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Why can't C++ deduce T in a call to Foo<T>::Foo(T&&)?

By : whosle01
Date : July 29 2020, 09:00 AM
This might help you The issue here is that, since the class is templated on T, in the constructor Foo(T&&) we are not performing type deduction; We always have an r-value reference. That is, the constructor for Foo actually looks like this:
code :
template<typename T>
struct Foo {
    Foo(T&&) {}

template<typename T>
Foo<T> MakeFoo(std::add_rvalue_reference_t<T> value)
   return Foo<T>(std::move(value));

auto f = MakeFoo(x);
template<typename T>
Foo(T&&) -> Foo<T>;
template<class U>
Foo<U> MakeFoo(U&& u)
   return Foo<U>(std::forward<U>(u));

// ...
auto f = MakeFoo(x);

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"Could not deduce template argument" error when using Winsock bind() call with Boost

By : user1845987
Date : March 29 2020, 07:55 AM
wish of those help There's a Boost function called bind() which is a totally different thing from Winsock's bind().
You have two options if you need both functions available in a given module:

Deduce template argument from std::function call signature

By : Rodolphe LOUIS-SIDNE
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Consider this template function:
code :
std::function<bool()> bar;

foo(bar); // works just fine

Could not deduce (Eq a) from (Num a) or from (Floating a). But Can deduce (Eq a) from (Integral a). Why?

By : Bernard Gressing
Date : March 29 2020, 07:55 AM
seems to work fine Short answer
Because that's the definition of Num in the prelude:
code :
class Num a where
class (Real a, Enum a) => Integral a where
class (Num a, Ord a) => Real a where
class Eq a => Ord a where
newtype Sequence = Sequence (Integer -> Integer)

instance Num Sequence where
  (Sequence x) + (Sequence y) = Sequence $ \pt -> x pt + y pt
  (Sequence x) - (Sequence y) = Sequence $ \pt -> x pt - y pt
  (Sequence x) * (Sequence y) = Sequence $ \pt -> x pt * y pt
  negate (Sequence x) = Sequence $ \pt -> -pt
  abs (Sequence x) = Sequence $ \pt -> abs pt
  signum (Sequence x) = Sequence $ \pt -> signum pt
  fromInteger = Sequence . const

-- Ignore the fact that you'd implement these methods using Applicative.
instance Eq Sequence where
  -- This will never return True, ever.
  (Sequence x) == (Sequence y) =
      and [x pt == y pt | pt <- [0..]] &&
      and [x pt == y pt | pt <- [-1,-2..]]

Deduce template parameter pack from function call

By : user3654520
Date : March 29 2020, 07:55 AM
This might help you Template deduction is not possible, but maybe you can restructure your code in a way that MyClass defines the all necessary types and then you have a check function that takes MyClass as a template argument. That way, the checking function has access to all the necessary types.
code :
template <typename... Types> struct OtherClass {};

template <typename... Types>
struct MyClass
    typedef OtherClass<Types...> OtherClass_t;
    typedef int result_t;

    enum SomeEnum { value0 = -1 };

// version 1
template < typename C >
struct Checker {
    typename C::result_t operator()(typename C::SomeEnum value)
        typename C::OtherClass_t obj;
        typename C::result_t result;
        // calculate result from obj;
        return result;

// version 2
template < typename C >
typename C::result_t check_fun(typename C::SomeEnum value)
    typename C::OtherClass_t obj;
    typename C::result_t result;
    // calculate result from obj;
    return result;

int main() {
    typedef MyClass< int, bool > myclass_t;
    auto value = myclass_t::value0;
    // ... 
    Checker< myclass_t > check;
    int t = check(value);
    auto s = check_fun<myclass_t>(value);

how to deduce the size of a multidimensional array in a curly brace expression of a function call with templates

By : J. Lu
Date : March 29 2020, 07:55 AM
around this issue You cannot deduce multi dimensional array bounds from nested {{}}.
You could deduce it by adding some tokens.
code :
using namespace std;
auto arr = array{ array{1,2}, array{3,4} };

for (auto row : arr) {
    for (auto e : row) {
        std::cout << e << ",";
    std::cout << "\n";
tensor{ tensor{1,2}, tensor{1,2} }
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