ft_itoa

Info

The function converts an integer into its string representation. It handles both positive and negative numbers, allocates memory for the resulting string (including the null terminator), and returns the string.

Walkthrough

Here the logic is pretty straightforward. Our steps should be:

1. get the len of the integer N
   • +1 if N=0, still the len of 0 is 1
   • +1 if N<0, space for the minus sign.
2. allocate memory for string “result” → (len+1)*sizeof(char)
   • memory should be initialized, since we should loop until the negative sign - when filling the result; garbage values may interfere. Therefore I’ll use ft_calloc().
3. (loop) assign last digit of (N%10)+'0' to result[len--]

Number Length

static int	numlen(int n)
{
	int	length;
 
	length = 0;
	if (n <= 0)
	{
		// if N is 0 -> make it 1
		// if N is negative -> we need +1 byte for the sign
		length++;
		n = -n;
	}
	while (n != 0)
	{
		// count and substract each digit of N
		n /= 10;
		length++;
	}
	return (length);
}

Memory Allocation

char	*ft_itoa(int n)
{
	int		len;
	char	*result;
	long	num;
 
	num = n;
	len = numlen(n);
	result = (char *)ft_calloc(sizeof(char), (len + 1));
	if (!result)
		return (NULL);
	// With num len known, add null terminator directly
	result[len] = '\0';

Why we are using long for num, even though we accept int as a function parameter?

Let me explain:

• The range of int is from -2147483648 to 2147483647.
• When n is -2147483648 (which is INT_MIN), converting it to positive would exceed the range of int because 2147483648 is out of the intrange, therefore it would cause an Integer Overflow
• By using long, which has a larger range, we can safely handle this conversion.

Filling the Result

// ...
 
char	*ft_itoa(int n)
{
	int		len;
	char	*result;
	long	num;
 
	num = n;
	len = numlen(n);
	result = (char *)ft_calloc(sizeof(char), (len + 1));
	if (!result)
		return (NULL);
	// With num len known, add null terminator directly
	result[len] = '\0';
	if (num < 0)
	{
	    result[0] = '-'; // Add '-' at the start for negative numbers
	    num = -num;      // Convert num to positive
	}
	while (--len >= 0 && result[len] != '-')
	{
	    result[len] = (num % 10) + '0'; // Add digits from the end
	    num /= 10;
	}
	return result;
}

Diagram