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Angel Angel
Asked: 2020-12-07 00:02:03 +0800 CST 2020-12-07 00:02:03 +0800 CST

Are class and typename the same in the template context?

  • 772

I use:

template<class T>
Arbol<T>::Arbol(){

}

when Tyou are going to do a class.

and use:

template<typename T>
Arbol<T>::Arbol(){

}

when it is going Tto be a type, for example bool, that is, charfundamental types.

But looking at a code fragment of a program (not created by me) I saw that it used typenamewhere I usually use class, and until now I had not noticed that. if now you ask me why, well, I wouldn't be able to say one hundred percent why, but perhaps somehow I (imagine) that the compiler differentiated when it was a type included in the language and acted accordingly.

My question is:

Is it the same to use one or the other?

Well, now I've tested both of them and they work fine, but I don't know if I'm missing something, or maybe there's some subtle difference between classand typenamein this context.

c++

3 Answers

  1. Best Answer

    In this context, It is the same , there is no difference.

    They can be used interchangeably and exist for historical reasons.

    In this blog post by Stan Lippman, he explains that Stroustrup initially didn't want to introduce a new keyword and reused class. Until the ISO-C++ standard it was the only way to declare it.

    During the standardization process, it was discovered that this caused syntactic ambiguities (see in the blog) and it was decided to introduce a new keyword to solve it. this is typename. It was finally overloaded classto work as typenameto maintain backwards compatibility.

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  2. Regarding the use of typenameor classto declare the type-parameter of a template, in most cases using one or the other of the keywords is indifferent, but there are certain cases in which it is required to use one or the other.

    For example, using is requiredclass for the type of a template-parameter ( but this could change in C++17 ) and using is required typenameto help the compiler distinguish between a type or a value (when the type is dependent).

    Excerpt from Paper n4051 (the translation is mine):

    Allow typename in a template-parameter (template template)

    Since the introduction of template aliases, C++ has had templates that are not template classes, in particular it now has template-parameters that are not template classes. However, the syntax for template-parameters still requires the keyword class:

    template<typename T> struct A {};
template<typename T> using B = int;

template<template<typename> class X> struct C;
C<A> ca; // bien
C<B> cb; // bien, no es una clase plantilla
template<template<typename> typename X> struct D; // error, no se puede usar typename aqui

    In short, for the vast majority of cases classor typenameis indifferent, but typenameit is the only option to disambiguate dependent types and classis required for the template-parameter type; they are the only two cases where both keywords are not interchangeable, but they are not the most common uses.

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  3. When declaring a templatesimple it is indifferent to use classor typename. Both are exactly identical. However, for certain uses we will not have as much freedom of choice.

    If we need to make use of a type provided by the class/struct used to specialize the templatewe will have to redeclare the type in our template. In this case we will have to use typename:

    template<typename T>
class Foo
{
    typedef typename T::type newType;
};

template<class T>
class Foo
{
    typedef typename T::type newType;
};

    If one of the types of the templateturns out to be another template(for example a container) we will have to use classif we want the templatecompile:

    template< template<class> class Container, class NestedType>
void func(const Container<NestedType> &value)
{
  // ...
}
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