++;
    }
    if ((16 - len) % 2)
        ft_putchar(' ');
    j = 0;
    while (j < len)
    {
        if (*(addr + i / 16 * 16 + j) >= 32 && *(addr + i / 16 * 16 + j) <= 126)
            ft_putchar(*(addr + i / 16 * 16 + j));
        else
            ft_putchar('.');
        j++;
    }
    ft_putchar('\n');
}

void    print_hex(unsigned char value, int index)
{
    if (index < 2)
    {
        print_hex(value / 16, index + 1);
        if (value % 16 >= 10)
            ft_putchar('a' + value % 16 % 10);
        else
            ft_putchar('0' + value % 16);
    }
}

void    print_memory(const void *addr, size_t size)
{
    char    *ptr;
    size_t  i;

    if (addr && size > 0)
    {
        ptr = (char*)addr;
        i = 0;
        while (i < size)
        {
            print_hex(*(ptr + i), 0);
            if (i % 2)
                ft_putchar(' ');
            if ((i + 1) % 16 == 0 || (i + 1) == size)
                print_ascii(addr, i);
            i++;
        }
    }
}

This code works for your exam.