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 1 hour ago
DunsDuns
604
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 1 hour ago
DunsDuns
604
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 1 hour ago
DunsDuns
604
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 1 hour ago
DunsDuns
604
asked 1 hour ago
DunsDuns
604
asked 1 hour ago
DunsDuns
604
asked 1 hour ago
DunsDuns
604
asked 1 hour ago
DunsDuns
604
604
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 56 mins ago
StoryTellerStoryTeller
96.5k12196261
great explanation, thank you very much
– Duns
42 mins ago
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 56 mins ago
StoryTellerStoryTeller
96.5k12196261
great explanation, thank you very much
– Duns
42 mins ago
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 56 mins ago
StoryTellerStoryTeller
96.5k12196261
great explanation, thank you very much
– Duns
42 mins ago
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 56 mins ago
StoryTellerStoryTeller
96.5k12196261
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 56 mins ago
StoryTellerStoryTeller
96.5k12196261
answered 56 mins ago
StoryTellerStoryTeller
96.5k12196261
answered 56 mins ago
StoryTellerStoryTeller
96.5k12196261
answered 56 mins ago
StoryTellerStoryTeller
96.5k12196261
96.5k12196261
great explanation, thank you very much
– Duns
42 mins ago
add a comment |
great explanation, thank you very much
– Duns
42 mins ago
great explanation, thank you very much
– Duns
42 mins ago
great explanation, thank you very much
– Duns
42 mins ago
add a comment |
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