Issue with calling static template method inside a template class
I have a template class and a static template method inside. I also have a simple function g, that just allows user to call this method in an appropriate way. This code gives me an error "expected primary-expression before '>' token" inside g function. But if I call a method inside g function like this:
return A<types::a>::f<T>();
In that case code compiles fine and gives no error. How can I possibly fix this issue and what is the problem?
enum class types : uint8_t
{
a, b
};
template<types type>
struct A {
template<typename T>
static T f();
};
template<>
template<typename T>
T A<types::a>::f() {
cout << "a" << endl;
return T{};
}
template<>
template<typename T>
T A<types::b>::f() {
cout << "b" << endl;
return T{};
}
template<types type, typename T>
T g() {
return A<type>::f<T>();
}
int main() {
g<types::a, int>();
}
c++
asked Jan 18 at 12:50
Никита ЛарионовНикита Ларионов
254
New contributor
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I have a template class and a static template method inside. I also have a simple function g, that just allows user to call this method in an appropriate way. This code gives me an error "expected primary-expression before '>' token" inside g function. But if I call a method inside g function like this:
return A<types::a>::f<T>();
In that case code compiles fine and gives no error. How can I possibly fix this issue and what is the problem?
enum class types : uint8_t
{
a, b
};
template<types type>
struct A {
template<typename T>
static T f();
};
template<>
template<typename T>
T A<types::a>::f() {
cout << "a" << endl;
return T{};
}
template<>
template<typename T>
T A<types::b>::f() {
cout << "b" << endl;
return T{};
}
template<types type, typename T>
T g() {
return A<type>::f<T>();
}
int main() {
g<types::a, int>();
}
c++
asked Jan 18 at 12:50
Никита ЛарионовНикита Ларионов
254
New contributor
Никита Ларионов is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Istypesan enum?
– Max Langhof
Jan 18 at 12:51
Yes, it's an enum. enum class types : uint8_t { a, b };
– Никита Ларионов
Jan 18 at 12:54
add a comment |
I have a template class and a static template method inside. I also have a simple function g, that just allows user to call this method in an appropriate way. This code gives me an error "expected primary-expression before '>' token" inside g function. But if I call a method inside g function like this:
return A<types::a>::f<T>();
In that case code compiles fine and gives no error. How can I possibly fix this issue and what is the problem?
enum class types : uint8_t
{
a, b
};
template<types type>
struct A {
template<typename T>
static T f();
};
template<>
template<typename T>
T A<types::a>::f() {
cout << "a" << endl;
return T{};
}
template<>
template<typename T>
T A<types::b>::f() {
cout << "b" << endl;
return T{};
}
template<types type, typename T>
T g() {
return A<type>::f<T>();
}
int main() {
g<types::a, int>();
}
c++
asked Jan 18 at 12:50
Никита ЛарионовНикита Ларионов
254
New contributor
Никита Ларионов is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a template class and a static template method inside. I also have a simple function g, that just allows user to call this method in an appropriate way. This code gives me an error "expected primary-expression before '>' token" inside g function. But if I call a method inside g function like this:
return A<types::a>::f<T>();
In that case code compiles fine and gives no error. How can I possibly fix this issue and what is the problem?
enum class types : uint8_t
{
a, b
};
template<types type>
struct A {
template<typename T>
static T f();
};
template<>
template<typename T>
T A<types::a>::f() {
cout << "a" << endl;
return T{};
}
template<>
template<typename T>
T A<types::b>::f() {
cout << "b" << endl;
return T{};
}
template<types type, typename T>
T g() {
return A<type>::f<T>();
}
int main() {
g<types::a, int>();
}
c++
c++
asked Jan 18 at 12:50
Никита ЛарионовНикита Ларионов
254
New contributor
Никита Ларионов is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 18 at 12:50
Никита ЛарионовНикита Ларионов
254
New contributor
Никита Ларионов is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 18 at 12:55
Никита Ларионов
asked Jan 18 at 12:50
Никита ЛарионовНикита Ларионов
254
New contributor
Никита Ларионов is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 18 at 12:50
Никита ЛарионовНикита Ларионов
254
asked Jan 18 at 12:50
Никита ЛарионовНикита Ларионов
254
254
New contributor
Никита Ларионов is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Никита Ларионов is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Никита Ларионов is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Istypesan enum?
– Max Langhof
Jan 18 at 12:51
Yes, it's an enum. enum class types : uint8_t { a, b };
– Никита Ларионов
Jan 18 at 12:54
add a comment |
Istypesan enum?
