How to get the tail of a std::string? - c++

Question

How to retrieve the tail of a std::string?
If wishes could come true, it would work like that:
string tailString = sourceString.right(6);
But this seems to be too easy, and doesn't work...
Any nice solution available?
Optional question: How to do it with the Boost string algorithm library?
ADDED:
The method should be save even if the original string is smaller than 6 chars.

Answer 1

There is one caveat to be aware of: if substr is called with a position past the end of the array (superior to the size), then an out_of_range exception is thrown.
Therefore:
std::string tail(std::string const& source, size_t const length) {
if (length >= source.size()) { return source; }
return source.substr(source.size() - length);
} // tail
You can use it as:
std::string t = tail(source, 6);

Answer 2

Using the substr() method and the size() of the string, simply get the last part of it:
string tail = source.substr(source.size() - 6);
For handling case of a string smaller than the tail size see Benoit's answer (and upvote it, I don't see why I get 7 upvotes while Benoit provides a more complete answer!)

Answer 3

You could do:
std::string tailString = sourceString.substr((sourceString.length() >= 6 ? sourceString.length()-6 : 0), std::string::npos);
Note that npos is the default argument, and might be omitted. If your string has a size that 6 exceeds, then this routine will extract the whole string.

Answer 4

This should do it:
string str("This is a test");
string sub = str.substr(std::max<int>(str.size()-6,0), str.size());
or even shorter, since subst has string end as default for second parameter:
string str("This is a test");
string sub = str.substr(std::max<int>(str.size()-6,0));

Answer 5

You can use iterators to do this:
#include <iostream>
#include <string>
using namespace std;
int main ()
{
char *line = "short line for testing";
// 1 - start iterator
// 2 - end iterator
string temp(line);
if (temp.length() >= 8) { // probably want at least one or two chars
// otherwise exception is thrown
int cut_len = temp.length()-6;
string cut (temp.begin()+cut_len,temp.end());
cout << "cut is: " << cut << endl;
} else {
cout << "Nothing to cut!" << endl;
}
return 0;
}
Output:
cut is: esting

Answer 6

Since you also asked for a solution using the boost library:
#include "boost/algorithm/string/find.hpp"
std::string tail(std::string const& source, size_t const length)
{
boost::iterator_range<std::string::const_iterator> tailIt = boost::algorithm::find_tail(source, length);
return std::string(tailIt.begin(), tailIt.end());
}

Answer 7

Try substr method.

Answer 8

I think, using iterators is C++ way
Something like that:
#include <string>
#include <iostream>
#include <algorithm>
#include <iterator>
using namespace std;
std::string tail(const std::string& str, size_t length){
string s_tail;
if(length < str.size()){
std::reverse_copy(str.rbegin(), str.rbegin() + length, std::back_inserter(s_tail));
}
return s_tail;
}
int main(int argc, char* argv[]) {
std::string s("mystring");
std::string s_tail = tail(s, 6);
cout << s_tail << endl;
s_tail = tail(s, 10);
cout << s_tail << endl;
return 0;
}

Answer 9

Try the following:
std::string tail(&source[(source.length() > 6) ? (source.length() - 6) : 0]);

Answer 10

string tail = source.substr(source.size() - min(6, source.size()));