How can I print out C++ map values?

I have a map like this:

map<string, pair<string,string> > myMap;

And I've inserted some data into my map using:

myMap.insert(make_pair(first_name, make_pair(middle_name, last_name)));

How can I now print out all the data in my map?

Solution 1:

for(map<string, pair<string,string> >::const_iterator it = myMap.begin();
    it != myMap.end(); ++it)
{
    std::cout << it->first << " " << it->second.first << " " << it->second.second << "\n";
}

In C++11, you don't need to spell out map<string, pair<string,string> >::const_iterator. You can use auto

for(auto it = myMap.cbegin(); it != myMap.cend(); ++it)
{
    std::cout << it->first << " " << it->second.first << " " << it->second.second << "\n";
}

Note the use of cbegin() and cend() functions.

Easier still, you can use the range-based for loop:

for(const auto& elem : myMap)
{
   std::cout << elem.first << " " << elem.second.first << " " << elem.second.second << "\n";
}

Solution 2:

If your compiler supports (at least part of) C++11 you could do something like:

for (auto& t : myMap)
    std::cout << t.first << " " 
              << t.second.first << " " 
              << t.second.second << "\n";

For C++03 I'd use std::copy with an insertion operator instead:

typedef std::pair<string, std::pair<string, string> > T;

std::ostream &operator<<(std::ostream &os, T const &t) { 
    return os << t.first << " " << t.second.first << " " << t.second.second;
}

// ...
std:copy(myMap.begin(), myMap.end(), std::ostream_iterator<T>(std::cout, "\n"));

Solution 3:

Since C++17 you can use range-based for loops together with structured bindings for iterating over your map. This improves readability, as you reduce the amount of needed first and second members in your code:

std::map<std::string, std::pair<std::string, std::string>> myMap;
myMap["x"] = { "a", "b" };
myMap["y"] = { "c", "d" };

for (const auto &[k, v] : myMap)
    std::cout << "m[" << k << "] = (" << v.first << ", " << v.second << ") " << std::endl;

Output:

m[x] = (a, b)
m[y] = (c, d)

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