Re: header file for template functions/classes

From:

"Bo Persson" <bop@gmb.dk>

Newsgroups:

comp.lang.c++

Date:

Sat, 19 May 2007 13:27:31 +0200

Message-ID:

<5b858uF2riml5U1@mid.individual.net>

Jess wrote:
:: On May 19, 8:50 pm, John Harrison <john_androni...@hotmail.com>
:: wrote:
:::: I tried to put the declarations in a .h file, with "export"
:::: keyword. When I compiled the code, my compiler said "export"
:::: wasn't implemented and would be ignored. However, the .h file
:::: plus the separate .cpp file actually worked together. If the
:::: compiler doesn't implement "export", how could all these work
:::: out?
:::
:::: Jess
:::
::: Without seeing your code who can say? Maybe you are doing explicit
::: instantiation (or whatever it's called).
:::
::: john
::
::
:: The header file is (test.h):
::
:: export template<class T> class A{
:: public:
:: void f() const;
:: A(T a):x(a){}
:: private:
:: T x;
:: };
::
:: The source code is:
::
:: #include<iostream>
:: #include "test.h"
::
:: using namespace std;
::
:: template<class T>
:: void A<T>::f() const{
:: cout << x << endl;
:: };
::
::
:: int main(){
:: A<int> a(10);
:: a.f();
:: return 0;
:: }
::

This works because the implementation of f() is present in the same
file as its use. The effect is the same as if it was put in the header
file.

The including of files is performed by the preprocessor, long before
the compiler starts to look at the code. When you reach that stage, it
appears as one large file with all includes and preprocessor macros
already resolved.

If you were to put f() in a separate A.cpp file, the compiler would
not find it (when ignoring the export keyword).

Bo Persson