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cdo-loopsitoa

Trying to understand itoa function

I tried to re-write the itoa() function from K&R exercises, but I failed to define it. I see the answer of the function in the library , but I can't understand what is inside the do block. Please explain it to me. Thanks!

 /* itoa:  convert n to characters in s */
 void itoa(int n, char s[])
 {
     int i, sign;

     if ((sign = n) < 0)  /* record sign */
         n = -n;          /* make n positive */
     i = 0;
     do {       /* generate digits in reverse order */
         s[i++] = n % 10 + '0';   /* get next digit */
     } while ((n /= 10) > 0);     /* delete it */
     if (sign < 0)
         s[i++] = '-';
     s[i] = '\0';
     reverse(s);
 }
Solution

  • s[i++] = n % 10 + '0'; Means:

    1. s[i++] means char number i in char array s and increment i by 1
    2. n % 10 means that you take only the last digit, for example in 123, 123 % 10 returns 3 which is a digit, this is done to cut your number into digits so you can create a char with each of them.
    3. + '0' means that you add the ascii value of the char '0', for example the int 1 + '0' returns the character '1' so you can get a char array.
    4. n /= 10 means that we remove the last digit, indeed, we added it to our char array so we can remove it, for example 123 / 10 = 12 then you can do 12 % 10 to get the second digit: 2

    This gives us a char array inverted, for example for 123 we got an array like {'3', '2', '1'} so at the end we call reverse s to (thx captain obvious) reverse our char array

    For a negative n you can just add a '-' to your char array and multiply your n by -1 so it become positive and you can do your job as always :)

    Hope it helps you :)