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Number Of Digits

On the previous page we talked about assignments like i = i - 1, x = x + 10, or a = a * 5. Let's look at one specific assignment of this kind:

int x = 1234;

x = x / 10;
x = x / 10;
x = x / 10;
x = x / 10;

What would the value of x be after these divisions?

1234 is not divisible by 10, so if we divided them as real numbers, we would've got 123.4; but we only use the int whole numbers type for now, so the integer division will drop the fractional part, giving us 123 as a result. The remaining 4 can be found by calculating the remainder: 1234 % 10 will be 4.

So, 1234 / 10 is 123, so after x = x / 10 the value of x will become 123.

Do it once more, x = x / 10, and x will become 12.

Once more, and it's 1.

Finally, since 1 / 10 is 0, after one more x = x / 10 it will become zero.

Let's solve the problem where you will need to use a while loop (even though later we'll see how we can do it without any loops! – but for now, let's use it).

Write a function which returns number of digits in a number x.

#include <stdio.h>

int number_of_digits(int x) {
	/* your code goes here */
	return 0;
}

int main() {
	printf("Number of digits in 1 is %d\n",     number_of_digits(1));
	printf("Number of digits in 9 is %d\n",     number_of_digits(9));
	printf("Number of digits in 10 is %d\n",    number_of_digits(10));
	printf("Number of digits in 99 is %d\n",    number_of_digits(99));
	printf("Number of digits in 100 is %d\n",   number_of_digits(100));
	printf("Number of digits in 999 is %d\n",   number_of_digits(999));
	printf("Number of digits in 1000 is %d\n",  number_of_digits(1000));
	printf("Number of digits in 9999 is %d\n",  number_of_digits(9999));
	printf("Number of digits in 10000 is %d\n", number_of_digits(10000));
	printf("Number of digits in 99999 is %d\n", number_of_digits(99999));
	printf("Number of digits in 0 is %d\n",     number_of_digits(0));

	return 0;
}

If you feel stuck, go back to the previous page, read about the counters (c = c + 1), and see if you could use one here!

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