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auto fail to deduce correct return type


auto fail to deduce correct return type

By : BobbyDole
Date : October 22 2020, 06:10 PM
it helps some times Your code violates the following rule from the C++ Standard [dcl.spec.auto.8]:
code :


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Compiler Issue with auto? Error: in a declarator-list 'auto' must always deduce to the same type

Compiler Issue with auto? Error: in a declarator-list 'auto' must always deduce to the same type


By : NumBee
Date : March 29 2020, 07:55 AM
like below fixes the issue Compiles fine in g++ on Linux, so it appears to be a compiler bug. Probably this one.
Auto-deduce method's return type

Auto-deduce method's return type


By : babak Alizadeh
Date : March 29 2020, 07:55 AM
it fixes the issue I have the following functions:
code :
#include <type_traits>

template <class A1, class... Ax>
auto getValue()
    -> Value<typename std::conditional<sizeof...(Ax) == 0
                                     , FooSingle, FooMany>::type>
{
    typename std::conditional<sizeof...(Ax) == 0
                            , FooOne<Item> 
                            , FooAll<Item>>::type
              hairyStructure("abc", { Foo<A1>::getIndex(), Foo<Ax>::getIndex()... } );
    return Value<typename std::conditional<sizeof...(Ax) == 0
                                         , FooSingle, FooMany>::type>(hairyStructure);
}
How to allow compiler to deduce correct return type for templated get function?

How to allow compiler to deduce correct return type for templated get function?


By : Wari Wari
Date : March 29 2020, 07:55 AM
To fix the issue you can do Don't use template function specialization, it is almost always a bad idea.
Instead use overloading and forwarding.
code :
template<int idx>
using index_t = std::integral_constant<int, idx>;

// compile time error if it doesn't match
template<int i, class K>
void special_get(index_t<i>, const K&) = delete;

template<int idx, typename K> 
decltype(special_get( index_t<idx>{}, std::declval<K const&>() ) )
get(const K& k) { 
  return special_get( index_t<idx>{}, k );
}
inline std::string special_get(index_t<0>, const myExample& k) {
  return k.elem0;
}
inline int special_get(index_t<1>, const myExample& k) {
  return k.elem1;
}
How can I make templated operator deduce correct return type?

How can I make templated operator deduce correct return type?


By : Nail Loutfi Cheikh
Date : March 29 2020, 07:55 AM
wish helps you You can implement it as two (non-template) free functions. If the implementation is exactly the same, these can specify the return type for a shared implementation.
code :
namespace detail
{
    template<typename Out, typename In>
    MyClass<Out> times( const MyClass<In> & lhs, const MyClass<In> & rhs)
    {
        // shared code here
    }
}

myClass<isA> operator * ( const myClass<isB>& lhs, const myClass<isB>& rhs )
{ return detail::times<isA>(lhs, rhs); }

myClass<isB> operator * ( const myClass<isA>& lhs, const myClass<isA>& rhs )
{ return detail::times<isB>(lhs, rhs); }
decltype(auto) function return type does not deduce && types

decltype(auto) function return type does not deduce && types


By : Mihir Sabnis
Date : October 25 2020, 07:10 AM
seems to work fine This seems to be a bug in GCC. Since decltype(ret) is int&&, foo1 should have return type int&&. However, this immediately renders foo1 ill-formed, since a function that returns int&& cannot have its return value initialized from ret, which is an lvalue (you would need std::move to make it work properly). Note that Clang gets this right (see link in comments to the question).
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