C++: cannot define member function
Could you please advise why I am getting the error in the code below?
error: cannot define member function ‘Test<int>::Printer::Print’ within ‘Test<int>’
I am using gcc version 8.1.1 and compile the code as g++ -std=c++11.
Although, if I move the definition of function Print under the definition of struct Printer (i.e. making it inline implicitly), the compiler does not produce any error.
#include <iostream>
template <typename Type>
struct TestBase {
struct Printer {
template <typename T>
void Print(const T& t) {
std::cout << t << std::endl;
}
};
};
template <typename Type>
struct Test;
template<>
struct Test<int> : public TestBase<int> {
struct Printer : public TestBase<int>::Printer {
template <typename T>
void Print(int i, const T& t);
};
template <typename T>
void Printer::Print(int i, const T& t) {
std::cout << i << t << std::endl;
}
};
int main() {
Test<int> t;
}
UPDATE:
Brian pointed out the exact reason why it is the case: "... A member function definition that appears outside of the class definition shall appear in a namespace scope enclosing the class definition..."
Brian not only answered the main question that started this topic but also an additional question that I asked in the comment to the accepted answer of him.
c++ c++11
asked Nov 15 '18 at 20:51
TruLaTruLa
358212
add a comment |
Could you please advise why I am getting the error in the code below?
error: cannot define member function ‘Test<int>::Printer::Print’ within ‘Test<int>’
I am using gcc version 8.1.1 and compile the code as g++ -std=c++11.
Although, if I move the definition of function Print under the definition of struct Printer (i.e. making it inline implicitly), the compiler does not produce any error.
#include <iostream>
template <typename Type>
struct TestBase {
struct Printer {
template <typename T>
void Print(const T& t) {
std::cout << t << std::endl;
}
};
};
template <typename Type>
struct Test;
template<>
struct Test<int> : public TestBase<int> {
struct Printer : public TestBase<int>::Printer {
template <typename T>
void Print(int i, const T& t);
};
template <typename T>
void Printer::Print(int i, const T& t) {
std::cout << i << t << std::endl;
}
};
int main() {
Test<int> t;
}
UPDATE:
Brian pointed out the exact reason why it is the case: "... A member function definition that appears outside of the class definition shall appear in a namespace scope enclosing the class definition..."
Brian not only answered the main question that started this topic but also an additional question that I asked in the comment to the accepted answer of him.
c++ c++11
asked Nov 15 '18 at 20:51
TruLaTruLa
358212
Read the error message carefully. The compiler is telling you the truth. If you want to know why, it's because that's specified by the C++ standard.
– paddy
Nov 15 '18 at 20:55
I think I understand the compiler's message but I don't understand which rule is violated. Could you please help me to point that out?
– TruLa
Nov 15 '18 at 20:58
add a comment |
Could you please advise why I am getting the error in the code below?
error: cannot define member function ‘Test<int>::Printer::Print’ within ‘Test<int>’
I am using gcc version 8.1.1 and compile the code as g++ -std=c++11.
Although, if I move the definition of function Print under the definition of struct Printer (i.e. making it inline implicitly), the compiler does not produce any error.
#include <iostream>
template <typename Type>
struct TestBase {
struct Printer {
template <typename T>
void Print(const T& t) {
std::cout << t << std::endl;
}
};
};
template <typename Type>
struct Test;
template<>
struct Test<int> : public TestBase<int> {
struct Printer : public TestBase<int>::Printer {
template <typename T>
void Print(int i, const T& t);
};
template <typename T>
void Printer::Print(int i, const T& t) {
std::cout << i << t << std::endl;
}
};
int main() {
Test<int> t;
}
UPDATE:
Brian pointed out the exact reason why it is the case: "... A member function definition that appears outside of the class definition shall appear in a namespace scope enclosing the class definition..."
Brian not only answered the main question that started this topic but also an additional question that I asked in the comment to the accepted answer of him.
c++ c++11
asked Nov 15 '18 at 20:51
TruLaTruLa
358212
Could you please advise why I am getting the error in the code below?
error: cannot define member function ‘Test<int>::Printer::Print’ within ‘Test<int>’
I am using gcc version 8.1.1 and compile the code as g++ -std=c++11.
Although, if I move the definition of function Print under the definition of struct Printer (i.e. making it inline implicitly), the compiler does not produce any error.
