Define boost variant type to pass explicitly empty values

Question

I want to have boost::variant with empty state. So I define a boost::variant with boost::blank as the first alternative. But then I want to pass this as function parameter:

void f(Variant v);
...
void g()
{
   f(boost::blank{});
}

It does not look nice due to braces. Seem to be better if it accepted boost::none:

void g()
{
   f(boost::none);
}

But I don't think I have seen boost::variant<boost::none_t, ...> anywhere. boost::none_t is a satellite of boost::optional. Is it fine to use with boost::variant?

@rturrado that's not the same and also can be less efficient — sehe, Dec 16 '21 at 23:37

score 2 · Accepted Answer · answered Dec 16 '21 at 23:44

You can just default construct, which will initialize the first element type.

Alternatively you can define your own constant of type boost::blank:

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

using Variant = boost::variant<
    boost::blank,
    int,
    std::string>;

void foo(Variant v = {}) {
    std::cout << "foo: " << v.which() << "\n";
}

int main()
{
    foo();
    foo({});
    foo(Variant{});
    foo(boost::blank{});
    
    static constexpr boost::blank blank;
    foo(blank);
    
    foo(42);
    foo("LtUaE");
}

Prints

foo: 0
foo: 0
foo: 0
foo: 0
foo: 0
foo: 1
foo: 2

Define boost variant type to pass explicitly empty values

1 Answers1