Voided Check Template

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If you're writing a function template, prefer to write it as a single function template that should never be specialized or overloaded, and implement the function template entirely in terms of a. Here's roughly what i'm thinking: To add to those answers, the specification says (c++11 §8.3.5[dcl.func]/4): You can use overloading, but if a large part of the code would.

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I need a way to check if a templated class's type is void. To add to those answers, the specification says (c++11 §8.3.5[dcl.func]/4): This really sounds like a good idea though, if someone doesn't want to use type_traits. Was wondering if it's possible to have a template function that can branch depending on whether the type is derived from a.

17 Voided Check Templates [Free] TemplateArchive

Several answers already explain the rationale. I need a way to check if a templated class's type is void. It shows is_convertible is false from string_view to string.however, looking at how the stl does it in the header, they define a template typedef (of type _stringviewish) using _is_string_view_ish = enable_if_t<conjunction_v<is_convertiable<const _stringviewish&, basic_string_view<_elem, _traits>>, negation<is_convertable<const _stringviewish&, const. @gmannickg it's item 66.

17 Voided Check Templates [Free] TemplateArchive

I'm trying to detect whether a type (or more precisely, an identifier as technically a template is not a type) is a template or not, eg to have a function istemplate which behaves the following way: Here's roughly what i'm thinking: Compare vartype and typeid().name / typename. Jessen already mentioned in the comments , you have to look this oid.

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Like someone mentioned the main logic can be done in a different function, which accepts an extra flag to indicate the type, and this specialized declaration can just set the flag accordingly and directly pass on all the other arguments without touching anything. Here's roughly what i'm thinking: This really sounds like a good idea though, if someone doesn't want.

Voided Check Template - It does contain the extension certificate template information though, but i cannot parse a name from it, because it does only contain the oid of the template. This really sounds like a good idea though, if someone doesn't want to use type_traits. How to check if template type is of another template type? Compare vartype and typeid().name / typename. @gmannickg it's item 66 of sutter & alexandrescu's coding standards: Like someone mentioned the main logic can be done in a different function, which accepts an extra flag to indicate the type, and this specialized declaration can just set the flag accordingly and directly pass on all the other arguments without touching anything.

I need a way to check if a templated class's type is void. // should return true istemplate(); Jessen already mentioned in the comments , you have to look this oid up in your ad to get the template name. Compare vartype and typeid().name / typename. To add to those answers, the specification says (c++11 §8.3.5[dcl.func]/4):

If You're Writing A Function Template, Prefer To Write It As A Single Function Template That Should Never Be Specialized Or Overloaded, And Implement The Function Template Entirely In Terms Of A.

Jessen already mentioned in the comments , you have to look this oid up in your ad to get the template name. Compare vartype and typeid().name / typename. You can use overloading, but if a large part of the code would work for any type, you might consider extracting the differing part into a separate function and overload that. This really sounds like a good idea though, if someone doesn't want to use type_traits.

To Add To Those Answers, The Specification Says (C++11 §8.3.5[Dcl.func]/4):

Template <typename target_type, typename start_function_type, typename end_function_type> @gmannickg it's item 66 of sutter & alexandrescu's coding standards: Was wondering if it's possible to have a template function that can branch depending on whether the type is derived from a particular class. // should return true istemplate();

How To Check If Template Type Is Of Another Template Type?

Several answers already explain the rationale. Like someone mentioned the main logic can be done in a different function, which accepts an extra flag to indicate the type, and this specialized declaration can just set the flag accordingly and directly pass on all the other arguments without touching anything. Hmm because i had a large portion of same code until the 'specification' part. I'm trying to detect whether a type (or more precisely, an identifier as technically a template is not a type) is a template or not, eg to have a function istemplate which behaves the following way:

It Does Contain The Extension Certificate Template Information Though, But I Cannot Parse A Name From It, Because It Does Only Contain The Oid Of The Template.

I need a way to check if a templated class's type is void. It shows is_convertible is false from string_view to string.however, looking at how the stl does it in the header, they define a template typedef (of type _stringviewish) using _is_string_view_ish = enable_if_t<conjunction_v<is_convertiable<const _stringviewish&, basic_string_view<_elem, _traits>>, negation<is_convertable<const _stringviewish&, const. Here's roughly what i'm thinking: