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Finding the number of digits of an integer

Asked 2011-07-11T15:05:27.337
67

What is the best method to find the number of digits of a positive integer?

I have found this 3 basic methods:

  • conversion to string

    String s = new Integer(t).toString(); 
int len = s.length();

  • for loop

    for(long long int temp = number; temp >= 1;)
{
    temp/=10;
    decimalPlaces++;
}

  • logaritmic calculation

    digits = floor( log10( number ) ) + 1;

where you can calculate log10(x) = ln(x) / ln(10) in most languages.

First I thought the string method is the dirtiest one but the more I think about it the more I think it's the fastest way. Or is it?

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2 Answers

32

Well the correct answer would be to measure it - but you should be able to make a guess about the number of CPU steps involved in converting strings and going through them looking for an end marker

Then think how many FPU operations/s your processor can do and how easy it is to calculate a single log.

edit: wasting some more time on a monday morning :-)

String s = new Integer(t).toString(); 
int len = s.length();

One of the problems with high level languages is guessing how much work the system is doing behind the scenes of an apparently simple statement. Mandatory Joel link

This statement involves allocating memory for a string, and possibly a couple of temporary copies of a string. It must parse the integer and copy the digits of it into a string, possibly having to reallocate and move the existing memory if the number is large. It might have to check a bunch of locale settings to decide if your country uses "," or ".", it might have to do a bunch of unicode conversions.
Then finding the length has to scan the entire string, again considering unicode and any local specific settings such as - are you in a right->left language?.

Alternatively:

digits = (number == 0) ? 1 : floor(log10(number)) + 1;

Just because this would be harder for you to do on paper doesn't mean it's hard for a computer! In fact a good rule in high performance computing seems to have been - if something is hard for a human (fluid dynamics, 3d rendering) it's easy for a computer, and if it's easy for a human (face recognition, detecting a voice in a noisy room) it's hard for a computer!

You can generally assume that the builtin maths functions log/sin/cos etc - have been an important part of computer design for 50years. So even if they don't map directly into a hardw

answered 2011-07-11T15:14:11.367
1

Keep it simple:

long long int a = 223452355415634664;

int x;
for (x = 1; a >= 10; x++)
{
   a = a / 10;
}

printf("%d", x);
answered 2011-07-11T17:52:35.147

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