Boost.Variant provides a class called boost::variant
that resembles union
. You can store values of different types in a boost::variant
variable. At any point only one value can be stored. When a new value is assigned, the old value is overwritten. However, the new value may have a different type from the old value. The only requirement is that the types must have been passed as template parameters to boost::variant
so they are known to the boost::variant
variable.
boost::variant
supports any type. For example, it is possible to store a std::string
in a boost::variant
variable – something that wasn’t possible with union
before C++11. With C++11, the requirements for union
were relaxed. Now a union
can contain a std::string
. Because a std::string
must be initialized with placement new and has to be destroyed by an explicit call to the destructor, it can still make sense to use boost::variant
, even in a C++11 development environment.
boost::variant
#include <boost/variant.hpp>
#include <string>
int main()
{
boost::variant<double, char, std::string> v;
v = 3.14;
v = 'A';
v = "Boost";
}
boost::variant
is defined in boost/variant.hpp
. Because boost::variant
is a template, at least one parameter must be specified. One or more template parameters specify the supported types. In Example 24.1, v can store values of type double
, char
, or std::string
. However, if you tried to assign a value of type int
to v, the resulting code would not compile.
boost::variant
with boost::get()
#include <boost/variant.hpp>
#include <string>
#include <iostream>
int main()
{
boost::variant<double, char, std::string> v;
v = 3.14;
std::cout << boost::get<double>(v) << '\n';
v = 'A';
std::cout << boost::get<char>(v) << '\n';
v = "Boost";
std::cout << boost::get<std::string>(v) << '\n';
}
To display the stored values of v, use the free-standing function boost::get()
(see Example 24.2).
boost::get()
expects one of the valid types for the corresponding variable as a template parameter. Specifying an invalid type will result in a run-time error because validation of types does not take place at compile time.
Variables of type boost::variant
can be written to streams such as the standard output stream, bypassing the hazard of run-time errors (see Example 24.3).
boost::variant
on a stream#include <boost/variant.hpp>
#include <string>
#include <iostream>
int main()
{
boost::variant<double, char, std::string> v;
v = 3.14;
std::cout << v << '\n';
v = 'A';
std::cout << v << '\n';
v = "Boost";
std::cout << v << '\n';
}
For type-safe access, Boost.Variant provides a function called boost::apply_visitor()
.
boost::variant
#include <boost/variant.hpp>
#include <string>
#include <iostream>
struct output : public boost::static_visitor<>
{
void operator()(double d) const { std::cout << d << '\n'; }
void operator()(char c) const { std::cout << c << '\n'; }
void operator()(std::string s) const { std::cout << s << '\n'; }
};
int main()
{
boost::variant<double, char, std::string> v;
v = 3.14;
boost::apply_visitor(output{}, v);
v = 'A';
boost::apply_visitor(output{}, v);
v = "Boost";
boost::apply_visitor(output{}, v);
}
As its first parameter, boost::apply_visitor()
expects an object of a class derived from boost::static_visitor
. This class must overload operator()
for every type used by the boost::variant
variable it acts on. Consequently, the operator is overloaded three times in Example 24.4 because v supports the types double
, char
, and std::string
.
boost::static_visitor
is a template. The type of the return value of operator()
must be specified as a template parameter. If the operator does not have a return value, a template parameter is not required, as seen in the example.
The second parameter passed to boost::apply_visitor()
is a boost::variant
variable.
boost::apply_visitor()
automatically calls the operator()
for the first parameter that matches the type of the value currently stored in the second parameter. This means that the sample program uses different overloaded operators every time boost::apply_visitor()
is invoked – first the one for double
, followed by the one for char
, and finally the one for std::string
.
The advantage of boost::apply_visitor()
is not only that the correct operator is called automatically. In addition, boost::apply_visitor()
ensures that overloaded operators have been provided for every type supported by boost::variant
variables. If one of the three overloaded operators had not been defined, the code could not be compiled.
If overloaded operators are equivalent in functionality, the code can be simplified by using a template (see Example 24.5).
boost::variant
#include <boost/variant.hpp>
#include <string>
#include <iostream>
struct output : public boost::static_visitor<>
{
template <typename T>
void operator()(T t) const { std::cout << t << '\n'; }
};
int main()
{
boost::variant<double, char, std::string> v;
v = 3.14;
boost::apply_visitor(output{}, v);
v = 'A';
boost::apply_visitor(output{}, v);
v = "Boost";
boost::apply_visitor(output{}, v);
}
Because boost::apply_visitor()
ensures code correctness at compile time, it should be preferred over boost::get()
.