How to ignore non-letters from std::cin - c++

Question

So, I am trying to read a string from cin, then loop through the string to count which characters in that string are actually letters in the English alphabet. I have wrote a program that works just fine, but I want to know if there is a more efficient way of doing this, without looping through the entire English alphabet.
#include <iostream>
#include <string>
using namespace std;
int main() {
string my_str; //will use to store user input
getline(cin, my_str); //read in user input to my_str
int countOfLetters = 0; //begine count at 0
string alphabet = "abcdefghijklmnopqrstuwxyz"; //the entire english alphabet
for(int i = 0; i < my_str.length(); i++){
for(int j = 0; j < alphabet.length(); j++){
if (tolower(my_str.at(i)) == alphabet.at(j)){ //tolower() function used to convert all characters from user inputted string to lower case
countOfLetters += 1;
}
}
}
cout << countOfLetters;
return 0;
}
EDIT: Here is my new and improved code:
#include <iostream>
#include <string>
using namespace std;
int main() {
string my_str; //will use to store user input
getline(cin, my_str); //read in user input to my_str
int countOfLetters = 0; //begine count at 0
string alphabet = "abcdefghijklmnopqrstuwxyz"; //the entire english alphabet
for(unsigned int i = 0; i < my_str.length(); i++){
if (isalpha(my_str.at(i))){ //tolower() function used to convert all characters from user inputted string to lower case
countOfLetters += 1;
}
}
cout << countOfLetters;
return 0;
}
enter code here

Answer 1

Use isalpha() to see which characters are letters and exclude them.
So, you could modify your code like this:
#include <iostream>
#include <string>
using namespace std;
int main() {
string my_str;
getline(cin, my_str);
int countOfLetters = 0;
for (size_t i = 0; i < my_str.length(); i++) { // int i produced a warning
if (isalpha(my_str.at(i))) { // if current character is letter
++countOfLetters; // increase counter by one
}
}
cout << countOfLetters;
return 0;
}

Answer 2

You could perhaps use isalpha:
for(int i = 0; i < my_str.length(); i++)
if (isalpha(my_str.at(i))
countOfLetters++;

Answer 3

You can use the std::count_if() algorithm along with the iterator interface to count characters for which some predicate returns true. The predicate can use std::isalpha() to check for alphabetical characters. For example:
auto count = std::count_if(std::begin(str), std::end(str),
[&] (unsigned char c) { return std::isalpha(c); });

Answer 4

You could also check if the int cast is between 65-90 or 97-122
at example
(int)'a'
Should give 97
This will be the most performant method without any doubt.
Its better than using isalpha().
Check http://www.asciitable.com/ for ASCI Numbers

Answer 5

isalpha works well for this particular problem, but there's a more general solution if the list of characters to accept wasn't so simple. For example if you wanted to add some punctuation.
std::set<char> good_chars;
good_chars.insert('a'); good_chars.insert('A');
good_chars.insert('b'); good_chars.insert('B');
// ...
good_chars.insert('z'); good_chars.insert('Z');
good_chars.insert('_');
// the above could be simplified by looping through a string of course
for(int i = 0; i < my_str.length(); i++){
countOfLetters += good_chars.count(my_str[i]);
}