Re: Another inline question

From:

"Victor Bazarov" <v.Abazarov@comAcast.net>

Newsgroups:

comp.lang.c++

Date:

Wed, 12 Mar 2008 18:25:59 -0400

Message-ID:

<fr9l9o$e55$1@news.datemas.de>

asterisc wrote:

Hi there,

Before getting a binary, there is the following sequence of steps:
- preprocessing
- compilation
- assembly
- linking

While 'preprocessing' is clearly separated into its own step in the
language Standard, the other ones are not really defined so well.
Especially the assembly portion - nowadays most compilers produce
native code in an object file form. Linking is still a step in
most cases.

Is there any rule at which level should the inline work?

No rule.

I mean, where should we see the effect of the inline keyword ?

During compilation 'inline' acts as a suggestion for the compiler
to substitute the call with the body. During linking 'inline' gives
an instruction to only consider a single definition of the function.
IOW, different meanings at different stages.

I have the following fairly simple piece of code:

//test.cpp
#include <iostream>

inline void increment( int& i )
{
   ++i;
}

#define INCREMENT(i) (++i)

int main()
{
   int i = 0;
   increment( i );
   std::cout << i << endl;
   INCREMENT(i);
   std::cout << i << endl;
}

The #define is replaced in the preprocessing step.

Right.

As far as I could check, the effect of the inline is only at the
assembling step, where we won't have a call to the increment()
function. (if we apply an optimization)
The preprocessed source looks exactly the same as the .cpp file, only
the #define was expanded.

Is that the normal behavior?

Probably. I've not seen separate assembling step for quite a while,
so I'll just take your word for it.

Another question, how is inline working with a virtual member
function? (if we call that function thru a base pointer which point to
a derived object)

It doesn't. If the function is not known until run-time, like in
the case with a virtual function, it cannot be inlined, can it?

V
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