Converting a string to an integer

[BeeOnee]

Trying to convert a username into how many characters it is (ie the name Bob = 3) to do further calculation further down to multiply by age. I feel Im so close

  #include <iostream>
#include <string>

using namespace std;
int main(){
    string fullname;
    int a = fullname.length();
    int age;
    int enrol;
    int pin;
    cout <<"Welcome to the our App\n";
    cout <<"Please enter your full name:\n";
    cin >> fullname;
    cout <<"Please enter your age:\n";
    cin >> age;
    cout <<"Please enter your enrolment number:\n";
    cin >> enrol;
    cout <<"Generating pin...\n";
    pin = (a*age)*enrol;
    cout <<"Hi " << fullname << ". Your account " << enrol << " has been created.\nYour unique pin number is " << pin;

    return 0;

[Repeater]

 string fullname;
int a = fullname.length();

I see that at this point in the code, the string is empty. Of size zero. So a will always be zero.

So you might as well just write

int a = 0;

Did you mean for a to always be zero? If not, I suspect you need to think again about this part of the code.

[BeeOnee]

I see.
No i didnt mean for it to be zero, it has to have full name put through

I have now moved the line to later on in the code.

Just trying to work out out to count 2 or 3 words if I have a fullname and ignore spaces

  cout <<"Please enter your first name:\n";
    cin >> firstname;
    int fname = firstname.length();

[seeplus]

Why not use getline(cin, fullname) instead of cin >> to allow the use to enter their full name including spaces?

[BeeOnee]

So basically the question/ input is asking for the users full name and find the length of name so if name entered is John my code converts that to integer 4 to be multiplied by age input.

Im having trouble when users name is John Doe this should output 7 (ignoring the space) so while trying to get this to work I was cheating a little bit and asking for first name first, then asking for surname and adding together

But I also need it to work if someone had 3 names or a double barrelled name etc.

so your code helps but it includes the space which I dont want

Edit:
I have it working with this to subtract spaces but doesnt work with characters (John Bob-Doe returns 11, counting the - which im not to bothered with at the moment)

 cout <<"Please enter your full name:\n";
    getline(cin, fullname);
    //cin >> firstname;
    fullname.erase(remove(fullname.begin(), fullname.end(), ' '), fullname.end());
    int fname = fullname.length();

[George P]

The <cctype> library has isalpha(), useful on individual elements of a string.

https://en.cppreference.com/w/cpp/string/byte/isalpha

#include <iostream>
#include <string>
#include <cctype>

int CountLetters(const std::string&);

int main()
{
   std::string str1 { "John Doe" };
   std::string str2 { "John Bob-Doe" };

   std::cout << "\'" << str1 << "\' has " << CountLetters(str1) << " letters.\n";

   std::cout << "\'" << str2 << "\' has " << CountLetters(str2) << " letters.\n";
}

int CountLetters(const std::string& str)
{
   int num_letters { };

   for (const auto& itr : str)
   {
      if (::isalpha(itr))
      {
         num_letters++;
      }
   }

   return num_letters;
}

'John Doe' has 7 letters.
'John Bob-Doe' has 10 letters.

[seeplus]

Do you mean something like this which only counts alpha chars:

#include <iostream>
#include <string>
#include <algorithm>
#include <cstring>
using namespace std;

int main()
{
	std::string fullname;

	cout << "Please enter your full name: ";
	getline(cin, fullname);

	fullname.erase(remove_if(fullname.begin(), fullname.end(), [](unsigned char c) {return !std::isalpha(c); }), fullname.end());

	const auto namsze{ fullname.length() };

	cout << namsze << '\n';
}

Please enter your full name: John Bob-Doe
10

[BeeOnee]

Thanks guys, this helps alot.

i was looking more for Seeplus answer, but knowing it was alpha chars I needed was the key.
I dont think Ill forget that, Thanks

[Duthomhas]

You don’t need to erase anything, just count the things that matter.

#include <algorithm>
#include <ciso646>
#include <cctype>
#include <string>

int main()
{
  std::string fullname;
  std::cout << "Please enter your full name: ";
  getline( std::cin, fullname );

  auto name_length = std::count_if( fullname.begin(), fullname.end(), []( char c )
  {
    return !(c & 0x80) and !std::isspace( c );
  } );

  std::cout << "You have " << name_length << " letters in your name.\n";
}

The weird []( char c ) { ... } thing is a lambda — I could have used a normal function instead but this just keeps all relevant code in one spot.

The !(c & 0x80) clause is for UTF-8 awareness — making your non-whitespace letter counting thing able to count a multibyte UTF-8 sequence as a single code-point value (which is how it is displayed, even though it is made up of multiple characters).

If that doesn’t matter to you, just leave it out:

[]( char c )
{
  return !std::isspace( c );
}

Hope this helps.