absolute
lazarus / 1-Linux

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Kamau · 2012-12-22 06:24

Hello, I am an old Delphi-programmer and I discovered Lazarus just a
few days ago.

Installed it on an XP computer. It works fantastic!

Yesterday tried to install i on CentOS. Installation no problem. I
used yum and IDE version 0.9.26 was installed. The fpc version is
2.4.0

Had to add unit-directories.

Now when compiling a "program" containing no own code, just a form, I
get the error: "absolute can only be associated with a var or const"

Which is quite logical BUT.....

The units included in Lazarus have several absolute declarations
referring to types, not variables.

How do I get around this?

Is there a compiler switch to use in order to ignore this error?

/Kamau

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irfanbagus · 2012-12-22 09:06

wow 0.9.26

why you don't try newest version (1.0.4) ? if package for your distro
not available, it's easy and fast to build lazarus from source as long
as you aready have working fpc (fpc-2.6.0 for lazarus-1.0.4) and
lazarus dependencies. btw there is rpm package in here

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Kamau · 2012-12-22 09:50

I can surely try that...

On the other hand I suppose that the "problem" is not an IDE-problem
but a compile-problem and as such, the solution must have something to
do with fpc?

/Kamau

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theo · 2012-12-22 10:05

Quote from: Kamau on December 22, 2012, 10:50:00 amI can surely try
that...

Yes, try that. It's probably a configuration problem.
And version 0.9.26 is really old (2008)
http://en.wikipedia.org/wiki/Lazarus_%28IDE%29#Versions

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Kamau · 2012-12-22 10:49

One problem....

The version fpc-2.6.0 seems only to be available for 686 and I need to
use it on a 386 machine.

/Kamau

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Leledumbo · 2012-12-22 11:01

QuoteThe version fpc-2.6.0 seems only to be available for 686 and I
need to use it on a 386 machine.FPC supports as old as i386What
processor is your computer running on?

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Kamau · 2012-12-22 11:21

It is an Intel Celeron in an ASUS Eee PC 900.

/Kamau

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Leledumbo · 2012-12-22 12:35

Celeron should be i686

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Kamau · 2012-12-22 15:31

The CentOS 5 operating system which I have installed on that machine
is 32 bit. (i386)

/Kamau

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BigChimp · 2012-12-22 15:38

Yes, that's nice.

the i386 in CentOS means it runs on any processor that supports the
i386 instruction set, including i386, i486, pentium, pentium pro,
pentium II, pentium 3, celeron, xeon, Pentium IV, core, core duo etc
processors.

https://en.wikipedia.org/wiki/I686

So you can install FPC 2.6 on that very well.

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marcov · 2012-12-22 15:59

First, update to the newest version of course,

but second, could you give some examples of absolute usage?


If you mean absolute usage like

var videobuffer : array [0..43] of byte absolute [ $0b800:0] ;

or something, that is a syntax reserved for DOS. It will only work in
TP dialect mode on the dos platform (and of course, the addresses used
have no meaning on non-dos operating systems to begin with)

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Kamau · 2012-12-22 16:48

As I already said, it is not my code. It concerns surce wich is part
of Lazarus/fpc

Example: In the file "custombitmap.inc" included in Lazarus/fpc.

procedure TCustomBitmap.SetSize(AWidth, AHeight: integer);
var
  SCB: TSharedCustomBitmap absolute FSharedImage;
  CurIntfImage, NewIntfImage: TLazIntfImage;
  NewRawImage: TRawImage;
begin
..... 

The "FSharedImage" is not a variable in same file but a type.

There are several occurences of this type of declarations in more
files belonging to Lazarus/fpc.

So, logically, there must be a way to work around this error-message.

/Kamau

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theo · 2012-12-22 17:04

Quote from: Kamau on December 22, 2012, 05:48:07 pmSo, logically,
there must be a way to work around this error-message.

Upgrade, please.

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marcov · 2012-12-22 20:09

I think Theo is right.

First, fsharedimage is a field, not a type.

Apparently very old FPC's didn't support absolute to a field, only to
global and local variables.

So the recommendation is to upgrade FPC to 2.6.0 and lazarus 1.0.4

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Kamau · 2012-12-22 20:24

All nice and well, and I get the message.

BUT! As it is I would need an FPC 2.6.0 suitable for i386.

At this point in time I can't upgrade the CentOS to a 64 bit system.

I'ts all right.

Thanks.

/Kamau

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theo · 2012-12-22 20:37

i686 is not x86_64. Only the latter means 64bit.

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Leledumbo · 2012-12-23 04:16

Pentium II is already i686, FYI. And of course, it's not 64-bit in any
way

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Kamau · 2012-12-23 04:45

Just want to round off this topic with a summary. Starting with again
thanking the Lazarus and FPC teams for the fantastic job they have
done and are doing!

Here follows the solution to my initial question. "How to run Lazarus
on a Centos 5 system?"

Preparation:

Be sure to uninstall both any previous Lazarus and any previous FPC
before following the instruction below.

rpm -e lazarus
rpm -e fpc

1. FPC 2.6.0 can be installed on Centos 5, but I found no RPM's for
i386.

Solution: Use fpc-2.6.0.i386-linux.tar which can be downloaded from
FPC's site.

There is a binary.i386-linux.tar included. When you unpack it you find
an install.sh which executes and installs fpc-2.6.0

NOTE: The RPM's provided on the Lazarus site, ex:
fpc-2.6.0-2.fc17.i686.rpm, can not be used on CentOS 5.

The reason is that they require epel6 which is not compatible with
CentOS 5.

2. After installing FPC 2.6.0 the Lazarus RPM lazarus-1.0.4-0.i386.rpm
on the Lazarus site can be used.

It can not be installed with yum as it specifies dependencies wich
require files non-compatible to CentOS 5.

It can be installed using rpm: rpm -i --nodeps
lazarus-1.0.4-0.i386.rpm

It installs perfectly, without any problems.

3. Before running Lazarus

If you had a previous installation of Lazarus you might want to get
rid of your old configuration which can be found in your home-
directory.

If you run lazarus as root this is what you can do:

Log in as root in a terminal.

Remove the .lazarus directory by rm -rf .lazarus

You are now ready for a fresh start and as already said...

It runs perfectly.

/Kamau