How would you know the length of a long long

Question

I want to know how to get the length of a long long. I have a simple function where I am trying to get the length of a long long and I also get an error.

while (numberArray[i].length() > MAX_INPUT)
{
    cout << "ERROR: Program can only take in a number length " << MAX_INPUT << " or less. Try again ->";

    cin >> numberArray[i];
}

NOTE: numberArray is a static that was made an int. The call to .length() worked perfectly but I wanted to change the type to long long and now .length() does not seem to work anymore.

This is the type of message I get from the compiler every time I try and run the code:

In function 'void takeNumbers(std::__cxx11::string&, long long int (&)[11], int)':
main.cpp:48:26: error: request for member 'length' in 'numberArray[i]', which is of non-class type 'long long int'
    while (numberArray[i].length() > MAX_INPUT){

You can't just get the length of an integer like that in C++. You could convert the number to a string, and then take its length, using [`std::to_string`](http://en.cppreference.com/w/cpp/string/basic_string/to_string). — Arnav Borborah, Feb 28 '18 at 13:49
In C++ integers can't have member functions, because they are not classes. What do you think that `length()` function should do? What is the "length" of a integer? Are you trying to find the maximum value of a long long? — Jonathan Wakely, Feb 28 '18 at 13:49
N.B. there is no way it worked when the type was `int`, because you can't call a member function on an `int` either, it's not a class so doesn't have member functions. — Jonathan Wakely, Feb 28 '18 at 13:51
"Length" is an odd term to use for a number. Any number can be expressed in C++ in a decimal, binary, hexadecimal, or octal representation. I'm guessing by "length," you mean the number of digits in the decimal representation? — Mike Borkland, Feb 28 '18 at 13:59

Arnav Borborah · Accepted Answer · 2018-02-28T14:02:14.210

3

In C++, plain old data types such as long long int do not have member functions for obtaining their length. If you want the length of an integer, you can convert the number to a string, and then take it's length. For example:

while (std::to_string(numberArray[i]).length() > MAX_INPUT)

This makes use of the std::to_string function which is used for converting non-string POD types to the std::string class.

But suppose you needed to handle negative integers as well. The above wouldn't work since the leading minus sign would be counted as part of the length (Unless you want that). Another approach would be to use a loop along with division to find the length of an integer and place this loop in a function, as follows:

std::size_t numDigits(long long int n)
{
    std::size_t length = 0;

    do
    {
        ++length; 
        n /= 10;
    } while (n);

    return length;
}

Then in your loop:

while (numDigits(numberArray[i]) > MAX_INPUT)

Depending on your goal, you may pick either approach.

edited Feb 28 '18 at 14:02

answered Feb 28 '18 at 13:52

Arnav Borborah

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The function should be called `numDigits` or `numDecimalDigits` or something. An integer doesn't have a "length". – Jonathan Wakely Feb 28 '18 at 13:59
@JonathanWakely I have followed your suggestion. – Arnav Borborah Feb 28 '18 at 14:00
Also you forgot to account for a `-` sign. – Jonathan Wakely Feb 28 '18 at 14:00
1

@JonathanWakely That was exactly the reason that I suggested the second approach, I just forgot to list the reason. – Arnav Borborah Feb 28 '18 at 14:02

Bathsheba · Answer 2 · 2018-02-28T14:14:28.087

You'll need to roll your own function: the length of a number is arbitrary in that it assumes a radix, and assumptions need to be made about the treatment of non positive numbers. As a starting point, consider

unsigned length(unsigned long long n, unsigned radix = 10)
{
    unsigned l;
    for (l = 0; n; n /= radix, ++l);
    return l;
}

where you can extend this for negative numbers (should they increase the length by one?) to suit personal taste.

Note that this will 0 for the number 0: that's how I think of the length of 0, but that's not to everyone's taste.

score 0 · Answer 3 · answered Feb 28 '18 at 14:27

Just for completeness, another approach could be using a sort of lookup table as the following:

unsigned num_digits(long long x)  
{  
    x = std::abs(x);      
    if(x < 10)
       return 1;
    else if(x < 100)
       return 2;
    else if(x < 1000)
       return 3;
    //keep going ....
}

How would you know the length of a long long

3 Answers3