Overloading of a non-member operator for a templated inner class
I cannot figure out why the operator<< ( std::ostream & os, typename A<T>::B const& b ) can't be seen or used by the compiler.
#include <vector>
#include <iostream>
template <typename T>
struct A
{
struct B
{
std::vector<T> innervec;
};
std::vector<B> outervec;
};
template <typename T>
auto operator<< ( std::ostream & os, typename A<T>::B const& b ) -> std::ostream&
{
for (auto e : b.innvervec)
os << "nt" << e;
return os;
}
template <typename T>
auto operator<< ( std::ostream & os, A<T> const& a ) -> std::ostream&
{
for (auto e : a.outervec)
os << 'n' << e;
return os;
}
int main()
{
A<int> a;
A<int>::B b1;
A<int>::B b2;
b1.innervec.push_back( 11 );
b1.innervec.push_back( 12 );
b2.innervec.push_back( 21 );
b2.innervec.push_back( 22 );
a.outervec.push_back( b1 );
a.outervec.push_back( b2 );
std::cout << a << std::endl;
}
gives me the following error on VC++15:
error C2679: binary '<<': no operator found which takes a right-hand operand of type 'A<int>::B' (or there is no acceptable conversion)
And it doesn't compile on GCC as well (only tried with an online compiler though).
I suppose that a rule about scope and deduction is involved, but I couldn't find out which precisely.
c++ templates
asked Nov 25 '18 at 19:51
TarquiscaniTarquiscani
12119
add a comment |
I cannot figure out why the operator<< ( std::ostream & os, typename A<T>::B const& b ) can't be seen or used by the compiler.
#include <vector>
#include <iostream>
template <typename T>
struct A
{
struct B
{
std::vector<T> innervec;
};
std::vector<B> outervec;
};
template <typename T>
auto operator<< ( std::ostream & os, typename A<T>::B const& b ) -> std::ostream&
{
for (auto e : b.innvervec)
os << "nt" << e;
return os;
}
template <typename T>
auto operator<< ( std::ostream & os, A<T> const& a ) -> std::ostream&
{
for (auto e : a.outervec)
os << 'n' << e;
return os;
}
int main()
{
A<int> a;
A<int>::B b1;
A<int>::B b2;
b1.innervec.push_back( 11 );
b1.innervec.push_back( 12 );
b2.innervec.push_back( 21 );
b2.innervec.push_back( 22 );
a.outervec.push_back( b1 );
a.outervec.push_back( b2 );
std::cout << a << std::endl;
}
gives me the following error on VC++15:
error C2679: binary '<<': no operator found which takes a right-hand operand of type 'A<int>::B' (or there is no acceptable conversion)
And it doesn't compile on GCC as well (only tried with an online compiler though).
I suppose that a rule about scope and deduction is involved, but I couldn't find out which precisely.
c++ templates
asked Nov 25 '18 at 19:51
TarquiscaniTarquiscani
12119
add a comment |
I cannot figure out why the operator<< ( std::ostream & os, typename A<T>::B const& b ) can't be seen or used by the compiler.
#include <vector>
#include <iostream>
template <typename T>
struct A
{
struct B
{
std::vector<T> innervec;
};
std::vector<B> outervec;
};
template <typename T>
auto operator<< ( std::ostream & os, typename A<T>::B const& b ) -> std::ostream&
{
for (auto e : b.innvervec)
os << "nt" << e;
return os;
}
template <typename T>
auto operator<< ( std::ostream & os, A<T> const& a ) -> std::ostream&
{
for (auto e : a.outervec)
os << 'n' << e;
return os;
}
int main()
{
A<int> a;
A<int>::B b1;
A<int>::B b2;
b1.innervec.push_back( 11 );
b1.innervec.push_back( 12 );
b2.innervec.push_back( 21 );
b2.innervec.push_back( 22 );
a.outervec.push_back( b1 );
a.outervec.push_back( b2 );
std::cout << a << std::endl;
}
gives me the following error on VC++15:
error C2679: binary '<<': no operator found which takes a right-hand operand of type 'A<int>::B' (or there is no acceptable conversion)
And it doesn't compile on GCC as well (only tried with an online compiler though).
I suppose that a rule about scope and deduction is involved, but I couldn't find out which precisely.
c++ templates
asked Nov 25 '18 at 19:51
TarquiscaniTarquiscani
12119
I cannot figure out why the operator<< ( std::ostream & os, typename A<T>::B const& b ) can't be seen or used by the compiler.
