You tie it up
First you take to pieces of string ( make sure they are both flat ) and cross them like an x.start braiding.pull your finger after you have made the first stiching.pull loose ends.
To create a slipknot when braiding, start by forming a loop with your working end of the rope or cord. Then, take the working end and wrap it around the standing part, passing it through the loop you created. Pull the knot tight while adjusting the size of the loop as needed. This knot will allow for easy adjustments in tension when braiding.
you start with the A string and follow to the other strings
first string are the better players they start, second string backs them up.
1880 and continued until 1910.
10
A good start is to watch how to video's on youtube.
substr(string, position [, count]) It extract substring starting from start and going for count characters. If count is not specified, the string is clipped from the start till the end
#include<iostream> #include<vector> #include<string> std::vector<std::string> parse (const std::string& s, const char delim) { std::vector<std::string> result {}; auto start = 0U; auto end = s.find (delim); while (end != s.npos) { result.push_back (s.substr(start, end - start)); start = ++end; end = s.find (delim, start); } result.push_back (s.substr (start, s.npos - start)); return result; } std::vector<std::string> parse (const std::string& s, const std::string& delim) { std::vector<std::string> result {}; auto start = 0U; auto end = s.find (delim); while (end != s.npos) { result.push_back (s.substr(start, end - start)); start = end + delim.length(); end = s.find (delim, start); } result.push_back (s.substr (start, s.npos - start)); return result; } int main() { std::string str1 = "This is a string that will be parsed by a single-space delimiter."; std::string str2 = "This==is==a==string==that==will==be==parsed==by==equal==operator."; std::string str3 = "This string has no delimiter."; std::cout << str1 << std::endl; std::vector<std::string> v1 = parse (str1, ' '); for (auto i : v1 ) std::cout << i << std::endl; std::cout << std::endl; std::cout << str2 << std::endl; std::vector<std::string> v2 = parse (str2, "=="); for (auto i : v2 ) std::cout << i << std::endl; std::cout << std::endl; std::cout << str3 << std::endl; std::vector<std::string> v3 = parse (str3, '\\'); for (auto i : v3 ) std::cout << i << std::endl; std::cout << std::endl; }
you die
there is no such thing as a 1 string lanyard, only two or more, therefore, you can't start one.- Matthew Smith
It is best to start a child off on a string instrument. Stings are harder to play than brass instruments so, mastering a string early is the best bet.