Why can't you define a template with a struct?

Question

Pretty straight-forward:

template <class T>
void foo() {}  //compiles

template <struct T>
void goo() {}  //doesn't

Why?

score 6 · Accepted Answer · edited May 23 '17 at 12:19

6

class is only a keyword to indicate that T is a type. It does not actually mean T must be a class. (e.g. you can call foo<int>().)

The class keyword was used because typename didn't exist when the template feature was added. See Templates in c++,typename and class.

edited May 23 '17 at 12:19

Community

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answered Mar 12 '12 at 18:26

kennytm

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Ah. So you're saying class and typename are interchangeable with no side-effects whatsoever? +1 from me. – Luchian Grigore Mar 12 '12 at 18:29
@LuchianGrigore: Almost. You have to use `class` in a template-template parameter `template – kennytm Mar 12 '12 at 18:31
` – Luchian Grigore Mar 12 '12 at 18:33
@LuchianGrigore: Well if you don't know what it is, then yes `class` and `typename` are interchangeable inside the `<...>` ;). See http://stackoverflow.com/a/6485439/224671 for an explanation of template-template parameters. – kennytm Mar 12 '12 at 18:40

score 2 · Answer 2 · answered Mar 12 '12 at 18:26

2

Because a template has to be declared with either class or...

template <typename T>
void foo() {}

answered Mar 12 '12 at 18:26

DevSolar

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I'm looking more for the reasoning behind this specification, rather than just "because that's what the standard says". – Luchian Grigore Mar 12 '12 at 18:28
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@LuchianGrigore: There is not much to reason about... there is a need for a keyword that identifies a placeholder for a *type*, and while defining the language they could have decided basically *anything*, for example `template `, but they chose to use `class` (reuse of an existing keyword to avoid incrementing the number of reserved keywords). Note that this use of `class` is not synonym for `class` in a type declaration, which is the use of `class` that is in turn similar to `struct`. – David Rodríguez - dribeas Mar 12 '12 at 18:31
@DavidRodríguez-dribeas Actually Kenny's answer provides a pretty good explanation. Even your comment provides more explanation. – Luchian Grigore Mar 12 '12 at 18:33

score 1 · Answer 3 · answered Mar 12 '12 at 18:37

Because, the grammar forbids it:

template-declaration: 

export_opt template< template-parameter-list > declaration

template-parameter-list:
template-parameter
parameter-declaration

type-parameter:

class identifier
class identifier = type-id
typename identifier
typename identifier = type-id
template < template-parameter-list > class identifier
template < template-parameter-list > class identifier = template-name

Alternatively , you can think that struct is a special case of a class where the members are all public. Thus, allowing it in the template-parameter-list would not be as general.

Why can't you define a template with a struct?

3 Answers3