#include <vector>
#include <iostream>
template <typename T>
struct A
{
struct B
{
std::vector<T> innervec;
};
std::vector<B> outervec;
};
template <typename T>
auto operator<< ( std::ostream & os, typename A<T>::B const& b ) -> std::ostream&
{
for (auto e : b.innvervec)
os << "nt" << e;
return os;
}
template <typename T>
auto operator<< ( std::ostream & os, A<T> const& a ) -> std::ostream&
{
for (auto e : a.outervec)
os << 'n' << e;
return os;
}
int main()
{
A<int> a;
A<int>::B b1;
A<int>::B b2;
b1.innervec.push_back( 11 );
b1.innervec.push_back( 12 );
b2.innervec.push_back( 21 );
b2.innervec.push_back( 22 );
a.outervec.push_back( b1 );
a.outervec.push_back( b2 );
std::cout << a << std::endl;
}
gives me the following error on VC++15:
error C2679: binary '<<': no operator found which takes a right-hand operand of type 'A<int>::B' (or there is no acceptable conversion)
And it doesn't compile on GCC as well (only tried with an online compiler though).
I suppose that a rule about scope and deduction is involved, but I couldn't find out which precisely.
c++ templates
c++ templates
asked Nov 25 '18 at 19:51
TarquiscaniTarquiscani
12119
asked Nov 25 '18 at 19:51
TarquiscaniTarquiscani
12119
asked Nov 25 '18 at 19:51
TarquiscaniTarquiscani
12119
asked Nov 25 '18 at 19:51
TarquiscaniTarquiscani
12119
asked Nov 25 '18 at 19:51
TarquiscaniTarquiscani
12119
12119
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The error is because the overload for the nested type contain a non-deduced context parameter typename A<T>::B const& b and for operators you can't provide the explicit template argument T, you need to define the operator as friend for the nested A<T>::B as:
template <typename T>
struct A
{
struct B
{
std::vector<T> innervec;
friend auto operator<< ( std::ostream & os, B const& b ) -> std::ostream&
{
for (auto e : b.innervec)
os << "nt" << e;
return os;
}
};
std::vector<B> outervec;
};
answered Nov 25 '18 at 19:57
JansJans
8,89422635
No need for friendship, just for membership
– Géza Török
Nov 25 '18 at 20:05
Thanks :) I didn't know about non-deduced contexts.
– Tarquiscani
Nov 25 '18 at 20:09
@GézaTörök stream operators can't be members.
– Jans
Nov 25 '18 at 20:09
add a comment |
The problematic part is A<T>::B in the parameter list, as the compiler cannot deduce type expressions of this form, since existence and type of member B depends on T. I would take the type of b argument as template argument T directly, and use SFINAE to constrain accepted types.
answered Nov 25 '18 at 20:00
Géza TörökGéza Török
1,222515
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The error is because the overload for the nested type contain a non-deduced context parameter typename A<T>::B const& b and for operators you can't provide the explicit template argument T, you need to define the operator as friend for the nested A<T>::B as:
template <typename T>
struct A
{
struct B
{
std::vector<T> innervec;
friend auto operator<< ( std::ostream & os, B const& b ) -> std::ostream&
{
for (auto e : b.innervec)
os << "nt" << e;
return os;
}
};
std::vector<B> outervec;
};
answered Nov 25 '18 at 19:57
JansJans
8,89422635
No need for friendship, just for membership
– Géza Török
Nov 25 '18 at 20:05
Thanks :) I didn't know about non-deduced contexts.
– Tarquiscani
Nov 25 '18 at 20:09
@GézaTörök stream operators can't be members.
– Jans
Nov 25 '18 at 20:09
add a comment |
The error is because the overload for the nested type contain a non-deduced context parameter typename A<T>::B const& b and for operators you can't provide the explicit template argument T, you need to define the operator as friend for the nested A<T>::B as:
template <typename T>
struct A
{
struct B
{
std::vector<T> innervec;
friend auto operator<< ( std::ostream & os, B const& b ) -> std::ostream&
{
for (auto e : b.innervec)
os << "nt" << e;
return os;
}
};
std::vector<B> outervec;
};
answered Nov 25 '18 at 19:57
JansJans
8,89422635
No need for friendship, just for membership
– Géza Török
Nov 25 '18 at 20:05
Thanks :) I didn't know about non-deduced contexts.