#include <iostream>
template <typename Type>
struct TestBase {
struct Printer {
template <typename T>
void Print(const T& t) {
std::cout << t << std::endl;
}
};
};
template <typename Type>
struct Test;
template<>
struct Test<int> : public TestBase<int> {
struct Printer : public TestBase<int>::Printer {
template <typename T>
void Print(int i, const T& t);
};
template <typename T>
void Printer::Print(int i, const T& t) {
std::cout << i << t << std::endl;
}
};
int main() {
Test<int> t;
}
UPDATE:
Brian pointed out the exact reason why it is the case: "... A member function definition that appears outside of the class definition shall appear in a namespace scope enclosing the class definition..."
Brian not only answered the main question that started this topic but also an additional question that I asked in the comment to the accepted answer of him.
c++ c++11
c++ c++11
asked Nov 15 '18 at 20:51
TruLaTruLa
358212
asked Nov 15 '18 at 20:51
TruLaTruLa
358212
edited Nov 15 '18 at 21:19
TruLa
asked Nov 15 '18 at 20:51
TruLaTruLa
358212
asked Nov 15 '18 at 20:51
TruLaTruLa
358212
asked Nov 15 '18 at 20:51
TruLaTruLa
358212
358212
Read the error message carefully. The compiler is telling you the truth. If you want to know why, it's because that's specified by the C++ standard.
– paddy
Nov 15 '18 at 20:55
I think I understand the compiler's message but I don't understand which rule is violated. Could you please help me to point that out?
– TruLa
Nov 15 '18 at 20:58
add a comment |
Read the error message carefully. The compiler is telling you the truth. If you want to know why, it's because that's specified by the C++ standard.
– paddy
Nov 15 '18 at 20:55
I think I understand the compiler's message but I don't understand which rule is violated. Could you please help me to point that out?
– TruLa
Nov 15 '18 at 20:58
Read the error message carefully. The compiler is telling you the truth. If you want to know why, it's because that's specified by the C++ standard.
– paddy
Nov 15 '18 at 20:55
Read the error message carefully. The compiler is telling you the truth. If you want to know why, it's because that's specified by the C++ standard.
– paddy
Nov 15 '18 at 20:55
I think I understand the compiler's message but I don't understand which rule is violated. Could you please help me to point that out?
– TruLa
Nov 15 '18 at 20:58
I think I understand the compiler's message but I don't understand which rule is violated. Could you please help me to point that out?
– TruLa
Nov 15 '18 at 20:58
add a comment |
1 Answer
1
active
oldest
votes
[class.mfct]/1, emphasis mine:
... A member function definition that appears outside of the class definition shall appear in a namespace scope enclosing the class definition. ...
An enclosing class scope is thus not an allowed location for the definition.
answered Nov 15 '18 at 21:03
BrianBrian
65.4k797185
Brian, thank you very much for such a prompt and precise answer. I have a suspicion that this might not be the end of a story. Let's move the member functionPrintdefinition inside thestruct Printerdefinition and change themainfunction asTest<int>::Printer p; p.Print(3);. I'll start getting theno matching function for call toerror message. Is it somehow related to your answer? Is it becausevoid Print(const T& t)is defined outside of the class definition?
– TruLa
Nov 15 '18 at 21:10
1
@TruLa This is becauseTest::Printer::Printtakes two arguments and it hidesTestBase::Printer::Print. See stackoverflow.com/questions/1896830/…
– Brian
Nov 15 '18 at 21:16
add a comment |
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1 Answer
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[class.mfct]/1, emphasis mine:
... A member function definition that appears outside of the class definition shall appear in a namespace scope enclosing the class definition. ...
An enclosing class scope is thus not an allowed location for the definition.
answered Nov 15 '18 at 21:03
BrianBrian
65.4k797185
Brian, thank you very much for such a prompt and precise answer. I have a suspicion that this might not be the end of a story. Let's move the member functionPrintdefinition inside thestruct Printerdefinition and change themainfunction asTest<int>::Printer p; p.Print(3);. I'll start getting theno matching function for call toerror message. Is it somehow related to your answer? Is it becausevoid Print(const T& t)is defined outside of the class definition?
– TruLa
Nov 15 '18 at 21:10
1
@TruLa This is becauseTest::Printer::Printtakes two arguments and it hidesTestBase::Printer::Print. See stackoverflow.com/questions/1896830/…
– Brian
Nov 15 '18 at 21:16
add a comment |
[class.mfct]/1, emphasis mine:
... A member function definition that appears outside of the class definition shall appear in a namespace scope enclosing the class definition. ...
