| Topic: |
DEVELOP > c-Plus-Plus |
| User: |
"Joe Hesse" |
| Date: |
06 Dec 2007 11:05:07 AM |
| Object: |
Define friend operator << for class template. |
Hi,
I have a template class and I want to define an operator << as a friend
function. For each instantiation of the class I want a corresponding
instantiation of operator <<.
The following example fails to compile with g++ version 4.1.2.
I would appreciate it if you could help me fix it or point me to a suitable
reference.
Thank you,
Joe Hesse
********************************************
#include <iostream>
// forward declaration
template <typename T>
class Foo;
// forward declaration
template <typename T>
std::ostream & operator << (std::ostream &, const Foo<T> &);
template <typename T>
class Foo {
private:
T value;
public:
Foo(const T & v) : value(v) {}
friend std::ostream & operator << <> (std::ostream &, const Foo<T> &);
};
// implement operator <<
template <typename T>
std::ostream & operator << (std::ostream &o, const Foo<T> &f) {
return o << f.value ;
}
int main() {
Foo<int> fi;
std::cout << fi;
return 0;
}
/* Here are the compiler error messages
Test.cpp: In function int main():
Test.cpp:26: error: no matching function for call to Foo<int>::Foo()
Test.cpp:16: note: candidates are: Foo<T>::Foo(const T&) [with T = int]
Test.cpp:12: note: Foo<int>::Foo(const Foo<int>&)
*/
********************************************
.
|
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| User: "terminator" |
|
| Title: Re: Define friend operator << for class template. |
06 Dec 2007 11:26:22 AM |
|
|
On Dec 6, 8:05 pm, "Joe Hesse" <joe_he...@actcx.com> wrote:
Hi,
I have a template class and I want to define an operator << as a friend
function. For each instantiation of the class I want a corresponding
instantiation of operator <<.
The following example fails to compile with g++ version 4.1.2.
I would appreciate it if you could help me fix it or point me to a suitable
reference.
Thank you,
Joe Hesse
********************************************
#include <iostream>
// forward declaration
template <typename T>
class Foo;
// forward declaration
template <typename T>
std::ostream & operator << (std::ostream &, const Foo<T> &);
template <typename T>
class Foo {
private:
T value;
public:
Foo(const T & v) : value(v) {}
friend std::ostream & operator << <> (std::ostream &, const Foo<T> &);
};
// implement operator <<
template <typename T>
std::ostream & operator << (std::ostream &o, const Foo<T> &f) {
return o << f.value ;
}
int main() {
Foo<int> fi;
std::cout << fi;
return 0;
}
/* Here are the compiler error messages
Test.cpp: In function int main():
Test.cpp:26: error: no matching function for call to Foo<int>::Foo()
Test.cpp:16: note: candidates are: Foo<T>::Foo(const T&) [with T = int]
Test.cpp:12: note: Foo<int>::Foo(const Foo<int>&)
*/
********************************************
Instein says "take it simple, as simple as posible but not simpler".
Why did you not use the simplest imaginable syntax?
I would write:
friend std::ostream & operator << (std::ostream &, const Foo &);
but this is not what the compiler complaigns about;Please learn to
read:
Test.cpp:26: error: no matching function for call to Foo<int>::Foo()
Test.cpp:16: note: candidates are: Foo<T>::Foo(const T&) [with T = int]
Test.cpp:12: note: Foo<int>::Foo(const Foo<int>&)
Since you have defined a constructor ,C++ will no more automatically
generate a default constructor. this is the errorneous line:
Foo<int> fi;
In order to resolve this add the following inside the braces for
declaration of your template class:
Foo(){};
regards,
FM.
.
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|
| User: "want.to.be.professer" |
|
| Title: Re: Define friend operator << for class template. |
07 Dec 2007 03:01:24 AM |
|
|
Once you not provide a constructor ,There will be a default
constructor;
When you provide, There will be your provided constructor only;
.
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|
| User: "Joe Hesse" |
|
| Title: Re: Define friend operator << for class template. |
06 Dec 2007 11:25:38 AM |
|
|
Please forgive me, I have an obvious error. No response is needed from the
newsgroup.
"Joe Hesse" <joe_hesse@actcx.com> wrote in message
news:13lgau67k1b9c6c@corp.supernews.com...
Hi,
I have a template class and I want to define an operator << as a friend
function. For each instantiation of the class I want a corresponding
instantiation of operator <<.
The following example fails to compile with g++ version 4.1.2.
I would appreciate it if you could help me fix it or point me to a
suitable reference.
Thank you,
Joe Hesse
********************************************
#include <iostream>
// forward declaration
template <typename T>
class Foo;
// forward declaration
template <typename T>
std::ostream & operator << (std::ostream &, const Foo<T> &);
template <typename T>
class Foo {
private:
T value;
public:
Foo(const T & v) : value(v) {}
friend std::ostream & operator << <> (std::ostream &, const Foo<T> &);
};
// implement operator <<
template <typename T>
std::ostream & operator << (std::ostream &o, const Foo<T> &f) {
return o << f.value ;
}
int main() {
Foo<int> fi;
std::cout << fi;
return 0;
}
/* Here are the compiler error messages
Test.cpp: In function int main():
Test.cpp:26: error: no matching function for call to Foo<int>::Foo()
Test.cpp:16: note: candidates are: Foo<T>::Foo(const T&) [with T = int]
Test.cpp:12: note: Foo<int>::Foo(const Foo<int>&)
*/
********************************************
.
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