使用数组初始化std :: string的向量

Question

I wish to initialise a vector using an array of std::string s.

I have the following solution, but wondered if there's a more elegant way of doing this?

std::string str[] = { "one", "two", "three", "four" };
vector< std::string > vec;
vec = vector< std::string >( str, str + ( sizeof ( str ) /  sizeof ( std::string ) ) );

I could, of course, make this more readable by defining the size as follows:

int size =  ( sizeof ( str ) /  sizeof ( std::string ) );

and replacing the vector initialisation with:

vec = vector< std::string >( str, str + size );

But this still feels a little "inelegant".

Answer 1

Well the intermediate step isn't needed:

std::string str[] = { "one", "two", "three", "four" };
vector< std::string > vec( str, str + ( sizeof ( str ) /  sizeof ( std::string ) ) );

In C++11 you'd be able to put the brace initialization in the constructor using the initializer list constructor.

Answer 2

In C++11, we have std::begin and std::end , which work for both STL-style containers and built-in arrays:

#include <iterator>

std::vector<std::string> vec(std::begin(str), std::end(str));

although, as mentioned in the comments, you usually won't need the intermediate array at all:

std::vector<std::string> vec {"one", "two", "three", "four"};

In C++03, you could use a template to deduce the size of the array, either to implement your own begin and end , or to initialise the array directly:

template <typename T, size_t N>
std::vector<T> make_vector(T &(array)[N]) {
    return std::vector<T>(array, array+N);
}

std::vector<std::string> vec = make_vector(str);