Question

I started looking at std::common_type and am not exactly sure about its purpose and its functionality.
A few things still strike me as odd:

• Order of arguments is important: common_type<Foo, Bar, Baz> might be different from common_type<Baz, Foo, Bar>. Either might compile, the other might not. While this is clear from the way common_type is defined, it feels strange and unintuitive. Is this for lack of a universal solution or intended?
• Instantiation can result in a compiler error instead of something I can handle. How to check if common_type will actually compile? is_convertible is not enough as common_type might be specialized?

• There is still no way to figure out the common type in a situation like this:

  struct Baz;
  struct Bar { int m; };
  struct Foo { int m; }; 
  struct Baz { Baz(const Bar&); Baz(const Foo&); };
  

  The recommended solution would be to specialize common_type which is tedious. Is there a better solution?

For reference see §20.9.7.6 Table 57 in N3242.

Answer 1

std::common_type was introduced for use with std::duration — if you add a std::duration<int> and a std::duration<short> then the result should be std::duration<int>. Rather than specifying an endless stream of allowed pairings, the decision was made to delegate to a separate template which found the result using the core language rules applicable to the ?: arithmetic-if operator.

People then saw that this template might be generally useful, and it was added as std::common_type, and extended to handle an arbitrary number of types. In the C++0x library it is only used for pairs of types though.

You should be able to use the new SFINAE rules to detect whether or not some instantiation of std::common_type is valid. I haven’t tried though. In most cases if there isn’t a “common type” then there isn’t anything meaningful you can do anyway, so a compile error is reasonable.

std::common_type is not magic — it follows the rules of ?:. If true?a:b will compile, std::common_type<decltype(a),decltype(b)>::type will give you the type of the result.