Default template parameter with class
I've just found out about a strange syntax for default template parameters
template<class T = class Z>
struct X
{};
What does the second "class" keyword mean in this context?
c++ templates default-value
asked Jan 27 at 13:02
DunsDuns
905
add a comment |
I've just found out about a strange syntax for default template parameters
template<class T = class Z>
struct X
{};
What does the second "class" keyword mean in this context?
c++ templates default-value
asked Jan 27 at 13:02
DunsDuns
905
add a comment |
I've just found out about a strange syntax for default template parameters
template<class T = class Z>
struct X
{};
What does the second "class" keyword mean in this context?
c++ templates default-value
asked Jan 27 at 13:02
DunsDuns
905
I've just found out about a strange syntax for default template parameters
template<class T = class Z>
struct X
{};
What does the second "class" keyword mean in this context?
c++ templates default-value
c++ templates default-value
asked Jan 27 at 13:02
DunsDuns
905
asked Jan 27 at 13:02
DunsDuns
905
asked Jan 27 at 13:02
DunsDuns
905
asked Jan 27 at 13:02
DunsDuns
905
asked Jan 27 at 13:02
DunsDuns
905
905
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
It's nothing special really. C++ allows you to refer to a class via an elaborated type specifier. E.g.
void foo(class bar*);
This declares a function foo that accepts an argument of the type bar*. If bar was not declared previously, this elaborate type specifier constitutes a declaration of bar in the namespace containing foo. I.e. as if you had written:
class bar;
void foo(bar*);
Back to your example, X is a class template that expects a single type parameter, denoted by class T, but could have been denoted just the same as typename T. Said type parameter has a default argument, named by the elaborated class specifier class Z. That declaration can be rewritten just like the function above:
class Z;
template<class T = Z>
struct X
{};
answered Jan 27 at 13:08
StoryTellerStoryTeller
100k12204273
great explanation, thank you very much
– Duns
Jan 27 at 13:23
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It's nothing special really. C++ allows you to refer to a class via an elaborated type specifier. E.g.
void foo(class bar*);
This declares a function foo that accepts an argument of the type bar*. If bar was not declared previously, this elaborate type specifier constitutes a declaration of bar in the namespace containing foo. I.e. as if you had written:
class bar;
void foo(bar*);
Back to your example, X is a class template that expects a single type parameter, denoted by class T, but could have been denoted just the same as typename T. Said type parameter has a default argument, named by the elaborated class specifier class Z. That declaration can be rewritten just like the function above:
class Z;
template<class T = Z>
struct X
{};
answered Jan 27 at 13:08
StoryTellerStoryTeller
100k12204273
great explanation, thank you very much
– Duns
Jan 27 at 13:23
add a comment |
It's nothing special really. C++ allows you to refer to a class via an elaborated type specifier. E.g.
void foo(class bar*);
This declares a function foo that accepts an argument of the type bar*. If bar was not declared previously, this elaborate type specifier constitutes a declaration of bar in the namespace containing foo. I.e. as if you had written:
class bar;
void foo(bar*);
Back to your example, X is a class template that expects a single type parameter, denoted by class T, but could have been denoted just the same as typename T. Said type parameter has a default argument, named by the elaborated class specifier class Z. That declaration can be rewritten just like the function above:
class Z;
template<class T = Z>
struct X
{};
answered Jan 27 at 13:08
StoryTellerStoryTeller
100k12204273
great explanation, thank you very much
– Duns
Jan 27 at 13:23
add a comment |
It's nothing special really. C++ allows you to refer to a class via an elaborated type specifier. E.g.
void foo(class bar*);
This declares a function foo that accepts an argument of the type bar*. If bar was not declared previously, this elaborate type specifier constitutes a declaration of bar in the namespace containing foo. I.e. as if you had written:
class bar;
void foo(bar*);
Back to your example, X is a class template that expects a single type parameter, denoted by class T, but could have been denoted just the same as typename T. Said type parameter has a default argument, named by the elaborated class specifier class Z. That declaration can be rewritten just like the function above:
class Z;
template<class T = Z>
struct X
{};
answered Jan 27 at 13:08
StoryTellerStoryTeller
100k12204273
It's nothing special really. C++ allows you to refer to a class via an elaborated type specifier. E.g.
void foo(class bar*);
This declares a function foo that accepts an argument of the type bar*. If bar was not declared previously, this elaborate type specifier constitutes a declaration of bar in the namespace containing foo. I.e. as if you had written:
class bar;
void foo(bar*);
Back to your example, X is a class template that expects a single type parameter, denoted by class T, but could have been denoted just the same as typename T. Said type parameter has a default argument, named by the elaborated class specifier class Z. That declaration can be rewritten just like the function above:
class Z;
template<class T = Z>
struct X
{};
answered Jan 27 at 13:08
StoryTellerStoryTeller
100k12204273
answered Jan 27 at 13:08
StoryTellerStoryTeller
100k12204273
answered Jan 27 at 13:08
StoryTellerStoryTeller
100k12204273
answered Jan 27 at 13:08
StoryTellerStoryTeller
100k12204273
100k12204273
great explanation, thank you very much
– Duns
Jan 27 at 13:23
add a comment |
great explanation, thank you very much
– Duns
Jan 27 at 13:23
great explanation, thank you very much
– Duns
Jan 27 at 13:23
great explanation, thank you very much
– Duns
Jan 27 at 13:23
add a comment |
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