Hi, I made the following test program:
//////////////////////// classes_1.cpp
#include <iostream>
#include "classes_1. h"
using namespace std;
A::A():i(0){cou t <<"const A: i =" << i <<endl;}
A::~A(){cout <<"destr A"<<endl;}
inline void A::showA() {cout << "show A: i =" << i <<endl;};
//////////////////////// classes_1.h
#ifndef A_H
#define A_H
class A{
int i;
public:
void set(int ii){i=ii;}
inline void showA();
A();
~A();};
#endif
////////////////////////
and a simple "main.cpp" file which makes an A object , sets an i and shows
it.
It compiles without problems, but I have the following link-error:
->error LNK2001: unresolved external symbol "public: void __thiscall
A::showA(void)" (?showA@A@@QAEX XZ)
when I remove both the inlines I don't have a link-error .
What am I doing wrong ??
Thanks in advance. 9 2055
John Rambo wrote: //////////////////////// classes_1.cpp #include <iostream> #include "classes_1. h" using namespace std;
A::A():i(0){cou t <<"const A: i =" << i <<endl;} A::~A(){cout <<"destr A"<<endl;} inline void A::showA() {cout << "show A: i =" << i <<endl;}; //////////////////////// classes_1.h #ifndef A_H #define A_H class A{ int i; public: void set(int ii){i=ii;} inline void showA(); A(); ~A();}; #endif //////////////////////// and a simple "main.cpp" file which makes an A object , sets an i and shows it.
It compiles without problems, but I have the following link-error:
[...]
Move the implementation of the inline function from classes_1.cpp to
classes_1.h
HTH,
Niels Dekker www.xs4all.nl/~nd/dekkerware
Thanks, but isn't it possible to place the implementation in the cpp file
(otherwise I have problems with the "using namespace std" which may not be
placed in the header file...) ???
"Niels Dekker - no reply address" <un*****@this.i s.invalid> wrote in message
news:41******** *******@this.is .invalid... John Rambo wrote: //////////////////////// classes_1.cpp #include <iostream> #include "classes_1. h" using namespace std;
A::A():i(0){cou t <<"const A: i =" << i <<endl;} A::~A(){cout <<"destr A"<<endl;} inline void A::showA() {cout << "show A: i =" << i <<endl;}; //////////////////////// classes_1.h #ifndef A_H #define A_H class A{ int i; public: void set(int ii){i=ii;} inline void showA(); A(); ~A();}; #endif //////////////////////// and a simple "main.cpp" file which makes an A object , sets an i and
shows it.
It compiles without problems, but I have the following link-error: [...]
Move the implementation of the inline function from classes_1.cpp to classes_1.h
HTH,
Niels Dekker www.xs4all.nl/~nd/dekkerware
John Rambo wrote: #include <iostream> #include "classes_1. h" using namespace std;
inline void A::showA() {cout << "show A: i =" << i <<endl;};
I wrote: Move the implementation of the inline function from classes_1.cpp to classes_1.h
John Rambo wrote: Thanks, but isn't it possible to place the implementation in the cpp file
Not if you want your member function to be "inline".
(otherwise I have problems with the "using namespace std" which may not be placed in the header file...) ???
That's a good point. But you don't need the "using namespace std", if
you do "std::"! As follows:
#include <iostream>
inline void A::showA() {std::cout << "show A: i =" << i <<
std::endl;};
Niels Dekker www.xs4all.nl/~nd/dekkerware
Ok. Thanks a lot; I have still a question, however:
how can I tell the compiler that it is the std-version of << I want to use
?? "using std::<<;" doesn't works .
"Niels Dekker - no reply address" <un*****@this.i s.invalid> wrote in message
news:41******** *******@this.is .invalid... John Rambo wrote: #include <iostream> #include "classes_1. h" using namespace std;
inline void A::showA() {cout << "show A: i =" << i <<endl;}; I wrote: Move the implementation of the inline function from classes_1.cpp to classes_1.h
John Rambo wrote: Thanks, but isn't it possible to place the implementation in the cpp
file Not if you want your member function to be "inline".
(otherwise I have problems with the "using namespace std" which may not
be placed in the header file...) ???
That's a good point. But you don't need the "using namespace std", if you do "std::"! As follows:
#include <iostream> inline void A::showA() {std::cout << "show A: i =" << i << std::endl;};
Niels Dekker www.xs4all.nl/~nd/dekkerware
Please don't top-post. Rearranged.
John Rambo wrote: John Rambo wrote: > Thanks, but isn't it possible to place the implementation in the > cpp file
Not if you want your member function to be "inline".
> (otherwise I have problems with the "using namespace std" which > may not be placed in the header file...) ???
That's a good point. But you don't need the "using namespace std", if you do "std::"! As follows:
#include <iostream> inline void A::showA() {std::cout << "show A: i =" << i << std::endl;};
Ok. Thanks a lot; I have still a question, however: how can I tell the compiler that it is the std-version of << I want to use ?? "using std::<<;" doesn't works .
Why would you need that?
found it: "std::operator< <"
"Rolf Magnus" <ra******@t-online.de> wrote in message
news:cf******** *****@news.t-online.com... Please don't top-post. Rearranged.
John Rambo wrote:
John Rambo wrote: > Thanks, but isn't it possible to place the implementation in the > cpp file
Not if you want your member function to be "inline".
> (otherwise I have problems with the "using namespace std" which > may not be placed in the header file...) ???
That's a good point. But you don't need the "using namespace std", if you do "std::"! As follows:
#include <iostream> inline void A::showA() {std::cout << "show A: i =" << i << std::endl;};
Ok. Thanks a lot; I have still a question, however: how can I tell the compiler that it is the std-version of << I want to use ?? "using std::<<;" doesn't works .
Why would you need that?
On Sat, 14 Aug 2004 18:45:16 GMT, John Rambo <Ia*****@hotmai l.com> wrote: found it: "std::operator< <"
It's very unlikely that you would need that.
std::cout << "hello\n";
That works perfectly well because C++ has a rule (sometimes called Koenig
lookup) which says that if the type of an argument to a particular
function or operator is in a particular namespace then that namespace is
searched automatically. So type type of cout is in the std namespace
(std::ostream), so you don't need to tell the compiler to look for
operator<< in the std namesapce.
john
On Sat, 14 Aug 2004 15:28:49 GMT, "John Rambo" <Ia*****@hotmai l.com>
wrote: Ok. Thanks a lot; I have still a question, however: how can I tell the compiler that it is the std-version of << I want to use ?? "using std::<<;" doesn't works .
If you really need to do that then try:
using std::operator<<
rossum
--
The ultimate truth is that there is no Ultimate Truth
"John Harrison" <jo************ *@hotmail.com> wrote: On Sat, 14 Aug 2004 18:45:16 GMT, John Rambo <Ia*****@hotmai l.com> wrote:
found it: "std::operator< <"
It's very unlikely that you would need that.
std::cout << "hello\n";
That works perfectly well because C++ has a rule (sometimes called Koenig lookup) which says that if the type of an argument to a particular function or operator is in a particular namespace then that namespace is searched automatically.
I suppose he wanted to guard against something like:
// in header
namespace foo
{
std::ostream &operator<<(std ::ostream &, char const *);
};
// in non-header
using namespace foo;
int bar()
{
std::cout << "hello\n";
}
};
(but I would expect the compiler to complain about ambiguous
function) This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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