Thread: itoa explanation?

im having trouble understanding the code in itoa's do while loop. why is this finding the remainder of dividing n by 10 and adding '0'? and the condition while((n/=10)!=0). whats happening here?

Code:

/* reverse:	reverse string s in place */
void reverse(char s[])
{
	int c,i,j;

	for(i = 0,j = strlen(s)-1; i < j; i++,j--)
	{
		c = s[i];
		s[i] = s[j];
		s[j] = c;
	}
}

/* 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);
}
int main()
{
	char string[10];
	int x;

	x = 30;
	itoa(x,string);
	printf("%s\n",string);
	return 0;
}

It is getting the current digit in the tens place and then adding '0' to it to get it's ascii character value. Then divide by 10 and repeat.

Your numbers are ordered like so in the ASCII set:

Code:

'0', '1', '2' ... '9'

Thus, if you have an integer, and you add the ASCII value of '0' to it, it will move numericly down the line, giving you the corresponding ASCII value.

Example: You want '3'. You start with ASCII '0' and add a decimal 3 to it. This gives you:

Code:

'0' + 0 = '0'
'0' + 1 = '1'
'0' + 2 = '2'
'0' + 3 = '3'

Take a number, and keep dividing it by ten. Eventually, your result will be zero, and you'll have run out of number.

Code:

111 / 10 = 11
11 / 10 = 1
1 / 10 = 0

When you run out of number, you're done, so you stop looping then.

Quzah.

Code:

 (n % 10) + '0' = 0 ?    /* n = 3 now */

You're mixing up division with modulus. This gives you the remainder.

Quzah.

Did you even try it?

Code:

printf("3 %% 10 is %d", (3%10) );

Don't you remember basic math? Given a number N, divided by a bigger number, you get 0 remainder N. Surely you've done long division before?

Quzah.