This question already has answers here: Why is it disallowed for partial specialization in a non-type argument to use nested template parameters
I am using a SFINAE mechanism to deduce a type. Resolve<T>::type is deduced to T if class T doesn\'t contain yes and it\'s deduced to MyClass if it contains yes.
I have a template, template <typename T> class wrapper, that I would like to specialize based on the existence of typename T::context_type. If typename T::context_type is declared, then the cons
With curiosity, I was trying an alternate implementation of is_class construct using the sizeof() trick. Following is the code:
Can we have a SFINAE trick to know, if the class has certain subclass/type. Something like, template<typename TYPE> // searches for \"my_type\"
Edit: Posted an answer of my own, kept the original accepted answer... got me thinking about aliases.
I decided to try my own hand at a bit of Substitution Failure Is Not A Erro开发者_JAVA技巧r (SFINAE) code to test if the global operator<< is defined for a custom type.
I\'ve got a (relatively) brief code sample here. #include <type_traits> template&开发者_运维问答lt;typename T> class function;
A (somewhat) outdated article explores ways to use decltype along with SFINAE to detect if a type supports certain operators, such as == or <.
The std::vector<T> class is a model of the STL Container concept, and as such any proper implementation of vector has to include a nested typedef value_type as well as reference.This should be d
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