| Home | Libraries | People | FAQ | More

### common_type

Header: `#include <boost/type_traits/common_type.hpp>` or `#include <boost/type_traits.hpp>`

    namespace boost {
      template <class ...T>  struct common_type;
    }

common_type is a traits class used to deduce a type common to a several types, useful as the return type of functions operating on multiple input types such as in mixed-mode arithmetic..

The nested typedef `::type` could be defined as follows:

    template <class ...T>
    struct common_type;

    template <class T, class U, class ...V>
    struct common_type<T,U,...V> {
        typedef typename common_type<typename common_type<T, U>::type, V...>::type type;
    };

    template <class T>
    struct common_type<T> {
        typedef T type;
    };

    template <class T, class U>
    struct common_type<T, U> {
        typedef decltype(declval<bool>() ? declval<T>() : declval<U>()) type;
    };

All parameter types must be complete. This trait is permitted to be specialized by a user if at least one template parameter is a user-defined type. Note: Such specializations are required when only explicit conversions are desired among the common_type arguments.

Note that when the compiler does not support variadic templates (and the macro BOOST_NO_VARIADIC_TEMPLATES is defined) then the maximum number of template arguments is 3.

##### Configuration macros

When the compiler does not support static assertions then the user can select the way static assertions are reported. Define

- BOOST_COMMON_TYPE_USES_STATIC_ASSERT: define it if you want to use Boost.StaticAssert
- BOOST_COMMON_TYPE_USES_MPL_ASSERT: define it if you want to use Boost.MPL static assertions

The default behavior is to use mpl assertions in this case, but setting BOOST_COMMON_TYPE_USES_STATIC_ASSERT may reduce compile times and header dependencies somewhat.

Depending on the static assertion used you will have an hint of the failing assertion either through the symbol or through the text.

When possible common_type is implemented using `decltype`. Otherwise when BOOST_COMMON_TYPE_DONT_USE_TYPEOF is not defined it uses Boost.TypeOf.

##### Tutorial

In a nutshell, common_type is a trait that takes 1 or more types, and returns a type which all of the types will convert to. The default definition demands this conversion be implicit. However the trait can be specialized for user-defined types which want to limit their inter-type conversions to explicit, and yet still want to interoperate with the common_type facility.

Example:

    template <class T, class U>
    complex<typename common_type<T, U>::type>
    operator+(complex<T>, complex<U>);

In the above example, "mixed-mode" complex arithmetic is allowed. The return type is described by common_type. For example the resulting type of adding a `complex<float>` and `complex<double>` might be a `complex<double>`.

Here is how someone might produce a variadic comparison function:

    template <class ...T>
    typename common_type<T...>::type
    min(T... t);

This is a very useful and broadly applicable utility.

##### How to get the common type of types with explicit conversions?

Another choice for the author of the preceding operator could be

    template <class T, class U>
    typename common_type<complex<T>, complex<U> >::type
    operator+(complex<T>, complex<U>);

As the default definition of common_type demands the conversion be implicit, we need to specialize the trait for complex types as follows.

    template <class T, class U>
    struct common_type<complex<T>, complex<U> > {
        typedef complex< common_type<T, U> > type;
    };

##### How important is the order of the common_type<> template arguments?

The order of the template parameters is important.

`common_type<A,B,C>::type` is not equivalent to `common_type<C,A,B>::type`, but to `common_type<common_type<A,B>::type, C>::type`.

Consider

    struct A {};
    struct B {};
    struct C {
        C() {}
        C(A const&) {}
        C(B const&) {}
        C& operator=(C const&) {
            return *this;
        }
    };

The following doesn't compile

    typedef boost::common_type<A, B, C>::type ABC; // Does not compile

while

    typedef boost::common_type<C, A, B>::type ABC;

compiles.

Thus, as `common_type<A,B>::type` is undefined, `common_type<A,B,C>::type` is also undefined.

It is intended that clients who wish for `common_type<A, B>` to be well defined to define it themselves:

    namespace boost
    {

    template <>
    struct common_type<A, B> {typedef C type;};

    }

Now this client can ask for `common_type<A, B, C>` (and get the same answer).

Clients wanting to ask `common_type<A, B, C>` in any order and get the same result need to add in addition:

    namespace boost
    {

    template <> struct common_type<B, A>
    : public common_type<A, B> {};

    }

This is needed as the specialization of `common_type<A, B>` is not be used implicitly for `common_type<B, A>`.

##### Can the common_type of two types be a third type?

Given the preceding example, one might expect `common_type<A,B>::type` to be `C` without any intervention from the user. But the default `common_type<>` implementation doesn't grant that. It is intended that clients who wish for `common_type<A, B>` to be well defined to define it themselves:

    namespace boost
    {

    template <>
    struct common_type<A, B> {typedef C type;};

    template <> struct common_type<B, A>
    : public common_type<A, B> {};

    }

Now this client can ask for `common_type<A, B>`.

##### How common_type behaves with pointers?

Consider

    struct C { }:
    struct B : C { };
    struct A : C { };

Shouldn't `common_type<A*,B*>::type` be `C*`? I would say yes, but the default implementation will make it ill-formed.

The library could add a specialization for pointers, as

    namespace boost
    {

        template <typename A, typename B>
        struct common_type<A*, B*> {
            typedef common_type<A, B>* type;
        };
    }

But in the absence of a motivating use cases, we prefer not to add more than the standard specifies.

Of course the user can always make this specialization.

##### Can you explain the pros/cons of common_type against Boost.Typeof?

Even if they appear to be close, `common_type` and `typeof` have different purposes. You use `typeof` to get the type of an expression, while you use common_type to set explicitly the type returned of a template function. Both are complementary, and indeed common_type is equivalent to `decltype(declval<bool>() ? declval<T>() : declval<U>())`

common_type is also similar to promote_args<class ...T> in boost/math/tools/promotion.hpp, though it is not exactly the same as promote_args either. common_type<T1, T2>::type simply represents the result of some operation on T1 and T2, and defaults to the type obtained by putting T1 and T2 into a conditional statement.

It is meant to be customizable (via specialization) if this default is not appropriate.