How to pass std::map as a default constructor parameter

The correct expression for VAL is std::map<std::string, std::string>(). I think that looks long and ugly, so I'd probably add a public typedef member to the class:

class Foo {
public:
  typedef std::map<std::string, std::string> map_type;
  Foo( int arg1, int arg2, const map_type = map_type() );
  // ...
};

And by the way, did you mean for the last constructor argument to be a reference? const map_type& is probably better than just const map_type.

You create a value-initialized temporary. For example:

Foo::Foo(int arg1,
         int arg2,
         const std::map<std::string, std::string>& the_map =
             std::map<std::string, std::string>())
{
}

(A typedef might help to make this more readable in your code)

Since C++11 you can use aggregate initialization:

void foo(std::map<std::string, std::string> myMap = {});

Example:

#include <iostream>
#include <map>
#include <string>

void foo(std::map<std::string, std::string> myMap = {})
{
    for(auto it = std::cbegin(myMap); it != std::cend(myMap); ++it)
        std::cout << it->first << " : " << it->second << '\n';
}

int main(int, char*[])
{
    const std::map<std::string, std::string> animalKids = {
        { "antelope", "calf" }, { "ant", "antling" },
        { "baboon", "infant" }, { "bear", "cub" },
        { "bee", "larva" }, { "cat", "kitten" }
    };

    foo();
    foo(animalKids);

    return 0;
}

You can play around with this example at Godbolt.