I have declared a boost::variant
which accepts three types: string
, bool
and int
. The following code is showing that my variant accepts const char*
and converts it to bool
. Is it a normal behavior for boost::variant
to accept and convert types not on its list?
#include <iostream>
#include "boost/variant/variant.hpp"
#include "boost/variant/apply_visitor.hpp"
using namespace std;
using namespace boost;
typedef variant<string, bool, int> MyVariant;
class TestVariant
: public boost::static_visitor<>
{
public:
void operator()(string &v) const
{
cout << "type: string -> " << v << endl;
}
template<typename U>
void operator()(U &v)const
{
cout << "type: other -> " << v << endl;
}
};
int main(int argc, char **argv)
{
MyVariant s1 = "some string";
apply_visitor(TestVariant(), s1);
MyVariant s2 = string("some string");
apply_visitor(TestVariant(), s2);
return 0;
}
output:
type: other -> 1
type: string -> some string
If I remove the bool type from MyVariant and change it to this:
typedef variant<string, int> MyVariant;
const char*
is no more converted to bool
. This time it's converted to string
and this is the new output:
type: string -> some string
type: string -> some string
This indicates that variant
tries to convert other types first to bool
and then to string
. If the type conversion is something inevitable and should always happen, is there any way to give conversion to string
a higher priority?
I don't think this is anything particularly to do with boost::variant
, it's about which constructor gets selected by overload resolution. The same thing happens with an overloaded function:
#include <iostream>
#include <string>
void foo(bool) {
std::cout << "bool\n";
}
void foo(std::string) {
std::cout << "string\n";
}
int main() {
foo("hi");
}
output:
bool
I don't know of a way to change what constructors a Variant has [edit: as James says, you can write another class that uses the Variant in its implementation. Then you can provide a const char*
constructor that does the right thing.]
Maybe you could change the types in the Variant. Another overloading example:
struct MyBool {
bool val;
explicit MyBool(bool val) : val(val) {}
};
void bar(MyBool) {
std::cout << "bool\n";
}
void bar(const std::string &) {
std::cout << "string\n";
}
int main() {
bar("hi");
}
output:
string
Unfortunately now you have to write bar(MyBool(true))
instead of foo(true)
. Even worse in the case of your variant with string/bool/int
, if you just change it to a variant of string/MyBool/int
then MyVariant(true)
would call the int
constructor.