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static inline function not found during linking

Hi,

I have a problem with my code that the compiler does not find any inline
functions which are static.

The simple code example is written below, this is what the compiler
throws at me.

] $ g++ main.cpp foo.cpp
/home/klaas/tmp/cciAcYgl.o: In function `main':
inline.cpp:(.text+0x9): undefined reference to `foo::bar()'
collect2: ld returned 1 exit status

Is this normal behaviour??? I seriously doubt that.
If foo::bar(void) is not inline, it works.
When everything is written in one file, it works too.

regards,
This is the simple code

== foo.hpp ==
#ifndef _FOO_
#define _FOO_
class foo {
public:
static int a;
static int bar(void) ;
};
#endif
== foo.cpp ==
#include "foo.hpp"
int foo::a = 0;
inline int foo::bar(void) { return a; }
== main.cpp ==
#include "foo.hpp"
int main(void) {
int b = foo::bar();
return 0;
};
==========
Dec 21 '07 #1
5 4572
ciccio wrote:
I have a problem with my code that the compiler does not find any
inline functions which are static.
Since your function is never declared 'inline' when 'main.cpp' is
compiled, the compiler expects a definition of that function to exist
somewhere. Since the implementation of that function is declared
'inline' in 'foo.cpp', the body is essentially ignored because it is
not called anywhere in 'foo.cpp'.

Either put the function in the header (where declaring it 'inline'
makes sense), or remove the 'inline' modifier from the function
definition in 'foo.cpp'.
>
The simple code example is written below, this is what the compiler
throws at me.

] $ g++ main.cpp foo.cpp
/home/klaas/tmp/cciAcYgl.o: In function `main':
inline.cpp:(.text+0x9): undefined reference to `foo::bar()'
collect2: ld returned 1 exit status

Is this normal behaviour??? I seriously doubt that.
If foo::bar(void) is not inline, it works.
When everything is written in one file, it works too.

regards,
This is the simple code

== foo.hpp ==
#ifndef _FOO_
#define _FOO_
class foo {
public:
static int a;
static int bar(void) ;
};
#endif
== foo.cpp ==
#include "foo.hpp"
int foo::a = 0;
inline int foo::bar(void) { return a; }
== main.cpp ==
#include "foo.hpp"
int main(void) {
int b = foo::bar();
return 0;
};
==========
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Dec 21 '07 #2
Thanks for the quick response,

I just read it in stroustrup. I had not a clue!

Regards
Victor Bazarov wrote:
ciccio wrote:
>I have a problem with my code that the compiler does not find any
inline functions which are static.

Since your function is never declared 'inline' when 'main.cpp' is
compiled, the compiler expects a definition of that function to exist
somewhere. Since the implementation of that function is declared
'inline' in 'foo.cpp', the body is essentially ignored because it is
not called anywhere in 'foo.cpp'.

Either put the function in the header (where declaring it 'inline'
makes sense), or remove the 'inline' modifier from the function
definition in 'foo.cpp'.
>The simple code example is written below, this is what the compiler
throws at me.

] $ g++ main.cpp foo.cpp
/home/klaas/tmp/cciAcYgl.o: In function `main':
inline.cpp:(.text+0x9): undefined reference to `foo::bar()'
collect2: ld returned 1 exit status

Is this normal behaviour??? I seriously doubt that.
If foo::bar(void) is not inline, it works.
When everything is written in one file, it works too.

regards,
This is the simple code

== foo.hpp ==
#ifndef _FOO_
#define _FOO_
class foo {
public:
static int a;
static int bar(void) ;
};
#endif
== foo.cpp ==
#include "foo.hpp"
int foo::a = 0;
inline int foo::bar(void) { return a; }
== main.cpp ==
#include "foo.hpp"
int main(void) {
int b = foo::bar();
return 0;
};
==========
Dec 21 '07 #3
Okay I got this part now,

But ... when I use templates this is not the case.

Here is again an example (now without the static, since that does not
matter).

The compilation does not give any linking error here.
So what is the reason of this then? Does this mean that inline
functions of templates are just bogus?

Regards
== foo.hpp ==
#ifndef _FOO_
#define _FOO_

template<typename T>
class foo {
public:
int a;
int bar(void) ;
};
#endif
== foo.cpp ==
#include "foo.hpp"
template<typename T>
inline int foo<T>::bar(void) { return a; }
template class foo<int>;
== main.cpp ==
#include "foo.hpp"
int main(void) {
int b = foo<int>().bar();
return 0;
};

Dec 21 '07 #4
ciccio wrote:
Okay I got this part now,

But ... when I use templates this is not the case.
What is not the case?
Here is again an example (now without the static, since that does not
matter).

The compilation does not give any linking error here.
So what is the reason of this then? Does this mean that inline
functions of templates are just bogus?
Not "inline functions of templates", but "function templates declared
inline".

Right after defining your function and declaring it 'inline' (which
you're not really supposed to do, so your compiler _probably_ ignored
it), you _explicitly_ instantiate your template, which basically
introduces the definitions of all member functions (if it can, and it
should be able to, since you just defined your function), which are
of course _not_ inline.

IOW, whatever you declared 'inline' is *not* the function the linker
is looking for when resolving the reference from 'main'. It's looking
for 'foo<int>::bar', not for 'foo<T>::bar'. And your 'foo<int>::bar'
is non-inline (as the result of explicit instantiation).
>
Regards
== foo.hpp ==
#ifndef _FOO_
#define _FOO_

template<typename T>
class foo {
public:
int a;
int bar(void) ;
};
#endif
== foo.cpp ==
#include "foo.hpp"
template<typename T>
inline int foo<T>::bar(void) { return a; }
template class foo<int>;
== main.cpp ==
#include "foo.hpp"
int main(void) {
int b = foo<int>().bar();
return 0;
};
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Dec 21 '07 #5
ciccio wrote:
Okay I got this part now,

But ... when I use templates this is not the case.

Here is again an example (now without the static, since that does not
matter).

The compilation does not give any linking error here.
So what is the reason of this then? Does this mean that inline
functions of templates are just bogus?

Regards
== foo.hpp ==
#ifndef _FOO_
#define _FOO_

template<typename T>
class foo {
public:
int a;
int bar(void) ;
};
#endif
== foo.cpp ==
#include "foo.hpp"
template<typename T>
inline int foo<T>::bar(void) { return a; }
template class foo<int>;
== main.cpp ==
#include "foo.hpp"
int main(void) {
int b = foo<int>().bar();
return 0;
};
Irrelevant to your issue, but still a problem, is your include guard.

Any identifier with a leading underscore, followed by an uppercase
letter, is reserved to the implementation. Period. You may not use it
for your own purposes.

So your include guard of _FOO_ is illegal.
Dec 21 '07 #6

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