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error C2783: '_Ty &&std::forward(remove_reference<_Ty>::type &&) throw()' : could not deduce templ


error C2783: '_Ty &&std::forward(remove_reference<_Ty>::type &&) throw()' : could not deduce templ

By : Prashant Singh
Date : November 17 2020, 11:55 AM
I wish did fix the issue. Value categories like "lvalue" and "rvalue" are properties of expressions. Expressions that name variables are always lvalue-expressions, even if they name a variable that has the type rvalue reference to some_type.
We use lvalue-reference and rvalue-references to bind different categories of expressions: Per convention, we treat lvalue-references as being bound to lvalues, and rvalue-references as being bound to rvalues.
code :
int   i = 42;
int&  l = i;
int&& r = 21;

l // this expression is an lvalue-expression
r // this expression is an lvalue-expression, too (!)

std::forward<int& >(l) // this function-call expression is an lvalue-expression
std::forward<int&&>(r) // this function-call expression is an rvalue-expression
void push(const T& item)
void push(const T& item) {
    // ...
    queue.push_back(item); // pass the lvalue argument as an lvalue
    // ...
}
void push(T&& item)
void push(T&& item) {
    // ...
    queue.push_back(std::move(item)); // pass the rvalue argument as an rvalue
    // ...
}
template<typename U>
void push(forwarding_reference<U> item)
template<typename U>
void push(forwarding_reference<U> item) // not C++, read on
{
    // ...
    queue.push_back(std::forward<U>(item)); // pass lvalue arguments as lvalues
                                            // and rvalue arguments as rvalues
    // ...
}
template<typename U>
void push(U&& item) {
    // ...
    queue.push_back(std::forward<U>(item)); // pass lvalue arguments as lvalues
                                            // and rvalue arguments as rvalues
    // ...
}


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Strange Template error : error C2783: could not deduce template argument

Strange Template error : error C2783: could not deduce template argument


By : Shusei Kamihasu
Date : March 29 2020, 07:55 AM
will help you There is no way for the compiler to deduce t3 from the function arguments. You need to pass this argument explicitly. Change the order of the parameters to make this possible
code :
template <typename t3, typename t1, typename t2> 
t3 adder1 (t1  a , t2 b)
    {
        return t3(a + b); // use t3 instead of fixed "int" here!
    };
template <typename t1, typename t2> 
auto adder1 (t1  a , t2 b) -> decltype(a+b)
    {
        return a + b;
    };
int sum = (int) adder1(1,6.0); // cast from double to int
template <typename t1, typename t2> 
typename promote<t1, t2>::type adder1 (t1 a, t2 b)
    {
        return (a + b);
    };
error C2783 Could not deduce template arguments

error C2783 Could not deduce template arguments


By : Marcin Obolewicz
Date : March 29 2020, 07:55 AM
will help you I am stuck with this error . I have found a workaround too but it kind of kills the whole purpose of exercise. , The error is actually quite clear, in the following template:
code :
template <typename T1,typename T2>
double Sum(const typename std::map<T1,T2>::const_iterator& input_begin,
           const typename std::map<T1,T2>::const_iterator& input_end)
namespace detail {
template <typename T> T valueOf( T const & t ) { return t; }
template <typename K, typename V>
V valueOf( std::pair<const K, V> const & p ) {
   return p.second;
}
}
template <typename Iterator>
double Sum( Iterator begin, Iterator end ) {
  double result = 0;
  for (; begin != end; ++begin) {
     result += detail::valueOf(*begin);
  }
  return result;
}
error C2783: could not deduce template argument for structure argument

error C2783: could not deduce template argument for structure argument


By : lasq
Date : March 29 2020, 07:55 AM
hope this fix your issue This is a compiler bug. If you replace the contents of myfunc with valid code (as suggested), it still doesn't work. For a description, status (and acknowledgement) of the bug, see Microsoft Connect. You might try to use a helper type to get argument deduction (which works):
code :
namespace ns1 {
    template <class T> class class_1 {
    public: class_1 (int a, int b){}
    };
}

namespace ns2 {
    template<class T> struct deduction_helper{};

    using namespace ns1;
    template <typename T> inline ns1::class_1<T> myfunc(deduction_helper<T>);

    template<typename T>
    inline ns1::class_1<T> myfunc(deduction_helper<T>) {
        int a,b;
        std::cin>>a;
        std::cin>>b;
        ns1::class_1<T> c(a,b); return c;
    }

}


namespace ns3 {
    struct myStruct {
        ns1::class_1<double> var1;
        ns1::class_1<double> var2;

        myStruct ( const ns1::class_1<double>& cl0= ns2::myfunc(ns2::deduction_helper<double>()),
                   const ns1::class_1<double>& cl1= ns2::myfunc(ns2::deduction_helper<double>())
                 ): var1(cl0), var2(cl1) {};
    };
}

int main()
{
    ns3::myStruct x;
}
Why std::forward() doesn't deduce type?

Why std::forward() doesn't deduce type?


By : user2343674
Date : March 29 2020, 07:55 AM
To fix this issue Here is code below. Why if I replace typename remove_reference::type& with S& it won't work nicely? (I mean a type will be deduced wrong) , Consider the original use case of forward:
code :
template<class T>
void f(T&& t) { g(std::forward<T>(t)); }
template <class T> constexpr T&& forward(remove_reference_t<T>&& t) noexcept;
How does std::forward deduce the type of `_Ty`?

How does std::forward deduce the type of `_Ty`?


By : R.rani
Date : October 12 2020, 09:00 PM
fixed the issue. Will look into that further I'm learning about templates and particularly std::forward; when I check its implementation it uses another class template std::remove_reference in its arguments list: ,
how does the compiler figure out the type of _Ty?
code :
template<typename T>
void foo(T&& t) {
    std::forward<T>(t);
}
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