An enclosing class scope is thus not an allowed location for the definition.
answered Nov 15 '18 at 21:03
BrianBrian
65.4k797185
Brian, thank you very much for such a prompt and precise answer. I have a suspicion that this might not be the end of a story. Let's move the member functionPrintdefinition inside thestruct Printerdefinition and change themainfunction asTest<int>::Printer p; p.Print(3);. I'll start getting theno matching function for call toerror message. Is it somehow related to your answer? Is it becausevoid Print(const T& t)is defined outside of the class definition?
– TruLa
Nov 15 '18 at 21:10
1
@TruLa This is becauseTest::Printer::Printtakes two arguments and it hidesTestBase::Printer::Print. See stackoverflow.com/questions/1896830/…
– Brian
Nov 15 '18 at 21:16
add a comment |
[class.mfct]/1, emphasis mine:
... A member function definition that appears outside of the class definition shall appear in a namespace scope enclosing the class definition. ...
An enclosing class scope is thus not an allowed location for the definition.
answered Nov 15 '18 at 21:03
BrianBrian
65.4k797185
[class.mfct]/1, emphasis mine:
... A member function definition that appears outside of the class definition shall appear in a namespace scope enclosing the class definition. ...
An enclosing class scope is thus not an allowed location for the definition.
answered Nov 15 '18 at 21:03
BrianBrian
65.4k797185
answered Nov 15 '18 at 21:03
BrianBrian
65.4k797185
answered Nov 15 '18 at 21:03
BrianBrian
65.4k797185
answered Nov 15 '18 at 21:03
BrianBrian
65.4k797185
65.4k797185
Brian, thank you very much for such a prompt and precise answer. I have a suspicion that this might not be the end of a story. Let's move the member functionPrintdefinition inside thestruct Printerdefinition and change themainfunction asTest<int>::Printer p; p.Print(3);. I'll start getting theno matching function for call toerror message. Is it somehow related to your answer? Is it becausevoid Print(const T& t)is defined outside of the class definition?
– TruLa
Nov 15 '18 at 21:10
1
@TruLa This is becauseTest::Printer::Printtakes two arguments and it hidesTestBase::Printer::Print. See stackoverflow.com/questions/1896830/…
– Brian
Nov 15 '18 at 21:16
add a comment |
Brian, thank you very much for such a prompt and precise answer. I have a suspicion that this might not be the end of a story. Let's move the member functionPrintdefinition inside thestruct Printerdefinition and change themainfunction asTest<int>::Printer p; p.Print(3);. I'll start getting theno matching function for call toerror message. Is it somehow related to your answer? Is it becausevoid Print(const T& t)is defined outside of the class definition?
– TruLa
Nov 15 '18 at 21:10
1
@TruLa This is becauseTest::Printer::Printtakes two arguments and it hidesTestBase::Printer::Print. See stackoverflow.com/questions/1896830/…
– Brian
Nov 15 '18 at 21:16
Brian, thank you very much for such a prompt and precise answer. I have a suspicion that this might not be the end of a story. Let's move the member function
Print definition inside the struct Printer definition and change the main function as Test<int>::Printer p; p.Print(3);. I'll start getting the no matching function for call to error message. Is it somehow related to your answer? Is it because void Print(const T& t) is defined outside of the class definition?– TruLa
Nov 15 '18 at 21:10
Brian, thank you very much for such a prompt and precise answer. I have a suspicion that this might not be the end of a story. Let's move the member function
Print definition inside the struct Printer definition and change the main function as Test<int>::Printer p; p.Print(3);. I'll start getting the no matching function for call to error message. Is it somehow related to your answer? Is it because void Print(const T& t) is defined outside of the class definition?– TruLa
Nov 15 '18 at 21:10
1
1
@TruLa This is because
Test::Printer::Print takes two arguments and it hides TestBase::Printer::Print. See stackoverflow.com/questions/1896830/…– Brian
Nov 15 '18 at 21:16
@TruLa This is because
Test::Printer::Print takes two arguments and it hides TestBase::Printer::Print. See stackoverflow.com/questions/1896830/…– Brian
Nov 15 '18 at 21:16
add a comment |
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Read the error message carefully. The compiler is telling you the truth. If you want to know why, it's because that's specified by the C++ standard.
– paddy
Nov 15 '18 at 20:55
I think I understand the compiler's message but I don't understand which rule is violated. Could you please help me to point that out?
– TruLa
Nov 15 '18 at 20:58