Lazarus
Using the Lazarus IDE => Options => Topic started by: Ed on September 26, 2005, 08:00:55 pm
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I was migrating code from Turbo Pascal 7 for DOS to linux, and have most of it compiling using the command line fpc source -Mtp. I just started using Lazarus. I get new errors, and have now chosen ONLY the -So option for the project compiler options. It still gives a pointer to a procedure error that -Mtp does not. I see no other source code tricks I can use to sidestep the compiler error, so feel that I need to keep the pointer comparison such as:
IntHandler: pointer;
If (IntHandler <> @SomeIntHandler) then...
I tried the $T+ and $T-, but nothing changes. I get "expected longint but got ^procedure(Word,word...)
I see no way to force any other options, or specifically -Mtp.
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Please, give a more complete code sample, including variable and type declarations.
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Please, give a more complete code sample, including variable and type declarations.
I will do that, but the code is at home. Tthe main part of the question is why does the command line execution of fpc compile but the same code from Lazarus does not? Lazarus runs the same compiler, does it not?
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Yes, but the options passed might be different.
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Somehow, the forum forgot who I am.
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The code is
Procedure SetInterrupts(IntHandler: Pointer; InterruptFrequency: Single);
Begin
Asm
CLI; //Inline($FA); {CLI - disable interrupts during vector change}
end;
If (IntHandler <> @RUSHandler) then
I keep getting Error: Incompatible types: got "<address of procedure(Word, Word, Word, Word, Word, Word...); Register>" exptected "LongInt"
Thanks.
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I still can't even try to compile this code. Can you give a complete example including variable and type declarations?
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I can reproduce it with the following app:
program a;
procedure RUSHandler(a,b,c,d,e,f: Word);
begin
end;
Procedure SetInterrupts(IntHandler: Pointer; InterruptFrequency: Single);
Begin
Asm
CLI; //Inline($FA); {CLI - disable interrupts during vector change}
end;
If (IntHandler <> @RUSHandler) then
end;
begin
end.
When you compile like this, without specifying mode, fpc compiles in its native fpc mode. This mode is similar to tp.
Somehow the result of @ is typed on comparisations.
As a workaround you can typecast: -> Pointer(@RUSHandler)
For a better explanation, you can thry the fpc forum or the fpc mailinglist, since this is not realy a lazarus issue (and I'm not that deep into the fpc internals of different modi)
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Thanks. My point was that the snippet of code I gave (more complete code is like that proposed by Marc) compiles just fine with the command line
fpc -Mtp source.pas
but does not with Lazarus when I specify the -So compile option, which is supposed to be equivalent to -Mtp. I wanted to know how to perfectly replicate fpc -Mtp to successfully compile my source from within Lazarus.
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Thanks, Marc, for the complete example. Now, at least, I can reproduce the problem and see for myself that your example code only needs the -So parameter (and that it is equivalent to -Mtp). No need to change the code.
I cannot reproce the problems Ed is having, because I cannot compile his snippet
project1.pas(9,20) Error: Identifier not found "RUSHandler"
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Here is a distilled program:
Program TestHandler;
Procedure RUSHandler(Flags,CS,IP,AX,BX,CX,DX,SI,DI,DS,ES,BP: Word);
Begin
end; {RUSHandler}
Procedure SetInterrupts(IntHandler: Pointer; InterruptFrequency: Single);
Begin
If (IntHandler <> @RUSHandler) then
{do something}
end;
Begin {Main program}
SetInterrupts(@RUSHandler, 5000.0)
end.
I verified that Lazarus was using no optons but -So, and it still had a compiler error that I do not get when I invoke the compiler from the command line with -Mtp. I do see that the forced typecasting as suggested by Marc "Pointer((@RUSHandler)" did work, so this becomes less important. I still wonder why the results differ...
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All I can say is, that this sample app compiles here using fpc 2.0.0 and 2.0.1.