Question

I have three questions based on the following code fragments
I have a list of strings. It just happens to be a vector but could potentially be any source

vector<string> v1_names = boost::assign::list_of("Antigua and Barbuda")( "Brasil")( "Papua New Guinea")( "Togo");

The following is to store lengths of each name

vector<int> name_len;

the following is where I want to store the strings

std::vector<char> v2_names;

estimate memory required to copy names from v1_names

v2_names.reserve( v1_names.size()*20 + 4 );

Question: is this the best way to estimate storage? I fix the max len at 20 that is ok, then add space for null treminator
Now copy the names

for( std::vector<std::string>::size_type i = 0; i < v1_names.size(); ++i)
{
    std::string val( v1_names[i] );
    name_len.push_back(val.length());
    for(std::string::iterator it = val.begin(); it != val.end(); ++it)
    {
        v2_names.push_back( *it );
    }
    v2_names.push_back('\0');
}

Question: is this the most efficient way to copy the elements from v1_name to v2_names?

Main Question: How do I iterate over v2_names and print the country names contained in v2_names

Answer 1

To estimate storage, you should probably measure the strings, rather than rely on a hard-coded constant 20. For example:

size_t total = 0;
for (std::vector<std::string>::iterator it = v1_names.begin(); it != v1_names.end(); ++it) {
    total += it->size() + 1;
}

The main inefficiency in your loop is probably that you take an extra copy of each string in turn: std::string val( v1_names[i] ); could instead be const std::string &val = v1_names[i];.

To append each string, you can use the insert function:

v2_names.insert(v2_names.end(), val.begin(), val.end());
v2_names.push_back(0);

This isn’t necessarily the most efficient, since there’s a certain amount of redundant checking of available space in the vector, but it shouldn’t be too bad and it’s simple. An alternative would be to size v2_names at the start rather than reserving space, and then copy data (with std::copy) rather than appending it. But either one of them might be faster, and it shouldn’t make a lot of difference.

For the main question, if all you have is v2_names and you want to print the strings, you could do something like this:

const char *p = &v2_names.front();
while (p <= &v2_names.back()) {
    std::cout << p << "\n";
    p += strlen(p) + 1;
}

If you also have name_len:

size_t offset = 0;
for (std::vector<int>::iterator it = name_len.begin(); it != name_len.end(); ++it) {
    std::cout << &v2_names[offset] << "\n";
    offset += *it + 1;
}

Beware that the type of name_len is technically wrong – it’s not guaranteed that you can store a string length in an int. That said, even if int is smaller than size_t in a particular implementation, strings that big will still be pretty rare.