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

Convert it char by char, e.g. the last char of the string is '4', the previous one is '3' and so on. Use math to determine the chars, it might be easier to create "4321" string and then rotate it.

Answer 2

An "after accepted answer" that works for all int including INT_MIN.

static char *intTostring_helper(int i, char *s) {
  if (i < -9) {
    s = intTostring_helper(i/10, s);
  }
  *s++ = (-(i%10)) + '0' ;
  return s;
}

char *intTostring(int i, char *dest) {
  char *s = dest;
  if (i < 0) {  // If non 2s compliment, change to some IsSignBitSet() function.
    *s++ = '-';
  }
  else {
    i = -i;
  }
  s = intTostring_helper(i, s);
  *s = '\0';
  return dest;
}

Answer 3

To get the lowest digit, use num % 10. To convert a digit to a character, add '0'. To remove the lowest digit after you've handled it, divide by 10: num /= 10;. Repeat until done.

Answer 4

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';
}

Answer 5

Using C (not C++)

Assuming you're preallocating your buffer for str as in your question:

char *itostr(int num, char *str) {
    int len = 1;
    long tmp = num;
    int sign = num < 0;
    if (sign) {
        str[0] = '-';
        tmp = -tmp;
    }
    while (num/=10) ++len;
    str[len+sign] = 0;
    while (len--) {
        str[len+sign] = '0'+tmp%10;
        tmp /= 10;
    }
    return str;
}