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Templated operator[]... possible? Useful?


By : Amaan
Date : October 16 2020, 11:12 AM
I think the issue was by ths following , Could you have: , You can also do
code :


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How to call a templated operator overload without using the word 'operator'?


By : Gabriel Aldamiz-eche
Date : March 29 2020, 07:55 AM
This might help you The most comfortable thing is to let template argument deduction work for you:
code :
struct Results {   
    template<typename Act> int& operator()(Act) { /* ... */ }
};

results(RunAround()) = /* ... */;

variadic templated overload of operator [] in templated class


By : darrett
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further operator[] can only take one argument, the easiest solution is to overload operator() instead and access members via () instead of [].
The proper syntax would then be:
code :
    template <class T>
    template <class ... I>
    T& MultiDimArray<T>::operator()(I ... i)
    {
    }

Functors: templated struct vs templated operator()


By : Dariusz Skwarek
Date : March 29 2020, 07:55 AM
seems to work fine When the original functors were created none of the needed language facilities (return type deduction, perfect forwarding) did exist to solve the problem. The current design also has the benefit of allowing users to specialize the functors for their own types, even if this should strictly not be necessary.
C++1y introduces void specializations for all the functors (and uses void as a default argument) to make them easier to use. They will deduce and perfectly forward arguments and return types. This will also allow heterogeneous comparisons.
code :
std::sort(begin(v), end(v), less<>());

Unhide templated cast operator from templated base class


By : Roy Lo
Date : March 29 2020, 07:55 AM
help you fix your problem A using-declaration cannot refer to a template-id, to a namespace, to a scoped enumerator, to a destructor of a base class or to a specialization of a member template for a user-defined conversion function.

Can a non-commutative operator for nested templated class inside a templated class be defined?


By : బ్రహ్మ రెడ్డి
Date : March 29 2020, 07:55 AM
To fix this issue I have this code: , You can move them into class A as friends:
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