– Max Langhof
Jan 18 at 12:51
Yes, it's an enum. enum class types : uint8_t { a, b };
– Никита Ларионов
Jan 18 at 12:54
Is
types an enum?– Max Langhof
Jan 18 at 12:51
Is
types an enum?– Max Langhof
Jan 18 at 12:51
Yes, it's an enum. enum class types : uint8_t { a, b };
– Никита Ларионов
Jan 18 at 12:54
Yes, it's an enum. enum class types : uint8_t { a, b };
– Никита Ларионов
Jan 18 at 12:54
add a comment |
1 Answer
1
active
oldest
votes
You need to add a template keyword:
template<types type, typename T>
T g() {
return A<type>::template f<T>();
}
Demo
A<type>::f is a dependent name (depends on template parameter type). To correctly parse this statement when the template definition is encountered (as required by two-phase-lookup rules), the compiler has to know before instantiation whether f is a variable, a type or a template. It defaults to "it's a variable"; add template to denote a template or typename to denote a type.
Your other case works because in return A<types::a>::f<T>(); there is no dependent name.
answered Jan 18 at 12:53
Max LanghofMax Langhof
9,2751537
Thank you. That solved my problem.
– Никита Ларионов
Jan 18 at 12:57
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
You need to add a template keyword:
template<types type, typename T>
T g() {
return A<type>::template f<T>();
}
Demo
A<type>::f is a dependent name (depends on template parameter type). To correctly parse this statement when the template definition is encountered (as required by two-phase-lookup rules), the compiler has to know before instantiation whether f is a variable, a type or a template. It defaults to "it's a variable"; add template to denote a template or typename to denote a type.
Your other case works because in return A<types::a>::f<T>(); there is no dependent name.
answered Jan 18 at 12:53
Max LanghofMax Langhof
9,2751537
Thank you. That solved my problem.
– Никита Ларионов
Jan 18 at 12:57
add a comment |
You need to add a template keyword:
template<types type, typename T>
T g() {
return A<type>::template f<T>();
}
Demo
A<type>::f is a dependent name (depends on template parameter type). To correctly parse this statement when the template definition is encountered (as required by two-phase-lookup rules), the compiler has to know before instantiation whether f is a variable, a type or a template. It defaults to "it's a variable"; add template to denote a template or typename to denote a type.
Your other case works because in return A<types::a>::f<T>(); there is no dependent name.
answered Jan 18 at 12:53
Max LanghofMax Langhof
9,2751537
Thank you. That solved my problem.
– Никита Ларионов
Jan 18 at 12:57
add a comment |
You need to add a template keyword:
template<types type, typename T>
T g() {
return A<type>::template f<T>();
}
Demo
A<type>::f is a dependent name (depends on template parameter type). To correctly parse this statement when the template definition is encountered (as required by two-phase-lookup rules), the compiler has to know before instantiation whether f is a variable, a type or a template. It defaults to "it's a variable"; add template to denote a template or typename to denote a type.
Your other case works because in return A<types::a>::f<T>(); there is no dependent name.
answered Jan 18 at 12:53
Max LanghofMax Langhof
9,2751537
You need to add a template keyword:
template<types type, typename T>
T g() {
return A<type>::template f<T>();
}
Demo
A<type>::f is a dependent name (depends on template parameter type). To correctly parse this statement when the template definition is encountered (as required by two-phase-lookup rules), the compiler has to know before instantiation whether f is a variable, a type or a template. It defaults to "it's a variable"; add template to denote a template or typename to denote a type.
Your other case works because in return A<types::a>::f<T>(); there is no dependent name.
answered Jan 18 at 12:53
Max LanghofMax Langhof
9,2751537
edited Jan 18 at 12:57
answered Jan 18 at 12:53
Max LanghofMax Langhof
9,2751537
answered Jan 18 at 12:53
Max LanghofMax Langhof
9,2751537
answered Jan 18 at 12:53
Max LanghofMax Langhof
9,2751537
9,2751537
Thank you. That solved my problem.
– Никита Ларионов
Jan 18 at 12:57
add a comment |
Thank you. That solved my problem.
– Никита Ларионов
Jan 18 at 12:57
Thank you. That solved my problem.
– Никита Ларионов
Jan 18 at 12:57
Thank you. That solved my problem.
– Никита Ларионов
Jan 18 at 12:57
add a comment |
Никита Ларионов is a new contributor. Be nice, and check out our Code of Conduct.
Никита Ларионов is a new contributor. Be nice, and check out our Code of Conduct.
Никита Ларионов is a new contributor. Be nice, and check out our Code of Conduct.
Никита Ларионов is a new contributor. Be nice, and check out our Code of Conduct.
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Is
typesan enum?– Max Langhof
Jan 18 at 12:51
Yes, it's an enum. enum class types : uint8_t { a, b };
– Никита Ларионов
Jan 18 at 12:54