Template with sized return type in function C++?

Question

Is it possible to design a template function in such a way

template <typename T1, typename T2>
T3 f(T1 x, T2 y);

where if sizeof(T1) > sizeof(T2) then T3 = T1 , otherwise T3 = T2

Further request...

I tried to define something like:

template <int i>
struct A {
    int a;
};

template <int j>
struct B {
    int b;
};

template <int i, int j>
typename std::conditional< i > j , A<i>, B<j> >::type
func() {
    ;
}

But when i try to compile ...

Info: Internal Builder is used for build
g++ -O0 -g3 -Wall -c -fmessage-length=0 -std=c++11 -o main.o "..\\main.cc" 
In file included from ..\main.cc:8:0:
..\header.h:34:30: error: wrong number of template arguments (1, should be 3)
 typename std::conditional< i > j , A<i>, B<j> >::type
                              ^
In file included from c:\mingw\lib\gcc\mingw32\4.8.1\include\c++\bits\move.h:57:0,
                 from c:\mingw\lib\gcc\mingw32\4.8.1\include\c++\bits\stl_pair.h:59,
                 from c:\mingw\lib\gcc\mingw32\4.8.1\include\c++\bits\stl_algobase.h:64,
                 from c:\mingw\lib\gcc\mingw32\4.8.1\include\c++\bits\char_traits.h:39,
                 from c:\mingw\lib\gcc\mingw32\4.8.1\include\c++\ios:40,
                 from c:\mingw\lib\gcc\mingw32\4.8.1\include\c++\ostream:38,
                 from c:\mingw\lib\gcc\mingw32\4.8.1\include\c++\iostream:39,
                 from ..\header.h:11,
                 from ..\main.cc:8:
c:\mingw\lib\gcc\mingw32\4.8.1\include\c++\type_traits:77:12: error: provided for 'template<bool <anonymous>, class, class> struct std::conditional'
     struct conditional;
            ^
In file included from ..\main.cc:8:0:
..\header.h:34:32: error: 'j' in namespace 'std' does not name a type
 typename std::conditional< i > j , A<i>, B<j> >::type
                                ^
..\header.h:34:34: error: expected unqualified-id before ',' token
 typename std::conditional< i > j , A<i>, B<j> >::type

So basically doesn't work in that case... (non type templates).

Answer 1

You can use std::conditional :

template <typename T1, typename T2>
typename std::conditional< (sizeof(T1) > sizeof(T2)),
                  T1, T2 >::type
f(T1 x, T2 y);

If you wanted, you could factor this out like so:

template <typename T1, typename T2>
using largest_type = 
    typename std::conditional< (sizeof(T1) > sizeof(T2)),
                               T1, T2 >::type;

template <typename T1, typename T2>
largest_type<T1,T2>
f(T1 x, T2 y);

Writing largest_type as a variadic template is left as an exercise :D

Answer 2

It's a vexing parse!

typename std::conditional< i > j , A<i>, B<j> >::type
//                           ^
//                           Looks like the close of template params

Put parens around the expression:

typename std::conditional< (i > j) , A<i>, B<j> >::type

or reverse the condition:

typename std::conditional< j < i , A<i>, B<j> >::type