define implicit conversion from template class to primitive type based on other template

define implicit conversion from template class to primitive type based on other template

By : Tristan McCullen
Date : November 21 2020, 03:00 PM
it should still fix some issue You can use a static_assert to check if the conversion should be allowed:
code :
operator A() 
    static_assert(std::is_same<B, specific_B_type>::value, "No conversion possible");    
    return this->value;

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C++ Template : Choosing overloaded functions using implicit conversion to template instantiated type!

C++ Template : Choosing overloaded functions using implicit conversion to template instantiated type!

By : DME
Date : March 29 2020, 07:55 AM
Hope this helps Consider these overloaded functions, , You can do it a little differently:
code :
template<int N> void func();
template<> void func<1>(){/*the body*/}
template<> void func<2>(){/*the body*/}
Implicit type conversion with template

Implicit type conversion with template

By : Awais Shaikh
Date : March 29 2020, 07:55 AM
will be helpful for those in need The solution is already shown in this answer. Now, more about the problem...
The problem in your code is how overload resolution is performed. When a template function is considered for overload resolution the compiler will perform type deduction on the arguments and come up with a type substitution that matches the call or else it fails to apply that template, removes it from the set of potential candidates and continues over. The problem at this point is that type deduction only deduces exact matches (with possible extra const/volatile qualification). Because the matching is exact, the compiler will not use any conversion (again, other than cv).
code :
unsigned int i = 0;
std::min( i, 10 );    // Error! 
template <typename T>
class test {
    friend test operator+( test const & lhs, test const & rhs ) {  // [1]
        return test();
test<int> t;                                                       // [2]
test<int> operator+( test<int> const & lhs, test<int> const & rhs ) { 
   return test<int>();
How to condition a class template and primitive type template

How to condition a class template and primitive type template

By : hozour
Date : March 29 2020, 07:55 AM
seems to work fine I am trying to build a class template that inherits the class. If the class template is given a primitive type as template argument, then it gives an illegal inheritance error. I tried doing , This is where a template specialization comes in handy.
code :
#include <type_traits>

template <class Class_, bool = std::is_class<Class_>::value>
struct EndianGuard_ : public Class_ 
    // inherit if class

template <class Class_>
struct EndianGuard_<Class_, false> 
    // don't inherit if not a class

struct foo {};

int main() 
    EndianGuard_<int> f; //ok
    EndianGuard_<foo> f2; // ok
Implicit conversion between template type

Implicit conversion between template type

By : user2394269
Date : March 29 2020, 07:55 AM
wish help you to fix your issue User-defined conversions are never considered when attempting to match up function argument types with function parameter types for template argument deduction. So this issue is a somewhat more complicated version of this sort of error:
code :
template <int> struct X {};
struct Y {
    operator X<2> () const { return {}; }
template <int N>
void f(X<N>) {}

int main()
    Y y;
    f(y); // Error: Cannot deduce template argument.
#include <type_traits>

template<bool v>
struct bool_t {
    // (Add some appropriate SFINAE.)
    template<template<int> typename T>
    constexpr T<v ? 1 : 0> convert_template() const {
        return {};

template<typename T, template<int> class TT>
struct type_specializes : std::false_type {};

template<template<int> class TT, int N>
struct type_specializes<TT<N>, TT> : std::true_type {};

template<int N>
struct number_t {
    // Allow "conversion" to my own template:
    template<template<int> typename T>
    constexpr std::enable_if_t<type_specializes<number_t, T>::value, number_t>
    convert_template() const { return {}; }

    // Used only in decltype; no definition needed.
    template<int n1, int n2>
    static number_t<n1 + n2> sum_impl(number_t<n1>, number_t<n2>);

    template<typename T1, typename T2>
    friend auto operator+(T1&& x, T2&& y)
        -> decltype(number_t::sum_impl(
             x.template convert_template<number_t>(),
             y.template convert_template<number_t>()))
    { return {}; }

int main() {
    number_t<0>{} + bool_t<0>{};   
Implicit type conversion in c++ template

Implicit type conversion in c++ template

By : Jthomas0385
Date : March 29 2020, 07:55 AM
With these it helps I have a function template: ,
can a be converted to double automatically?
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