given a string s of type std::string I build another string s2 like so
std::string(s.begin(), s.begin() + s.find(" "))
but the following constructor doesn't work
std::string(s.begin(), s.find(" "))
anyone knows why ? I'm using g++ 4.8.1 under Ubuntu amd64
Both constructors have an iterator as second argument.
In this example I'm trying to build a string containing the first sub-string that ends where the first whitespace appears.
std::string::find() returns an offset, not an iterator (yes, it's largely a design inconsistency of the standard library). So to use this (and include proper error checking), do something like this:
size_t offset = s.find(" ");
std::string(s.begin(), (offset == std::string::npos ? s.end() : s.begin() + offset);
chars you expect something like that ...
std::find() algorithm, which returns an iterator?string:: namespace. Anyway thanks.std::string type needed a ::find(), since there is a perfectly good free function, the fact that it has a different return type just adds to the confusion.
Because find returns an index (size_type). Because a string is a contiguous container, s.begin() + s.find(" ") will return you an iterator, as it is equivalent to calling std::string::iterator operator+=(std::string::size_type s). The second one will of course be a compilation error.
O(n), that is.Perhaps you were thinking of the std::find() algorithm, which returns an iterator?
#include <iostream>
#include <algorithm>
int main()
{
std::string s = "Derek Smalls";
std::string s2(std::begin(s), std::find(std::begin(s), std::end(s), ' '));
std::cout << s2 << std::endl;
}