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Why is template type deduction failing here?


Why is template type deduction failing here?

By : user2954610
Date : November 22 2020, 10:31 AM
Hope that helps Template arguments for types cannot be deduced, period. This is because type specializations can provide a completely different set of members, so until you know the type's template arguments you don't even know what constructors are available.
This is why the standard library has helper functions, like std::make_pair. When constructing or referring to an std::pair you must specify the type arguments (std::pair(1, 2.0)) but in most cases they can be deduced when calling a function (std::make_pair(1, 2.0)). (Note that std::condition_variable::wait is a function, not a type.)
code :
template <typename Predicate>
lock_monitor_guard<Predicate> create_lock_monitor_guard(
    std::mutex & mutex,
    std::condition_variable & monitor,
    Predicate && predicate)
{
    return lock_monitor_guard<Predicate>(mutex, monitor, std::forward<Predicate>(predicate));
}

auto guard = create_lock_monitor_guard(
    jobs_mutex,
    jobs_monitor,
    [] { return true; });


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Template type deduction failing

Template type deduction failing


By : Shubham Singh
Date : March 29 2020, 07:55 AM
I hope this helps . I have the feeling you are misusing boost::is_base_of. The nested type will be either true_type or false_type. Neither of those two make sense to take as an argument and your pointers will not be convertible to those.
What you really want:
code :
#include <boost/type_traits/is_base_of.hpp>
#include <boost/utility/enable_if.hpp>

class MsgContent {};

class RecipientList
{
public:
    template<class T>
    typename boost::enable_if<
        typename boost::is_base_of<MsgContent, T>::type
      , void>::type
    fillMessageWithRecipients(T* t) { }
};

class SomeMsg : public MsgContent {};

int main()
{
  RecipientList recipients;
  SomeMsg m;
  recipients.fillMessageWithRecipients( &m );

  return 0;
}
Function pointer as template argument, type deduction failing

Function pointer as template argument, type deduction failing


By : Ian Howard
Date : March 29 2020, 07:55 AM
wish helps you The mistake that you provide partial specialization of template. There works rule all or nothing. So if you change signature as follow:
code :
template <class T, class R>
 R reduce(R (*Func)(T, R), T *in, R initial, int len) {
reduce(sum, data, 0.0, 5)
template type deduction failing (std::empty as a predicate)

template type deduction failing (std::empty as a predicate)


By : Bits Nibble Bytes
Date : March 29 2020, 07:55 AM
Hope that helps Unfortunately there is a problem to distinguish overloaded template functions as std::all_of is also a template function. Better explaination: std::function fails to distinguish overloaded functions
So to do this, a static_cast to the correct function type: bool ( * )( const std::vector< int >& ) is required:
code :
std::all_of( vv.begin(), vv.end(),
             static_cast< bool ( * )( const std::vector< int >& ) >( std::empty ) );
template< typename C >
inline auto overloaded_pred_for( const C&, bool ( *f )( const C& ) ) -> decltype( f )
{
    return f;
}
std::all_of( vv.begin(), vv.end(), 
             overloaded_pred_for( std::vector< int >(), std::empty ) );
What is the difference between auto deduction and template type deduction?

What is the difference between auto deduction and template type deduction?


By : Erin Poling
Date : March 29 2020, 07:55 AM
around this issue auto type deduction is usually the same as template type deduction, but auto type deduction assumes that a braced initializer represents a std::initializer_list, and template type deduction doesn’t.
When an auto–declared variable is initialized with a braced initializer, the deduced type is an instantiation of std::initializer_list. But if the corresponding template is passed the same initializer, type deduction fails, and the code is rejected:
code :
auto x = { 11, 23, 9 }; // x's type is     
                       //std::initializer_list<int>
template<typename T> // template with parameter
void f(T param); // template with parameter
template<typename T>
void f(std::initializer_list<T> initList);
f({ 11, 23, 9 }); // T deduced as int, and initList's
 // type is std::initializer_list<int>
Template Alias, Variable Template, and auto type deduction failing to deduce template argument

Template Alias, Variable Template, and auto type deduction failing to deduce template argument


By : user3004155
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further First of all, atTime is static method so only way to call it is to use scope resolution operator ::. atTime takes no arguments, so T cannot be deduced, and you need to put type in template arguments list explicitly:
code :
auto time1 = Signal::atTime<DiscreteTime>();
auto time2 = Signal::atTime<ContinuousTime>();
template<typename T>
explicit Signal( TimeSignal<T> time, double voltage = 0, double current = 0 ) :
    voltage_{voltage}, current_{current}
{ time_<T> = time; }
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