uintx_t to const char* in freestanding c++ using GNU compiler

Question

so I am trying to convert some integers in to character arrays that my terminal can write. so I can see the value of my codes calculations for debugging purposes when its running. as in if the int_t count = 57 I want the terminal to write 57. so char* would be an array of character of 5 and 7

The kicker here though is that this is in an freestanding environment so that means no standard c++ library. EDIT: this means No std::string, no c_str, no _tostring, I cant just print integers.

The headers I have access to are iso646,stddef,float,limits,stdint,stdalign, stdarg, stdbool and stdnoreturn

Ive tried a few things from casting the int as an const char*, witch just led to random characters being displayed. To feeding my compiler different headers from the GCC collection but they just keeped needing other headers that I continued feeding it until I did not know what header the compiler wanted.

so here is where the code needs to be used to be printed.

uint8_t count = 0;
while (true)
{
    terminal_setcolor(3);
    terminal_writestring("hello\n");

    count++;

    terminal_writestring((const char*)count);
    terminal_writestring("\n");
}

any advice with this would be greatly appreciated.

I am using an gnu, g++ cross compiler targeted at 686-elf and I guess I am using C++11 since I have access to stdnoreturn.h but it could be C++14 since I only just built the compiler with the latest gnu software dependencies.

Answer 1

Without C/C++ Standard Library you have no options except writing conversion function manually, e.g.:

template <int N>
const char* uint_to_string(
    unsigned int val,
    char (&str)[N],
    unsigned int base = 10)
{
    static_assert(N > 1, "Buffer too small");
    static const char* const digits = "0123456789ABCDEF";

    if (base < 2 || base > 16) return nullptr;

    int i = N - 1;
    str[i] = 0;

    do
    {
        --i;
        str[i] = digits[val % base];
        val /= base;
    }
    while (val != 0 && i > 0);

    return val == 0 ? str + i : nullptr;
}

template <int N>
const char* int_to_string(
    int val,
    char (&str)[N],
    unsigned int base = 10)
{
    // Output as unsigned.
    if (val >= 0) return uint_to_string(val, str, base);

    // Output as binary representation if base is not decimal.
    if (base != 10) return uint_to_string(val, str, base);

    // Output signed decimal representation.
    const char* res = uint_to_string(-val, str, base);

    // Buffer has place for minus sign
    if (res > str) 
    {
        const auto i = res - str - 1;
        str[i] = '-';
        return str + i;
    }
    else return nullptr;
}

Usage:

char buf[100];
terminal_writestring(int_to_string(42, buf));      // Will print '42'
terminal_writestring(int_to_string(42, buf, 2));   // Will print '101010'
terminal_writestring(int_to_string(42, buf, 8));   // Will print '52'
terminal_writestring(int_to_string(42, buf, 16));  // Will print '2A'
terminal_writestring(int_to_string(-42, buf));     // Will print '-42'
terminal_writestring(int_to_string(-42, buf, 2));  // Will print '11111111111111111111111111010110'
terminal_writestring(int_to_string(-42, buf, 8));  // Will print '37777777726'
terminal_writestring(int_to_string(-42, buf, 16)); // Will print 'FFFFFFD6'

Live example: http://cpp.sh/5ras

Answer 2

You could declare a string and get the pointer to it :

std::string str = std::to_string(count);
str += "\n";
terminal_writestring(str.c_str());