program that finds the # of digits of a number

[boy3005]

My teacher said there are two ways to find the # of digits of a number but i dont get either of them
what does both of the codes mean?

 log 10 ( abs(n)) + 1
if n=0 d =1
else 
d= log 10 ( abs(n))

the other one:

absn= abs(n);
d= 1;
while ( abs(n) / 10 > 10 )
d++;
absn = absn /10;
}

[Josue Molina]

In the first one, he's using the fact that the base 10 logarithm of a number besides 0 is one less than its number of digits.

In the second one, he's seeing how many times he can divide the number by 10; this is its number of digits.

Try it out and see for yourself.

Edit: here's a simple Python snippet that demonstrates this: http://ideone.com/iC8sPR

[giblit]

Those ways look a bit complicated to find the number of digits a number has...

while( number > 0 )
{
    number /= 10;
    ++count;
}

http://ideone.com/CWfvZQ