default visitor function for boost::variant

Question

suppose that I have such a variant definition:

typedef boost::variant <
v1,
v2,
v3,
...
vn
> v;

and I need to write a visitor class with visitor functions for each v1 to vn like this:

class myvisitor : public boost::static_visitor<bool> {
  bool operator()(v1) {}
  bool operator()(v2) {}
   ...
  bool operator()(vn) {}
}

So if all such functions are the same except the one for v1, then I would like to only define

 bool operator()(v1) {}

while leave all others to some default form to avoid writing lots of useless and duplicated code.

So if this is possible? or can the boost developer do this on his next version?

Answer 1

Just make the default "case" an open template member operator():

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#include <boost/variant.hpp>
#include <iostream>

struct MyStruct {
    int a, b, c;
};

using V = boost::variant<int, MyStruct, std::string, double>;

struct MyVisitor : boost::static_visitor<void> {
    void operator()(int) const                 { std::cout << __PRETTY_FUNCTION__ << "\n"; } 
    void operator()(std::string const &) const { std::cout << __PRETTY_FUNCTION__ << "\n"; } 

    // the default case:
    template <typename T> void operator()(T const &) const {
        std::cout << "FALLBACK: " << __PRETTY_FUNCTION__ << "\n"; 
    } 
};

int main() {
    V v;

    for (auto v : { V(42), V(3.14), V("hello world"), V( MyStruct{1,2,3} ) })
        boost::apply_visitor(MyVisitor(), v);
}

Output:

void MyVisitor::operator()(int) const
FALLBACK: void MyVisitor::operator()(const T&) const [with T = double]
void MyVisitor::operator()(const string&) const
FALLBACK: void MyVisitor::operator()(const T&) const [with T = MyStruct]