C++ - How to Find the Length of an Integer

Finding the length of an integer in C

C:

Why not just take the base-10 log of the absolute value of the number, round it down, and add one? This works for positive and negative numbers that aren't 0, and avoids having to use any string conversion functions.

The log10, abs, and floor functions are provided by math.h. For example:

int nDigits = floor(log10(abs(the_integer))) + 1;

You should wrap this in a clause ensuring that the_integer != 0, since log10(0) returns -HUGE_VAL according to man 3 log.

Additionally, you may want to add one to the final result if the input is negative, if you're interested in the length of the number including its negative sign.

Java:

int nDigits = Math.floor(Math.log10(Math.abs(the_integer))) + 1;

N.B. The floating-point nature of the calculations involved in this method may cause it to be slower than a more direct approach. See the comments for Kangkan's answer for some discussion of efficiency.

How do I determine the number of digits of an integer in C?

floor (log10 (abs (x))) + 1

http://en.wikipedia.org/wiki/Logarithm

How to find the length(number of digits)of a long in C?

The below program will help you, but its bettor to give some effort

     #include <stdio.h>
     #include <cs50.h>

        //prompt user for numb er

    int main(void)
    {
        long number = get_long("Enter card number: "); 
        int count = 0; 
        do{ 
            number = number / 10; 
            ++count; 
        }while (number != 0) ;

        printf("length of number = %d",count);

    }

C++ - how to find the length of an integer

The number of digits of an integer n in any base is trivially obtained by dividing until you're done:

unsigned int number_of_digits = 0;

do {
     ++number_of_digits; 
     n /= base;
} while (n);

How do I determine the size of my array in C?

Executive summary:

int a[17];
size_t n = sizeof(a)/sizeof(a[0]);

Full answer:

To determine the size of your array in bytes, you can use the sizeof
operator:

int a[17];
size_t n = sizeof(a);

On my computer, ints are 4 bytes long, so n is 68.

To determine the number of elements in the array, we can divide
the total size of the array by the size of the array element.
You could do this with the type, like this:

int a[17];
size_t n = sizeof(a) / sizeof(int);

and get the proper answer (68 / 4 = 17), but if the type of
a changed you would have a nasty bug if you forgot to change
the sizeof(int) as well.

So the preferred divisor is sizeof(a[0]) or the equivalent sizeof(*a), the size of the first element of the array.

int a[17];
size_t n = sizeof(a) / sizeof(a[0]);

Another advantage is that you can now easily parameterize
the array name in a macro and get:

#define NELEMS(x)  (sizeof(x) / sizeof((x)[0]))

int a[17];
size_t n = NELEMS(a);

Finding the number of digits of an integer

There's always this method:

n = 1;
if ( i >= 100000000 ) { n += 8; i /= 100000000; }
if ( i >= 10000     ) { n += 4; i /= 10000; }
if ( i >= 100       ) { n += 2; i /= 100; }
if ( i >= 10        ) { n += 1; }