Determine the size of long

[peter nn]

long tmp = 1;
int size_of_tmp_long_variable would be 1

long tmp = 12;
int size_of_tmp_long_variable would be 2

long tmp = 123;
int size_of_tmp_long_variable would be 3

Is there C++ function for that?

Best regards,
Peter

[wolfgang]

use cmath and log10.

#include <cmath>

// inside function:
long num = 29;
int length = (num == 0) ? 1 : log10(num) + 1;

This has flaws. No negative numbers, and it doesn't work with decimals.

[hanst99]

The size of a long always remains the same as long as you compile it the same way. You can use the sizeof operator to get the size of a datatype (or variable).

Oh, you meant like the number of characters? Sorry, misunderstood you. Like above.

[coder777]

you can convert it to a string (like itoa) and use the size of that string

[z05DSL3A]

#include <iostream>
#include <cmath>
#include <sstream>

int digit_count(long int number)
{
    if (number == 0) return 1;
    int count = 0;
    while( number != 0) 
    {
        number /= 10;
        ++count;
    }
    return count;
}

int digit_count_Wolfgang(long int number)
{
    return (number == 0)? 1: (int) log10((double)abs(number)) +1;;
}

int digit_count_coder777(long int number)
{
    std::stringstream stream;
    stream << abs(number);
    return stream.str().size();
}

int main ()
{
    long a_number =  -123456789;
    int number_of_digits = digit_count(a_number);
    int number_of_digits2 = digit_count_Wolfgang(a_number);
    int number_of_digits3 = digit_count_coder777(a_number);

    std::cout << "The number of digits in " << a_number << " is " << number_of_digits << std::endl;
    std::cout << "The number of digits in " << a_number << " is " << number_of_digits2 << std::endl;
    std::cout << "The number of digits in " << a_number << " is " << number_of_digits3 << std::endl;
    
    return 0;
}

[packetpirate]

This is the simplest way I can think of:

#include <iostream>

int main(int argc, char* argv[])
{
int x = 5467;
int size = sizeof(x) / sizeof(int);
std::cout << "x is " << size << " long." << std::endl;
return 0;
}

Size will first get the size of x, which is 32 bits, then divide by the size of an integer, which is 8. So the size will be 4.

[Bazzy]

sizeof(int)/sizeof(int) will always be 1...

[z05DSL3A]

packetpirate,

Although it is badly worded, it would appear that the OP wants to know how many digits there are in the number held in a particular long variable.

As Bazzy says your code would always give 1 for the size.

[packetpirate]

Read it again...
It does not say:

sizeof(int) / sizeof(int);

It says:

sizeof(x) / sizeof(int);

[Bazzy]

x is an int

[packetpirate]

Nevermind. I read that wrong.
I guess that method only works with arrays.

[moorecm]

int x = 123;

int digits = 0;
while( x > 0 ) {
    ++digits;
    x /= 10;
}

cout << digits << endl;

(untested...)

EDIT: Already implemented above by Grey Wolf.

[PanGalactic]

Here's my completely inappropriate for the beginner forum answer, using the Boost Conversion library:
size_t size_of_tmp_long_variable = boost::lexical_cast<std::string>(abs(tmp)).size();

http://www.boost.org/doc/libs/1_43_0/libs/conversion/lexical_cast.htm

[packetpirate]

@PanGalactic

My brain just exploded...

[PanGalactic]

@packetpirate

The brief explanation:

A lexical_cast is a conversion from numeric to string representation or vice versa.

boost::lexical_cast<std::string> // Convert the next parameter (see next line) to a string.
(abs(tmp))                       // The absolute value of the tmp numeric variable (same that everyone else is doing).
.size()                          // Return the size of the string. (The string itself is discarded.) 

It is doing essentially what Grey Wolf's digit_count_coder777() function does but using a lexical_cast instead of a stringstream.