– Tarquiscani
Nov 25 '18 at 20:09
@GézaTörök stream operators can't be members.
– Jans
Nov 25 '18 at 20:09
add a comment |
The error is because the overload for the nested type contain a non-deduced context parameter typename A<T>::B const& b and for operators you can't provide the explicit template argument T, you need to define the operator as friend for the nested A<T>::B as:
template <typename T>
struct A
{
struct B
{
std::vector<T> innervec;
friend auto operator<< ( std::ostream & os, B const& b ) -> std::ostream&
{
for (auto e : b.innervec)
os << "nt" << e;
return os;
}
};
std::vector<B> outervec;
};
answered Nov 25 '18 at 19:57
JansJans
8,89422635
The error is because the overload for the nested type contain a non-deduced context parameter typename A<T>::B const& b and for operators you can't provide the explicit template argument T, you need to define the operator as friend for the nested A<T>::B as:
template <typename T>
struct A
{
struct B
{
std::vector<T> innervec;
friend auto operator<< ( std::ostream & os, B const& b ) -> std::ostream&
{
for (auto e : b.innervec)
os << "nt" << e;
return os;
}
};
std::vector<B> outervec;
};
answered Nov 25 '18 at 19:57
JansJans
8,89422635
edited Nov 25 '18 at 20:02
answered Nov 25 '18 at 19:57
JansJans
8,89422635
answered Nov 25 '18 at 19:57
JansJans
8,89422635
answered Nov 25 '18 at 19:57
JansJans
8,89422635
8,89422635
No need for friendship, just for membership
– Géza Török
Nov 25 '18 at 20:05
Thanks :) I didn't know about non-deduced contexts.
– Tarquiscani
Nov 25 '18 at 20:09
@GézaTörök stream operators can't be members.
– Jans
Nov 25 '18 at 20:09
add a comment |
No need for friendship, just for membership
– Géza Török
Nov 25 '18 at 20:05
Thanks :) I didn't know about non-deduced contexts.
– Tarquiscani
Nov 25 '18 at 20:09
@GézaTörök stream operators can't be members.
– Jans
Nov 25 '18 at 20:09
No need for friendship, just for membership
– Géza Török
Nov 25 '18 at 20:05
No need for friendship, just for membership
– Géza Török
Nov 25 '18 at 20:05
Thanks :) I didn't know about non-deduced contexts.
– Tarquiscani
Nov 25 '18 at 20:09
Thanks :) I didn't know about non-deduced contexts.
– Tarquiscani
Nov 25 '18 at 20:09
@GézaTörök stream operators can't be members.
– Jans
Nov 25 '18 at 20:09
@GézaTörök stream operators can't be members.
– Jans
Nov 25 '18 at 20:09
add a comment |
The problematic part is A<T>::B in the parameter list, as the compiler cannot deduce type expressions of this form, since existence and type of member B depends on T. I would take the type of b argument as template argument T directly, and use SFINAE to constrain accepted types.
answered Nov 25 '18 at 20:00
Géza TörökGéza Török
1,222515
add a comment |
The problematic part is A<T>::B in the parameter list, as the compiler cannot deduce type expressions of this form, since existence and type of member B depends on T. I would take the type of b argument as template argument T directly, and use SFINAE to constrain accepted types.
answered Nov 25 '18 at 20:00
Géza TörökGéza Török
1,222515
add a comment |
The problematic part is A<T>::B in the parameter list, as the compiler cannot deduce type expressions of this form, since existence and type of member B depends on T. I would take the type of b argument as template argument T directly, and use SFINAE to constrain accepted types.
answered Nov 25 '18 at 20:00
Géza TörökGéza Török
1,222515
The problematic part is A<T>::B in the parameter list, as the compiler cannot deduce type expressions of this form, since existence and type of member B depends on T. I would take the type of b argument as template argument T directly, and use SFINAE to constrain accepted types.
answered Nov 25 '18 at 20:00
Géza TörökGéza Török
1,222515
edited Nov 25 '18 at 20:18
answered Nov 25 '18 at 20:00
Géza TörökGéza Török
1,222515
answered Nov 25 '18 at 20:00
Géza TörökGéza Török
1,222515
answered Nov 25 '18 at 20:00
Géza TörökGéza Török
1,222515
1,222515
add a comment |
add a comment |
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