Get number into a string C without stdio

Question

I want to know how to get number to string without standard C or C++ functions, for example:

char str[20];
int num = 1234;
// How to convert that number to strin         


        

Answer 1

A simplistic way to do this is to leave a lot of leading zeros. I like it because it uses only basic code, and doesn't require any dynamic memory allocation. It should consequently also be very fast:

char * convertToString(int num, str) {
    int val;

    val = num / 1000000000; str[0] = '0' + val; num -= val * 1000000000;
    val = num / 100000000;  str[1] = '0' + val; num -= val * 100000000;
    val = num / 10000000;   str[2] = '0' + val; num -= val * 10000000;
    val = num / 1000000;    str[3] = '0' + val; num -= val * 1000000;
    val = num / 100000;     str[4] = '0' + val; num -= val * 100000;
    val = num / 10000;      str[5] = '0' + val; num -= val * 10000;
    val = num / 1000;       str[6] = '0' + val; num -= val * 1000;
    val = num / 100;        str[7] = '0' + val; num -= val * 100;
    val = num / 10;         str[8] = '0' + val; num -= val * 10;
    val = num;              str[9] = '0' + val;
                            str[10] = '\0';

    return str;
}

Of course, there are tons of tweaks you could do to this - modifying the way the destination array gets created is possible, as is adding a boolean that says to trim leading 0s. And we could make this much more efficient using a loop. Here's in improved method:

void convertToStringFancier(int num, char * returnArrayAtLeast11Bytes, bool trimLeadingZeros) {
    int divisor = 1000000000;
    char str[11];
    int i;
    int val;

    for (i = 0; i < 10; ++i, divisor /= 10) {
        val = num / divisor;
        str[i] = '0' + val;
        num -= val * divisor;
    }
    str[i] = '\0';

    // Note that everything below here is just to get rid of the leading zeros and copy the array, which is longer than the actual number conversion.
    char * ptr = str;
    if (trimLeadingZeros) {
        while (*ptr == '0') { ++ptr; }
        if (*ptr == '\0') { // handle special case when the input was 0
            *(--ptr) = '0';
    }
    for (i = 0; i < 10 && *ptr != '\0'; ++i) {
    while (*ptr != '\0') {
        returnArrayAtLeast11Bytes[i] = *ptr;
    }
    returnArrayAtLeast11Bytes[i] = '\0';
}