Opened 7 years ago
Closed 6 years ago
#11602 closed Bugs (fixed)
boost.variant constructor accepts any type as parameter
| Reported by: | Owned by: | Antony Polukhin | |
|---|---|---|---|
| Milestone: | Boost 1.62.0 | Component: | variant |
| Version: | Boost 1.58.0 | Severity: | Problem |
| Keywords: | Cc: | raad@… |
Description
There is one "converting constructor" in boost.variant which accepts any type - even those which are not convertible to instantiation types of boost.variant. This causes that writing function accepting some specific variant instantiation cause that this function is on candidate list of any type in your program.
The problem is with this constructor:
template <typename T>
variant(const T& operand)
{
convert_construct(operand, 1L);
}
And the small code presenting the issue:
#include <iostream>
#include <boost/variant.hpp>
// types not related in any way
class A {};
class B {};
class C {};
class D {};
using ABC_variant = boost::variant<A,B,C>;
std::ostream& operator << (std::ostream& os, const ABC_variant&)
{
return os << "ABC";
}
int main() {
D d;
std::cout << d;
}
As described in http://stackoverflow.com/questions/32275725/boostvariant-construction-weirdness-its-ctor-accepts-everything
compiler (gcc4.9) tries to use ostream operator with this diagnostic:
/usr/include/boost/variant/variant.hpp:1591:38: error: no matching function for call to 'boost::variant<A, B, C>::initializer::initialize(void*, D&)'
initializer::initialize(
However it should simple report that ostream operator for D is missing.
One simple way is to declare this boost.variant constructor as explicit (what probably would impact to many client code) or to make some restriction in this constructor like this C++11 way:
template <typename T, typename ...C>
struct IsAnyOf;
template <typename T, typename ...C>
struct IsAnyOf<T,T,C...> : std::true_type {};
template <typename T>
struct IsAnyOf<T> : std::false_type {};
template <typename T, typename C1, typename ...C>
struct IsAnyOf<T,C1,C...> : IsAnyOf<T, C...> {};
template <typename T,
typename EnableIf = typename std::enable_if<IsAnyOf<VariantType...>::value>::type>
variant(const T& operand)
BR, Piotr Nycz (piotrwn1 @ gmail com)
Change History (6)
comment:1 by , 7 years ago
comment:2 by , 7 years ago
| Cc: | added |
|---|
comment:3 by , 7 years ago
comment:4 by , 6 years ago
| Milestone: | To Be Determined → Boost 1.62.0 |
|---|---|
| Owner: | changed from to |
| Status: | new → assigned |
comment:5 by , 6 years ago
Fixed in b3650685 develop, will be merged to the master branch as soon as the regression tests will cycle.
comment:6 by , 6 years ago
| Resolution: | → fixed |
|---|---|
| Status: | assigned → closed |

Maybe better would to use IsConvertibleToAnyOf instead of IsAnyOf:
template <typename T, typename ...C> struct IsConvertibleToAnyOf; template <typename T> struct IsConvertibleToAnyOf<T> : std::false_type {}; template <typename T, typename C1, typename ...C> struct IsConvertibleToAnyOf<T,C1,C...> : std::integral_constant<bool, std::is_convertible<T,C1>::value or IsConvertibleToAnyOf<T,C...>::value> {};BR,
Piotr Nycz