What is the purpose of the class = void in the following code snippets?
template< class, class = void >
struct has_type_member : false_type { };
template< class T >
struct has_type_member<T, void_t<typename T::type>> : true_type { };
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6it is a default for the parameter – 463035818_is_not_an_ai May 17 '17 at 12:49
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it will work even without `class = void`, no? – BЈовић May 17 '17 at 12:51
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1@BЈовић No it won't. – Barry May 17 '17 at 14:11
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@Barry Because it is called with 2 template arguments? The OP didn't put how it is used. – BЈовић May 18 '17 at 07:18
1 Answers
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template< class, class = void >
struct has_type_member : false_type { };
That's your default struct template, it asks for 2 template arguments but the second one is set to void as default, so this argument does not need to be specified explicitly, somewhat like a default function parameter.
Then :
template< class T >
struct has_type_member<T, void_t<typename T::type>> : true_type { };
Is a template specialization for your has_type_member struct, SFINAE will rule out this specialization if T::type doesn't exist (and thus, is invalid syntax), if it does exist it will pick this specialization otherwise.
The 2nd parameter is necessary to be used for template specialization, but we don't use it in our "fallback" struct so we just default to void.
Hatted Rooster
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1Strictly speaking, one could do `class = std::void_t<>` to hint at what goes on. – StoryTeller - Unslander Monica May 17 '17 at 13:04
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1@laike9m you need the `class` so that it is a 2 parameter template, for the specialisation. You want the `= void` so that users of the template only pass the type they are interested in. – Caleth Sep 20 '21 at 08:25