Efficient way to determine number of digits in an integer

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What is a very efficient way of determining how many digits there are in an integer in C++?

Well, the most efficient way, presuming you know the size of the integer, would be a lookup. Should be faster than the much shorter logarithm based approach. If you don't care about counting the '-', remove the + 1.

// generic solution
template <class T>
int numDigits(T number)
{
    int digits = 0;
    if (number < 0) digits = 1; // remove this line if '-' counts as a digit
    while (number) {
        number /= 10;
        digits++;
    }
    return digits;
}

// partial specialization optimization for 32-bit numbers
template<>
int numDigits(int32_t x)
{
    if (x == MIN_INT) return 10 + 1;
    if (x < 0) return numDigits(-x) + 1;

    if (x >= 10000) {
        if (x >= 10000000) {
            if (x >= 100000000) {
                if (x >= 1000000000)
                    return 10;
                return 9;
            }
            return 8;
        }
        if (x >= 100000) {
            if (x >= 1000000)
                return 7;
            return 6;
        }
        return 5;
    }
    if (x >= 100) {
        if (x >= 1000)
            return 4;
        return 3;
    }
    if (x >= 10)
        return 2;
    return 1;
}

// partial-specialization optimization for 8-bit numbers
template <>
int numDigits(char n)
{
    // if you have the time, replace this with a static initialization to avoid
    // the initial overhead & unnecessary branch
    static char x[256] = {0};
    if (x[0] == 0) {
        for (char c = 1; c != 0; c++)
            x[c] = numDigits((int32_t)c);
        x[0] = 1;
    }
    return x[n];
}

Efficient way to determine number of digits in an integer, Program to count digits in an integer (4 Different Methods) into string and then find the length of the string to get the number of digits in the original number. Perhaps the easiest way of getting the number of digits in an Integer is by converting it to String, and calling the length () method. This will return the length of the String representation of our number: 1 int length = String.valueOf (number).length ();

The simplest way is to do:

unsigned GetNumberOfDigits (unsigned i)
{
    return i > 0 ? (int) log10 ((double) i) + 1 : 1;
}

log10 is defined in <cmath> or <math.h>. You'd need to profile this to see if it's faster than any of the others posted here. I'm not sure how robust this is with regards to float point precision. Also, the argument is unsigned as negative values and log don't really mix.

Program to count digits in an integer (4 Different Methods , Learn different ways of getting the number of digits in an Integer in Java. different methods and will figure out which algorithm would best fit in  What is a very efficient way of determining how many digits there are in an integer in C++?Well, the most efficient way, presuming you know the size of the integer, would be a lookup. Should be faster than the much shorter logarithm based approach.

Perhaps I misunderstood the question but doesn't this do it?

int NumDigits(int x)  
{  
    x = abs(x);  
    return (x < 10 ? 1 :   
        (x < 100 ? 2 :   
        (x < 1000 ? 3 :   
        (x < 10000 ? 4 :   
        (x < 100000 ? 5 :   
        (x < 1000000 ? 6 :   
        (x < 10000000 ? 7 :  
        (x < 100000000 ? 8 :  
        (x < 1000000000 ? 9 :  
        10)))))))));  
}  

Number of Digits in an Integer in Java, Some problems ask you to find the number of digits of an integer or variable. For example, the number 23 has two digits, and the number 1337 has four. But how  The first method, GetIntegerDigitCountString, converts the integer to a string and then accesses the Length property on the string.String Length. And: The second method, GetIntegerDigitCount, uses logic to determine how many characters the integer would have if it were a string. Note: Both methods count the sign as a digit (character). So -1 is considered to be 2 digits.

int digits = 0; while (number != 0) { number /= 10; digits++; }

Note: "0" will have 0 digits! If you need 0 to appear to have 1 digit, use:

int digits = 0; do { number /= 10; digits++; } while (number != 0);

(Thanks Kevin Fegan)

In the end, use a profiler to know which of all the answers here will be faster on your machine...

Finding The Number of Digits, C++ program to count number of digits in a given integer. March 10 If you divide any number by 10 and store the result in an integer then it strips the last digit. So we C++ program to find product of digits in a number We use cookies to give you the best experience, as detailed in our privacy policy. Some problems ask you to find the number of digits of an integer or variable. For example, the number 23 has two digits, and the number 1337 has four. But how many digits does an integer x x x have, if x 2 x^2 x 2 has more than 2 but less than 4 digits?

Practical joke: This is the most efficient way (number of digits is calculated at compile-time):

template <unsigned long long N, size_t base=10>
struct numberlength
{
    enum { value = 1 + numberlength<N/base, base>::value };
};

template <size_t base>
struct numberlength<0, base>
{
    enum { value = 0 };
};

May be useful to determine the width required for number field in formatting, input elements etc.

C++ program to count number of digits in a given integer, In this example, you will learn to count the number of digits in an integer entered by the user with output and explanation The integer entered by the user is stored in variable n. Then the while loop is iterated until the test expression n != 0 is evaluated to 0 (false). After first iteration, the value of n will be 345 and the count is incremented to 1. After second iteration, the value of n will be 34 and the count is incremented to 2.

C Program to Count Number of Digits in an Integer, Example 1: Count Number of Digits in an Integer using while loop. public class NumberDigits { public static void main(String[] args) { int count = 0, num = 3452;  int length = 1; int x = 234567545; while ( x /= 10 ) length++; or use the log10 method from <math.h> . Note that log10 returns a double , so you'll have to adjust the result.

Java Program to Count Number of Digits in an Integer, How to find total digits in a given integer using loop in C programming. Logic to count digits in a given integer without using loop in C program. The beauty of this solution is the number of times it loops is equal to the number of set bits in a given integer. 1 Initialize count: = 0 2 If integer n is not zero (a) Do bitwise & with (n-1) and assign the value back to n n: = n&(n-1) (b) Increment count by 1 (c) go to step 2 3 Else return count

C program to count number of digits in an integer, We will show you two different ways to calculate total digits in a number. This method returns a string, we are wrapping this value with int() to get the integer value of the number user has entered. Solution 2 ( Using inbuilt methods ) :. Except for 0 and 1, the binary representation of an integer has more digits than its decimal counterpart. To find the number of binary digits (bits) corresponding to any given decimal integer, you could convert the decimal number to binary and count the bits. For example, the two-digit decimal integer 29 converts to the five-digit binary

Comments
  • In what base? 2? 10?
  • I would like to do it in base 10
  • I once asked a related question: How can you get the first digit in an int? Many of the same methodologies as below were used in people's answers. Here's the link in case it's relevant to your task [stackoverflow.com/questions/701322/]
  • Does inline assembly qualify?
  • While all these answers are in terms of base 10, it is pretty easy to change to compute the result for any desired base.
  • Probably faster than my answer, well done. For added efficiency, if you know that your input numbers will be mostly small ones (I'm guessing less than 100,000), then reverse the tests: if (x < 10) return 1; if (x < 100) return 2; etc., so that the function will do less tests and exit faster.
  • Or perhaps reorder and nest the if statements, to do a binary search instead of a linear search.
  • That's not a good idea. What happens when the architecture expands to 256 bit integers. You need to remember to come back and modify this code. In real life that will not happen and sice this is probably going to be used to build a buffer of the correct size you are now opening yourself to all sorts of buffer over run problems on larger architectres.
  • assuming a uniform distribution of numbers, the reverse linear search ( starting from max digits to 1 ) may be faster on average than the binary search as there are quite more numbers with N digits than with N-1 digits graphics.stanford.edu/~seander/…
  • I wouldn't worry very hard about 256 or 128 bit integers. Unless you need to count the number of electrons in the Universe (10^78 last time I did it), 64 bits will do pretty well. 32 bit machines lasted ~~15 years. I'd guess 64 bit machines will last a lot longer. For larger number, multiprecision arithmetic will be fine, and I doubt if efficiency of computing digit count will matter.
  • For 32 bit ints and 56 bit floats, this probably works. If the input is a long (64 bits), the 56 bits of double-precision log may cause this to produce the wrong answer in cases of values near large values of 10^n. Expect trouble above 2^50.
  • There's also the question of how accurate the log functions are. I haven't checked how accurate they are in modern libraries, and wouldn't be comfortable blindly trusting them to be good to one part in a billion.
  • @DavidThornley: log or other math functions are perfectly precise unless specified on the compiler command line. some will be converted to x86 intrinsics at compile time. some don't exist and will expand into formulas of existing intrinsics. for example if using -fpfast you could see usage of SSE instrinsics rather than x87, which yields less guarantee on the precision IIRC. but by default no problem.
  • @DavidThornley: It's more than precision. The question is whether it is guaranteed or not that log10 (10^k) ≥ k for all relevant k. That is is it guaranteed that any inevitable rounding error goes in the right direction. k + eps as a result works, k - eps doesn't. And "Perfectly precise" is naïve.